Class 12: Algebra of Matrices

$\displaystyle \textbf{Question 1: } \text{Compute the indicated products:}$
$\displaystyle \text{(i) }\begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}$
$\displaystyle \text{(ii) }\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\-3&2&1\end{bmatrix}$
$\displaystyle \text{(iii) }\begin{bmatrix}2&3&4\\3&4&5\\4&5&6\end{bmatrix}\begin{bmatrix}1&-3&5\\0&2&4\\3&0&5\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}=\begin{bmatrix}a\cdot a+b\cdot b&a(-b)+ba\\(-b)a+ab&(-b)(-b)+a\cdot a\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^2+b^2&-ab+ab\\-ab+ab&b^2+a^2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^2+b^2&0\\0&a^2+b^2\end{bmatrix}$
$\displaystyle \text{(ii)}$
$\displaystyle \begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\-3&2&1\end{bmatrix}=\begin{bmatrix}1\cdot1+(-2)(-3)&1\cdot2+(-2)\cdot2&1\cdot3+(-2)\cdot1\\2\cdot1+3(-3)&2\cdot2+3\cdot2&2\cdot3+3\cdot1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+6&2-4&3-2\\2-9&4+6&6+3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}7&-2&1\\-7&10&9\end{bmatrix}$
$\displaystyle \text{(iii)}$
$\displaystyle \begin{bmatrix}2&3&4\\3&4&5\\4&5&6\end{bmatrix}\begin{bmatrix}1&-3&5\\0&2&4\\3&0&5\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2\cdot1+3\cdot0+4\cdot3&2(-3)+3\cdot2+4\cdot0&2\cdot5+3\cdot4+4\cdot5\\3\cdot1+4\cdot0+5\cdot3&3(-3)+4\cdot2+5\cdot0&3\cdot5+4\cdot4+5\cdot5\\4\cdot1+5\cdot0+6\cdot3&4(-3)+5\cdot2+6\cdot0&4\cdot5+5\cdot4+6\cdot5\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2+0+12&-6+6+0&10+12+20\\3+0+15&-9+8+0&15+16+25\\4+0+18&-12+10+0&20+20+30\end{bmatrix}$
$\displaystyle =\begin{bmatrix}14&0&42\\18&-1&56\\22&-2&70\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: } \text{Show that }AB\neq BA\text{ in each of the following cases:}$
$\displaystyle \text{(i) }A=\begin{bmatrix}5&-1\\6&7\end{bmatrix}\text{ and }B=\begin{bmatrix}2&1\\3&4\end{bmatrix}$
$\displaystyle \text{(ii) }A=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\text{ and }B=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}$
$\displaystyle \text{(iii) }A=\begin{bmatrix}1&3&0\\1&1&0\\4&1&0\end{bmatrix}\text{ and }B=\begin{bmatrix}0&1&0\\1&0&0\\0&5&1\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle AB=\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}10-3&5-4\\12+21&6+28\end{bmatrix}=\begin{bmatrix}7&1\\33&34\end{bmatrix}$
$\displaystyle BA=\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}$
$\displaystyle =\begin{bmatrix}10+6&-2+7\\15+24&-3+28\end{bmatrix}=\begin{bmatrix}16&5\\39&25\end{bmatrix}$
$\displaystyle \therefore AB\neq BA.$

$\displaystyle \text{(ii)}$
$\displaystyle AB=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-1+0+0&-2+1+0&-3+0+0\\0+0+1&0-1+1&0+0+0\\2+0+4&4+3+4&6+0+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-1&-1&-3\\1&0&0\\6&11&6\end{bmatrix}$
$\displaystyle BA=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-1+0+6&1-2+9&0+2+12\\0+0+0&0-1+0&0+1+0\\-1+0+0&1-1+0&0+1+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}5&8&14\\0&-1&1\\-1&0&1\end{bmatrix}$
$\displaystyle \therefore AB\neq BA.$

$\displaystyle \text{(iii)}$
$\displaystyle AB=\begin{bmatrix}1&3&0\\1&1&0\\4&1&0\end{bmatrix}\begin{bmatrix}0&1&0\\1&0&0\\0&5&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0+3+0&1+0+0&0+0+0\\0+1+0&1+0+0&0+0+0\\0+1+0&4+0+0&0+0+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}3&1&0\\1&1&0\\1&4&0\end{bmatrix}$
$\displaystyle BA=\begin{bmatrix}0&1&0\\1&0&0\\0&5&1\end{bmatrix}\begin{bmatrix}1&3&0\\1&1&0\\4&1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0+1+0&0+1+0&0+0+0\\1+0+0&3+0+0&0+0+0\\0+5+4&0+5+1&0+0+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&1&0\\1&3&0\\9&6&0\end{bmatrix}$
$\displaystyle \therefore AB\neq BA.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: } \text{Compute }AB\text{ and }BA\text{, wherever they exist, in each of the following cases:}$
$\displaystyle \text{(i) }A=\begin{bmatrix}1&-2\\2&3\end{bmatrix}\text{ and }B=\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}$
$\displaystyle \text{(ii) }A=\begin{bmatrix}3&2\\-1&0\\-1&1\end{bmatrix}\text{ and }B=\begin{bmatrix}4&5&6\\0&1&2\end{bmatrix}$
$\displaystyle \text{(iii) }A=\begin{bmatrix}1&-1&2&3\end{bmatrix}\text{ and }B=\begin{bmatrix}0\\1\\3\\2\end{bmatrix}$
$\displaystyle \text{(iv) }\begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}c\\d\end{bmatrix}+\begin{bmatrix}a&b&c&d\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle AB=\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1-4&2-6&3-2\\2+6&4+9&6+3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-3&-4&1\\8&13&9\end{bmatrix}$
$\displaystyle \text{Since }B\text{ is of order }2\times3\text{ and }A\text{ is of order }2\times2,$
$\displaystyle BA\text{ does not exist, as the number of columns of }B\neq\text{ the number of rows of }A.$

$\displaystyle \text{(ii)}$
$\displaystyle AB=\begin{bmatrix}3&2\\-1&0\\-1&1\end{bmatrix}\begin{bmatrix}4&5&6\\0&1&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}12+0&15+2&18+4\\-4+0&-5+0&-6+0\\-4+0&-5+1&-6+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}12&17&22\\-4&-5&-6\\-4&-4&-4\end{bmatrix}$
$\displaystyle BA=\begin{bmatrix}4&5&6\\0&1&2\end{bmatrix}\begin{bmatrix}3&2\\-1&0\\-1&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}12-5-6&8+0+6\\0-1-2&0+0+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&14\\-3&2\end{bmatrix}$

$\displaystyle \text{(iii)}$
$\displaystyle AB=\begin{bmatrix}1&-1&2&3\end{bmatrix}\begin{bmatrix}0\\1\\3\\2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0-1+6+6\end{bmatrix}=\begin{bmatrix}11\end{bmatrix}$
$\displaystyle BA=\begin{bmatrix}0\\1\\3\\2\end{bmatrix}\begin{bmatrix}1&-1&2&3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&0&0\\1&-1&2&3\\3&-3&6&9\\2&-2&4&6\end{bmatrix}$

$\displaystyle \text{(iv)}$
$\displaystyle \begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}c\\d\end{bmatrix}+\begin{bmatrix}a&b&c&d\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$
$\displaystyle =\begin{bmatrix}ac+bd\end{bmatrix}+\begin{bmatrix}a^2+b^2+c^2+d^2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^2+b^2+c^2+d^2+ac+bd\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: } \text{Show that }AB\neq BA\text{ in each of the following cases:}$
$\displaystyle \text{(i) }A=\begin{bmatrix}1&3&-1\\2&-1&-1\\3&0&1\end{bmatrix}\text{ and }B=\begin{bmatrix}-2&3&-1\\-1&2&-1\\-6&9&-4\end{bmatrix}$
$\displaystyle \text{(ii) }A=\begin{bmatrix}10&-4&1\\-11&5&0\\9&-5&1\end{bmatrix}\text{ and }B=\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle AB=\begin{bmatrix}1&3&-1\\2&-1&-1\\3&0&1\end{bmatrix}\begin{bmatrix}-2&3&-1\\-1&2&-1\\-6&9&-4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-2-3+6&3+6-9&-1-3+4\\-4+1+6&6-2-9&-2+1+4\\-6+0-6&9+0+9&-3+0-4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&0&0\\3&-5&3\\-12&18&-7\end{bmatrix}$
$\displaystyle BA=\begin{bmatrix}-2&3&-1\\-1&2&-1\\-6&9&-4\end{bmatrix}\begin{bmatrix}1&3&-1\\2&-1&-1\\3&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-2+6-3&-6-3+0&2-3-1\\-1+4-3&-3-2+0&1-2-1\\-6+18-12&-18-9+0&6-9-4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&-9&-2\\0&-5&-2\\0&-27&-7\end{bmatrix}$
$\displaystyle \therefore AB\neq BA.$

$\displaystyle \text{(ii)}$
$\displaystyle AB=\begin{bmatrix}10&-4&1\\-11&5&0\\9&-5&1\end{bmatrix}\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}10-12+1&20-16+3&10-8+2\\-11+15+0&-22+20+0&-11+10+0\\9-15+1&18-20+3&9-10+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-1&7&4\\4&-2&-1\\-5&1&1\end{bmatrix}$
$\displaystyle BA=\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}\begin{bmatrix}10&-4&1\\-11&5&0\\9&-5&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}10-22+9&-4+10-5&1+0+1\\30-44+18&-12+20-10&3+0+2\\10-33+18&-4+15-10&1+0+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-3&1&2\\4&-2&5\\-5&1&3\end{bmatrix}$
$\displaystyle \therefore AB\neq BA.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: } \text{Evaluate the following:}$
$\displaystyle \text{(i) }\left(\begin{bmatrix}1&3\\-1&4\end{bmatrix}+\begin{bmatrix}3&-2\\-1&1\end{bmatrix}\right)\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}$
$\displaystyle \text{(ii) }\begin{bmatrix}1&2&3\end{bmatrix}\begin{bmatrix}1&0&2\\2&0&1\\0&1&2\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}$
$\displaystyle \text{(iii) }\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\left(\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2\end{bmatrix}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \left(\begin{bmatrix}1&3\\-1&4\end{bmatrix}+\begin{bmatrix}3&-2\\-1&1\end{bmatrix}\right)\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4&1\\-2&5\end{bmatrix}\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4+2&12+4&20+6\\-2+10&-6+20&-10+30\end{bmatrix}$
$\displaystyle =\begin{bmatrix}6&16&26\\8&14&20\end{bmatrix}$
$\displaystyle \text{(ii)}$
$\displaystyle \begin{bmatrix}1&2&3\end{bmatrix}\begin{bmatrix}1&0&2\\2&0&1\\0&1&2\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+4+0&0+0+3&2+2+6\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}5&3&10\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}5\times2+3\times4+10\times6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}82\end{bmatrix}$
$\displaystyle \text{(iii)}$
$\displaystyle \begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\left(\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2\end{bmatrix}\right)$
$\displaystyle =\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{bmatrix}1&-1&0\\1&0&-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1-1&-1+0&0+1\\0+2&0+0&0-2\\2+3&-2+0&0-3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&-1&1\\2&0&-2\\5&-2&-3\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: } \text{If }A=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ B=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\text{ and}$
$\displaystyle C=\begin{bmatrix}0&1\\1&0\end{bmatrix},\text{ then show that }A^2=B^2=C^2=I_2.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+0&0+0\\0+0&0+1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I_2$
$\displaystyle B^2=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+0&0+0\\0+0&0+1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I_2$
$\displaystyle C^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0+1&0+0\\0+0&1+0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I_2$
$\displaystyle \therefore A^2=B^2=C^2=I_2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: } \text{If }A=\begin{bmatrix}2&-1\\3&2\end{bmatrix}\text{ and }B=\begin{bmatrix}0&4\\-1&7\end{bmatrix},\text{ find}$
$\displaystyle 3A^2-2B+I.\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=\begin{bmatrix}2&-1\\3&2\end{bmatrix}\begin{bmatrix}2&-1\\3&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4-3&-2-2\\6+6&-3+4\end{bmatrix}=\begin{bmatrix}1&-4\\12&1\end{bmatrix}$
$\displaystyle \therefore 3A^2-2B+I$
$\displaystyle =3\begin{bmatrix}1&-4\\12&1\end{bmatrix}-2\begin{bmatrix}0&4\\-1&7\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}3&-12\\36&3\end{bmatrix}+\begin{bmatrix}0&-8\\2&-14\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4&-20\\38&-10\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: } \text{If }A=\begin{bmatrix}4&2\\-1&1\end{bmatrix},\text{ prove that }(A-2I)(A-3I)=O.$
$\displaystyle \text{Answer:}$
$\displaystyle (A-2I)(A-3I)$
$\displaystyle =\left(\begin{bmatrix}4&2\\-1&1\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}\right)\left(\begin{bmatrix}4&2\\-1&1\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}\right)$
$\displaystyle =\left(\begin{bmatrix}4&2\\-1&1\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}\right)\left(\begin{bmatrix}4&2\\-1&1\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}\right)$
$\displaystyle =\begin{bmatrix}2&2\\-1&-1\end{bmatrix}\begin{bmatrix}1&2\\-1&-2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2-2&4-4\\-1+1&-2+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: } \text{If }A=\begin{bmatrix}1&1\\0&1\end{bmatrix},\text{ show that }A^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}\text{ and}$
$\displaystyle A^3=\begin{bmatrix}1&3\\0&1\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=A\cdot A=\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+0&1+1\\0+0&0+1\end{bmatrix}=\begin{bmatrix}1&2\\0&1\end{bmatrix}$
$\displaystyle A^3=A^2\cdot A=\begin{bmatrix}1&2\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+0&1+2\\0+0&0+1\end{bmatrix}=\begin{bmatrix}1&3\\0&1\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: } \text{If }A=\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix},\text{ show that }A^2=O.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=A\cdot A=\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix}\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^2b^2-a^2b^2&ab^3-ab^3\\-a^3b+a^3b&-a^2b^2+a^2b^2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: } \text{If }A=\begin{bmatrix}\cos 2\theta&\sin 2\theta\\-\sin 2\theta&\cos 2\theta\end{bmatrix},\text{ find }A^2.$
$\displaystyle \hfill\text{[CBSE 2000C]}$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=A\cdot A$
$\displaystyle =\begin{bmatrix}\cos 2\theta&\sin 2\theta\\-\sin 2\theta&\cos 2\theta\end{bmatrix}\begin{bmatrix}\cos 2\theta&\sin 2\theta\\-\sin 2\theta&\cos 2\theta\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos^2 2\theta-\sin^2 2\theta&2\sin 2\theta\cos 2\theta\\-2\sin 2\theta\cos 2\theta&\cos^2 2\theta-\sin^2 2\theta\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos 4\theta&\sin 4\theta\\-\sin 4\theta&\cos 4\theta\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: } \text{If }A=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\text{ and}$
$\displaystyle B=\begin{bmatrix}-1&3&5\\1&-3&-5\\-1&3&5\end{bmatrix},\text{ show that }AB=BA=O_{3\times3}.$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\begin{bmatrix}-1&3&5\\1&-3&-5\\-1&3&5\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-2-3+5&6+9-15&10+15-25\\1+4-5&-3-12+15&-5-20+25\\-1-3+4&3+9-12&5+15-20\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O_{3\times3}$
$\displaystyle BA=\begin{bmatrix}-1&3&5\\1&-3&-5\\-1&3&5\end{bmatrix}\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-2-3+5&3+12-15&5+15-20\\2+3-5&-3-12+15&-5-15+20\\-2-3+5&3+12-15&5+15-20\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O_{3\times3}$
$\displaystyle \therefore AB=BA=O_{3\times3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: } \text{If }A=\begin{bmatrix}0&c&-b\\-c&0&a\\b&-a&0\end{bmatrix}\text{ and}$
$\displaystyle B=\begin{bmatrix}a^2&ab&ac\\ab&b^2&bc\\ac&bc&c^2\end{bmatrix},\text{ show that }AB=BA=O_{3\times3}.$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\begin{bmatrix}0&c&-b\\-c&0&a\\b&-a&0\end{bmatrix}\begin{bmatrix}a^2&ab&ac\\ab&b^2&bc\\ac&bc&c^2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0+abc-abc&0+b^2c-b^2c&0+bc^2-bc^2\\-a^2c+0+a^2c&-abc+0+abc&-ac^2+0+ac^2\\a^2b-a^2b+0&ab^2-ab^2+0&abc-abc+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O_{3\times3}$
$\displaystyle BA=\begin{bmatrix}a^2&ab&ac\\ab&b^2&bc\\ac&bc&c^2\end{bmatrix}\begin{bmatrix}0&c&-b\\-c&0&a\\b&-a&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0-abc+abc&a^2c+0-a^2c&-a^2b+a^2b+0\\0-b^2c+b^2c&abc+0-abc&-ab^2+ab^2+0\\0-bc^2+bc^2&ac^2+0-ac^2&-abc+abc+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O_{3\times3}$
$\displaystyle \therefore AB=BA=O_{3\times3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: } \text{If }A=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\text{ and}$
$\displaystyle B=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix},\text{ show that }AB=A\text{ and }BA=B.$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4+3-5&-4-9+10&-8-12+15\\-2-4+5&2+12-10&4+16-15\\2+3-4&-2-9+8&-4-12+12\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}=A$
$\displaystyle BA=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4+2-4&-6-8+12&-10-10+16\\-2-3+4&3+12-12&5+15-16\\2+2-3&-3-8+9&-5-10+12\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}=B$
$\displaystyle \therefore AB=A\text{ and }BA=B.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: } \text{If }A=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}\text{ and}$
$\displaystyle B=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix},\text{ compute }A^2-B^2.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2-B^2$
$\displaystyle =\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}-\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix}\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+3-5&-1-3-5&1+3-5\\-3-9+15&3+9+15&-3-9+15\\-5+15+25&5-15+25&-5+15+25\end{bmatrix}-\begin{bmatrix}0+4-3&0-12+12&0-12+12\\0-3+3&4+9-12&3+9-12\\0+4-4&-4-12+12&-3-12+16\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-1&-9&-1\\3&27&3\\35&15&35\end{bmatrix}-\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-2&-9&-1\\3&26&3\\35&15&34\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: } \text{For the following matrices verify the associativity of matrix}$
$\displaystyle \text{multiplication, i.e. }(AB)C=A(BC).$
$\displaystyle \text{(i) }A=\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix},\ B=\begin{bmatrix}1&0\\-1&2\\0&3\end{bmatrix}\text{ and }C=\begin{bmatrix}1\\-1\end{bmatrix}$
$\displaystyle \text{(ii) }A=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix},\ B=\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}\text{ and }C=\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$
$\displaystyle AB=\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix}\begin{bmatrix}1&0\\-1&2\\0&3\end{bmatrix}=\begin{bmatrix}-1&4\\-1&3\end{bmatrix}$
$\displaystyle (AB)C=\begin{bmatrix}-1&4\\-1&3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}-1-4\\-1-3\end{bmatrix}=\begin{bmatrix}-5\\-4\end{bmatrix}$
$\displaystyle BC=\begin{bmatrix}1&0\\-1&2\\0&3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}1\\-3\\-3\end{bmatrix}$
$\displaystyle A(BC)=\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix}\begin{bmatrix}1\\-3\\-3\end{bmatrix}=\begin{bmatrix}1-6+0\\-1+0-3\end{bmatrix}=\begin{bmatrix}-5\\-4\end{bmatrix}$
$\displaystyle \therefore (AB)C=A(BC).$

$\displaystyle \text{(ii)}$
$\displaystyle AB=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}=\begin{bmatrix}10&-5&11\\5&-2&5\\5&-4&4\end{bmatrix}$
$\displaystyle (AB)C=\begin{bmatrix}10&-5&11\\5&-2&5\\5&-4&4\end{bmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}=\begin{bmatrix}-5&20&-4\\-1&10&-2\\-7&10&-5\end{bmatrix}$
$\displaystyle BC=\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}=\begin{bmatrix}-2&2&-1\\3&0&3\\-1&4&-2\end{bmatrix}$
$\displaystyle A(BC)=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{bmatrix}-2&2&-1\\3&0&3\\-1&4&-2\end{bmatrix}=\begin{bmatrix}-5&20&-4\\-1&10&-2\\-7&10&-5\end{bmatrix}$
$\displaystyle \therefore (AB)C=A(BC).$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: } \text{For the following matrices verify the distributivity of matrix multiplication}$
$\displaystyle \text{over matrix addition, i.e. }A(B+C)=AB+AC.$
$\displaystyle \text{(i) }A=\begin{bmatrix}1&-1\\0&2\end{bmatrix},\ B=\begin{bmatrix}-1&0\\2&1\end{bmatrix}\text{ and }C=\begin{bmatrix}0&1\\1&-1\end{bmatrix}$
$\displaystyle \text{(ii) }A=\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix},\ B=\begin{bmatrix}0&1\\1&1\end{bmatrix}\text{ and }C=\begin{bmatrix}0&1\\1&-1\end{bmatrix}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$
$\displaystyle A(B+C)=\begin{bmatrix}1&-1\\0&2\end{bmatrix}\left(\begin{bmatrix}-1&0\\2&1\end{bmatrix}+\begin{bmatrix}0&1\\1&-1\end{bmatrix}\right)$
$\displaystyle =\begin{bmatrix}1&-1\\0&2\end{bmatrix}\begin{bmatrix}-1&1\\3&0\end{bmatrix}=\begin{bmatrix}-4&1\\6&0\end{bmatrix}$
$\displaystyle AB+AC=\begin{bmatrix}1&-1\\0&2\end{bmatrix}\begin{bmatrix}-1&0\\2&1\end{bmatrix}+\begin{bmatrix}1&-1\\0&2\end{bmatrix}\begin{bmatrix}0&1\\1&-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-3&-1\\4&2\end{bmatrix}+\begin{bmatrix}-1&2\\2&-2\end{bmatrix}=\begin{bmatrix}-4&1\\6&0\end{bmatrix}$
$\displaystyle \therefore A(B+C)=AB+AC.$

$\displaystyle \text{(ii)}$
$\displaystyle A(B+C)=\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\left(\begin{bmatrix}0&1\\1&1\end{bmatrix}+\begin{bmatrix}0&1\\1&-1\end{bmatrix}\right)$
$\displaystyle =\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\begin{bmatrix}0&2\\2&0\end{bmatrix}=\begin{bmatrix}-2&4\\2&2\\4&-2\end{bmatrix}$
$\displaystyle AB+AC=\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\begin{bmatrix}0&1\\1&1\end{bmatrix}+\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\begin{bmatrix}0&1\\1&-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-1&1\\1&2\\2&1\end{bmatrix}+\begin{bmatrix}-1&3\\1&0\\2&-3\end{bmatrix}=\begin{bmatrix}-2&4\\2&2\\4&-2\end{bmatrix}$
$\displaystyle \therefore A(B+C)=AB+AC.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: } \text{If }A=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},\ B=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}\text{ and}$
$\displaystyle C=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix},\text{ verify that }A(B-C)=AB-AC.$
$\displaystyle \text{Answer:}$
$\displaystyle A(B-C)=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\left(\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}-\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}\right)$
$\displaystyle =\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}-1&0&-6\\-1&0&3\\-1&1&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&-2&-8\\-2&0&-21\\0&1&16\end{bmatrix}$
$\displaystyle AB-AC=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}-\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2&5&-8\\2&14&-15\\-3&-9&13\end{bmatrix}-\begin{bmatrix}1&7&0\\4&14&6\\-3&-10&-3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&-2&-8\\-2&0&-21\\0&1&16\end{bmatrix}$
$\displaystyle \therefore A(B-C)=AB-AC.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: } \text{Compute the elements }a_{43}\text{ and }a_{22}\text{ of the matrix}$
$\displaystyle A=\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\begin{bmatrix}2&-1\\-3&2\\4&3\end{bmatrix}\begin{bmatrix}0&1&-1&2&-2\\3&-3&4&-4&0\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\left(\begin{bmatrix}2&-1\\-3&2\\4&3\end{bmatrix}\begin{bmatrix}0&1&-1&2&-2\\3&-3&4&-4&0\end{bmatrix}\right)$
$\displaystyle =\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\begin{bmatrix}-3&5&-6&8&-4\\6&-9&11&-14&6\\9&-5&8&-4&-8\end{bmatrix}$
$\displaystyle =\begin{bmatrix}6&-9&11&-14&6\\12&0&4&8&-24\\36&-37&49&-50&2\\24&0&8&16&-48\end{bmatrix}$
$\displaystyle \therefore a_{43}=8\text{ and }a_{22}=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: } \text{If }A=\begin{bmatrix}0&1&0\\0&0&1\\p&q&r\end{bmatrix}\text{ and }I\text{ is the identity matrix of order }3,\text{ show that}$
$\displaystyle A^3=pI+qA+rA^2.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=\begin{bmatrix}0&1&0\\0&0&1\\p&q&r\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&1\\p&q&r\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&1\\p&q&r\\rp&p+rq&q+r^2\end{bmatrix}$
$\displaystyle A^3=A^2A$
$\displaystyle =\begin{bmatrix}0&0&1\\p&q&r\\rp&p+rq&q+r^2\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&1\\p&q&r\end{bmatrix}$
$\displaystyle =\begin{bmatrix}p&q&r\\rp&p+rq&q+r^2\\pq+r^2p&rp+q^2+r^2q&p+2rq+r^3\end{bmatrix}$
$\displaystyle pI+qA+rA^2$
$\displaystyle =p\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+q\begin{bmatrix}0&1&0\\0&0&1\\p&q&r\end{bmatrix}+r\begin{bmatrix}0&0&1\\p&q&r\\rp&p+rq&q+r^2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}p&0&0\\0&p&0\\0&0&p\end{bmatrix}+\begin{bmatrix}0&q&0\\0&0&q\\pq&q^2&qr\end{bmatrix}+\begin{bmatrix}0&0&r\\rp&rq&r^2\\r^2p&rp+r^2q&qr+r^3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}p&q&r\\rp&p+rq&q+r^2\\pq+r^2p&rp+q^2+r^2q&p+2rq+r^3\end{bmatrix}$
$\displaystyle \therefore A^3=pI+qA+rA^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: } \text{If }\omega\text{ is a complex cube root of unity, show that}$
$\displaystyle \left(\begin{bmatrix}1&\omega&\omega^2\\\omega&\omega^2&1\\\omega^2&1&\omega\end{bmatrix}+\begin{bmatrix}\omega&\omega^2&1\\\omega^2&1&\omega\\\omega&\omega^2&1\end{bmatrix}\right)\begin{bmatrix}1\\\omega\\\omega^2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }\omega\text{ is a non-real cube root of unity,}$
$\displaystyle 1+\omega+\omega^2=0,\qquad \omega^3=1.$
$\displaystyle \text{LHS}=\left(\begin{bmatrix}1&\omega&\omega^2\\\omega&\omega^2&1\\\omega^2&1&\omega\end{bmatrix}+\begin{bmatrix}\omega&\omega^2&1\\\omega^2&1&\omega\\\omega&\omega^2&1\end{bmatrix}\right)\begin{bmatrix}1\\\omega\\\omega^2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+\omega&\omega+\omega^2&\omega^2+1\\\omega+\omega^2&\omega^2+1&1+\omega\\\omega^2+\omega&1+\omega^2&\omega+1\end{bmatrix}\begin{bmatrix}1\\\omega\\\omega^2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-\omega^2&-1&-\omega\\-1&-\omega&-\omega^2\\-1&-\omega&-\omega^2\end{bmatrix}\begin{bmatrix}1\\\omega\\\omega^2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-\omega^2-\omega-\omega^3\\-1-\omega^2-\omega^4\\-1-\omega^2-\omega^4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-(1+\omega+\omega^2)\\-(1+\omega+\omega^2)\\-(1+\omega+\omega^2)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0\\0\\0\end{bmatrix}=\text{RHS}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: } \text{If }A=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix},\text{ show that }A^2=I_3.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4+3-5&-6-12+15&-10-15+20\\-2-4+5&3+16-15&5+20-25\\2+3-4&-3-12+15&-5-15+16\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle =I_3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: } \text{If }A=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix},\text{ show that }A^2=I_3.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix}\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}16-3-12&-4+0+4&-16+4+12\\12+0-12&-3+0+4&-12+0+12\\12-3-9&-3+0+3&-12+4+9\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle =I_3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24:}$
$\displaystyle \text{(i) If }\begin{bmatrix}1&1&x\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&1&0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=0,\text{ find }x.$
$\displaystyle \text{(ii) If }\begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix},\text{ find }x.\hfill\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \begin{bmatrix}1&1&x\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&1&0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=0$
$\displaystyle \Rightarrow\begin{bmatrix}1+2x&2+x&3\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=0$
$\displaystyle \Rightarrow(1+2x)+(2+x)+3=0$
$\displaystyle \Rightarrow 3x+6=0$
$\displaystyle \Rightarrow x=-2.$
$\displaystyle \text{(ii)}$
$\displaystyle \begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}2-6&-6+12\\5-14&-15+28\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}-4&6\\-9&13\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}$
$\displaystyle \Rightarrow x=13.$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: } \text{If }\begin{bmatrix}x&4&1\end{bmatrix}\begin{bmatrix}2&1&2\\1&0&2\\0&2&-4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0,\text{ find }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{bmatrix}x&4&1\end{bmatrix}\begin{bmatrix}2&1&2\\1&0&2\\0&2&-4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0$
$\displaystyle \Rightarrow\begin{bmatrix}2x+4&x+2&2x+4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0$
$\displaystyle \Rightarrow x(2x+4)+4(x+2)-(2x+4)=0$
$\displaystyle \Rightarrow 2x^2+6x+4=0$
$\displaystyle \Rightarrow x^2+3x+2=0$
$\displaystyle \Rightarrow (x+1)(x+2)=0$
$\displaystyle \Rightarrow x=-1\text{ or }x=-2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: } \text{If }\begin{bmatrix}1&-1&x\end{bmatrix}\begin{bmatrix}0&1&-1\\2&1&3\\1&1&1\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=0,\text{ find }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{bmatrix}1&-1&x\end{bmatrix}\begin{bmatrix}0&1&-1\\2&1&3\\1&1&1\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=0$
$\displaystyle \Rightarrow\begin{bmatrix}x-2&x&x-4\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=0$
$\displaystyle \Rightarrow x+(x-4)=0$
$\displaystyle \Rightarrow 2x-4=0$
$\displaystyle \Rightarrow x=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: } \text{If }A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ prove that }A^2-A+2I=O.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2-A+2I=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}+2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&-2\\4&-4\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}+\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1-3+2&-2+2+0\\4-4+0&-4+2+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: } \text{If }A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ find }\lambda\text{ such that }A^2=5A+\lambda I.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\displaystyle 5A+\lambda I=5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+\lambda\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}15+\lambda&5\\-5&10+\lambda\end{bmatrix}$
$\displaystyle \text{Since }A^2=5A+\lambda I,$
$\displaystyle \begin{bmatrix}8&5\\-5&3\end{bmatrix}=\begin{bmatrix}15+\lambda&5\\-5&10+\lambda\end{bmatrix}$
$\displaystyle \Rightarrow 8=15+\lambda$
$\displaystyle \Rightarrow \lambda=-7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: } \text{If }A=\begin{bmatrix}3&1\\-1&2\end{bmatrix},\text{ show that }A^2-5A+7I_2=O.$
$\displaystyle \hfill\text{[CBSE 2003, 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle A^2-5A+7I_2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\displaystyle =\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: } \text{If }A=\begin{bmatrix}2&3\\-1&0\end{bmatrix},\text{ show that }A^2-2A+3I_2=O.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2-2A+3I_2=\begin{bmatrix}2&3\\-1&0\end{bmatrix}\begin{bmatrix}2&3\\-1&0\end{bmatrix}-2\begin{bmatrix}2&3\\-1&0\end{bmatrix}+3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&6\\-2&-3\end{bmatrix}-\begin{bmatrix}4&6\\-2&0\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1-4+3&6-6+0\\-2+2+0&-3-0+3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: } \text{Show that the matrix }A=\begin{bmatrix}2&3\\1&2\end{bmatrix}\text{ satisfies the equation}$
$\displaystyle A^3-4A^2+A=O.\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}=\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$\displaystyle A^3=A^2A=\begin{bmatrix}7&12\\4&7\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}=\begin{bmatrix}26&45\\15&26\end{bmatrix}$
$\displaystyle A^3-4A^2+A$
$\displaystyle =\begin{bmatrix}26&45\\15&26\end{bmatrix}-4\begin{bmatrix}7&12\\4&7\end{bmatrix}+\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}26&45\\15&26\end{bmatrix}-\begin{bmatrix}28&48\\16&28\end{bmatrix}+\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}26-28+2&45-48+3\\15-16+1&26-28+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: } \text{Show that the matrix }A=\begin{bmatrix}5&3\\12&7\end{bmatrix}\text{ is a root of the equation }A^2-12A-I=O.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2-12A-I=\begin{bmatrix}5&3\\12&7\end{bmatrix}\begin{bmatrix}5&3\\12&7\end{bmatrix}-12\begin{bmatrix}5&3\\12&7\end{bmatrix}-\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}61&36\\144&85\end{bmatrix}-\begin{bmatrix}60&36\\144&84\end{bmatrix}-\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}61-60-1&36-36-0\\144-144-0&85-84-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: } \text{If }A=\begin{bmatrix}3&-5\\-4&2\end{bmatrix},\text{ find }A^2-5A-14I.\hfill\text{[CBSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle A^2-5A-14I=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}-5\begin{bmatrix}3&-5\\-4&2\end{bmatrix}-14\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-\begin{bmatrix}15&-25\\-20&10\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}$
$\displaystyle =\begin{bmatrix}29-15-14&-25+25-0\\-20+20-0&24-10-14\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: } \text{If }A=\begin{bmatrix}3&1\\-1&2\end{bmatrix},\text{ show that }A^2-5A+7I=O.\text{ Use this to find }A^4.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\displaystyle A^2-5A+7I=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \therefore A^2=5A-7I$
$\displaystyle A^4=(A^2)^2=(5A-7I)^2$
$\displaystyle =25A^2-70A+49I$
$\displaystyle =25(5A-7I)-70A+49I$
$\displaystyle =55A-126I$
$\displaystyle =55\begin{bmatrix}3&1\\-1&2\end{bmatrix}-126\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}165&55\\-55&110\end{bmatrix}-\begin{bmatrix}126&0\\0&126\end{bmatrix}$
$\displaystyle =\begin{bmatrix}39&55\\-55&-16\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: } \text{If }A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix},\text{ find }k\text{ such that }A^2=kA-2I_2.\hfill\text{[CBSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix},\quad \text{we need to find }k\text{ such that }A^2=kA-2I_2.$
$\displaystyle A^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$
$\displaystyle kA-2I_2=k\begin{bmatrix}3&-2\\4&-2\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}3k-2&-2k\\4k&-2k-2\end{bmatrix}$
$\displaystyle \text{Equating the corresponding entries,}$
$\displaystyle 1=3k-2,\quad -2=-2k,\quad 4=4k,\quad -4=-2k-2$
$\displaystyle \Rightarrow k=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: } \text{If }A=\begin{bmatrix}1&0\\-1&7\end{bmatrix},\text{ find }k\text{ such that }A^2-8A+kI=O.\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}1&0\\-1&7\end{bmatrix}$
$\displaystyle A^2=\begin{bmatrix}1&0\\-1&7\end{bmatrix}\begin{bmatrix}1&0\\-1&7\end{bmatrix}=\begin{bmatrix}1&0\\-8&49\end{bmatrix}$
$\displaystyle A^2-8A+kI=\begin{bmatrix}1&0\\-8&49\end{bmatrix}-8\begin{bmatrix}1&0\\-1&7\end{bmatrix}+k\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&0\\-8&49\end{bmatrix}+\begin{bmatrix}-8&0\\8&-56\end{bmatrix}+\begin{bmatrix}k&0\\0&k\end{bmatrix}$
$\displaystyle =\begin{bmatrix}k-7&0\\0&k-7\end{bmatrix}$
$\displaystyle \text{Since }A^2-8A+kI=O,\ \begin{bmatrix}k-7&0\\0&k-7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\displaystyle \Rightarrow k=7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: } \text{If }A=\begin{bmatrix}1&2\\2&1\end{bmatrix}\text{ and }f(x)=x^2-2x-3,\text{ show that }f(A)=O.\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}1&2\\2&1\end{bmatrix},\qquad f(x)=x^2-2x-3$
$\displaystyle A^2=\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}=\begin{bmatrix}1+4&2+2\\2+2&4+1\end{bmatrix}=\begin{bmatrix}5&4\\4&5\end{bmatrix}$
$\displaystyle f(A)=A^2-2A-3I$
$\displaystyle =\begin{bmatrix}5&4\\4&5\end{bmatrix}-2\begin{bmatrix}1&2\\2&1\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}5&4\\4&5\end{bmatrix}-\begin{bmatrix}2&4\\4&2\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: } \text{If }A=\begin{bmatrix}2&3\\1&2\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ find }\lambda,\mu\text{ such that }A^2=\lambda A+\mu I.$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\displaystyle A^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}=\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$\displaystyle \lambda A+\mu I=\lambda\begin{bmatrix}2&3\\1&2\end{bmatrix}+\mu\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda\\\lambda&2\lambda+\mu\end{bmatrix}$
$\displaystyle \text{Equating the corresponding entries,}$
$\displaystyle 12=3\lambda,\qquad 4=\lambda,\qquad 7=2\lambda+\mu$
$\displaystyle \Rightarrow \lambda=4,\qquad 7=8+\mu$
$\displaystyle \Rightarrow \mu=-1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 39: } \text{Find the value of }x\text{ for which the matrix product}$
$\displaystyle \begin{bmatrix}2&0&7\\0&1&0\\1&-2&1\end{bmatrix}\begin{bmatrix}-x&14x&7x\\0&1&0\\x&-4x&-2x\end{bmatrix}\text{ is equal to an identity matrix.}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{bmatrix}2&0&7\\0&1&0\\1&-2&1\end{bmatrix}\begin{bmatrix}-x&14x&7x\\0&1&0\\x&-4x&-2x\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}-2x+7x&28x-28x&14x-14x\\0&1&0\\-x+x&14x-2-4x&7x-2x\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}5x&0&0\\0&1&0\\0&10x-2&5x\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle \text{Equating the corresponding entries,}$
$\displaystyle 5x=1\quad\text{and}\quad 10x-2=0$
$\displaystyle \Rightarrow x=\frac{1}{5}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 40: } \text{Solve the matrix equations:}$
$\displaystyle \text{(i) }\begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&0\\-2&3\end{bmatrix}\begin{bmatrix}x\\5\end{bmatrix}=0$
$\displaystyle \text{(ii) }\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0$
$\displaystyle \text{(iii) }\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$
$\displaystyle \text{(iv) }\begin{bmatrix}2x&3\end{bmatrix}\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}x\\8\end{bmatrix}=0$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$
$\displaystyle \begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&0\\-2&3\end{bmatrix}\begin{bmatrix}x\\5\end{bmatrix}=0$
$\displaystyle \Rightarrow\begin{bmatrix}x-2&3\end{bmatrix}\begin{bmatrix}x\\5\end{bmatrix}=0$
$\displaystyle \Rightarrow x(x-2)+15=0$
$\displaystyle \Rightarrow x^2-2x+15=0$
$\displaystyle \text{Since }(-2)^2-4(1)(15)=-56<0,\text{ there is no real solution.}$
$\displaystyle \text{Over the complex numbers, }x=\frac{2\pm\sqrt{-56}}{2}=1\pm i\sqrt{14}.$

$\displaystyle \text{(ii)}$
$\displaystyle \begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0$
$\displaystyle \Rightarrow\begin{bmatrix}6&2&4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0$
$\displaystyle \Rightarrow 4+4x=0$
$\displaystyle \Rightarrow x=-1.$

$\displaystyle \text{(iii)}$
$\displaystyle \begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$
$\displaystyle \Rightarrow\begin{bmatrix}x-2&-10&2x-8\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$
$\displaystyle \Rightarrow x(x-2)-40+(2x-8)=0$
$\displaystyle \Rightarrow x^2-48=0$
$\displaystyle \Rightarrow x^2=48$
$\displaystyle \Rightarrow x=\pm4\sqrt{3}.$

$\displaystyle \text{(iv)}$
$\displaystyle \begin{bmatrix}2x&3\end{bmatrix}\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}x\\8\end{bmatrix}=0$
$\displaystyle \Rightarrow\begin{bmatrix}2x-9&4x\end{bmatrix}\begin{bmatrix}x\\8\end{bmatrix}=0$
$\displaystyle \Rightarrow x(2x-9)+32x=0$
$\displaystyle \Rightarrow 2x^2+23x=0$
$\displaystyle \Rightarrow x(2x+23)=0$
$\displaystyle \Rightarrow x=0\text{ or }x=-\frac{23}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 41: } \text{If }A=\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix},\text{ compute }A^2-4A+3I_3.$
$\displaystyle \text{Answer:}$
$\displaystyle A^2-4A+3I_3$
$\displaystyle =\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}-4\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}+3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+6+0&2-8+0&0+10+0\\3-12+0&6+16-5&0-20+15\\0-3+0&0+4-3&0-5+9\end{bmatrix}-\begin{bmatrix}4&8&0\\12&-16&20\\0&-4&12\end{bmatrix}+\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}-\begin{bmatrix}4&8&0\\12&-16&20\\0&-4&12\end{bmatrix}+\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}7-4+3&-6-8+0&10-0+0\\-9-12+0&17+16+3&-5-20+0\\-3-0+0&1+4+0&4-12+3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}6&-14&10\\-21&36&-25\\-3&5&-5\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 42: } \text{If }f(x)=x^2-2x,\text{ find }f(A),\text{ where }A=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(A)=A^2-2A$
$\displaystyle =\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}-2\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0+4+0&0+5+4&0+0+6\\0+20+0&4+25+0&8+0+0\\0+8+0&0+10+6&0+0+9\end{bmatrix}-\begin{bmatrix}0&2&4\\8&10&0\\0&4&6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-\begin{bmatrix}0&2&4\\8&10&0\\0&4&6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4-0&9-2&6-4\\20-8&29-10&8-0\\8-0&16-4&9-6\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4&7&2\\12&19&8\\8&12&3\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 43: } \text{If }f(x)=x^3+4x^2-x,\text{ find }f(A),\text{ where }A=\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(A)=A^3+4A^2-A$
$\displaystyle A^2=\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0+2+2&0-3-2&0+0+0\\0-6+0&2+9+0&4+0+0\\0-2+0&1+3+0&2+0+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4&-5&0\\-6&11&4\\-2&4&2\end{bmatrix}$
$\displaystyle A^3=A^2A$
$\displaystyle =\begin{bmatrix}4&-5&0\\-6&11&4\\-2&4&2\end{bmatrix}\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0-10+0&4+15+0&8+0+0\\0+22+4&-6-33-4&-12+0+0\\0+8+2&-2-12-2&-4+0+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-10&19&8\\26&-43&-12\\10&-16&-4\end{bmatrix}$
$\displaystyle f(A)=\begin{bmatrix}-10&19&8\\26&-43&-12\\10&-16&-4\end{bmatrix}+4\begin{bmatrix}4&-5&0\\-6&11&4\\-2&4&2\end{bmatrix}-\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-10&19&8\\26&-43&-12\\10&-16&-4\end{bmatrix}+\begin{bmatrix}16&-20&0\\-24&44&16\\-8&16&8\end{bmatrix}-\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-10+16-0&19-20-1&8+0-2\\26-24-2&-43+44+3&-12+16-0\\10-8-1&-16+16+1&-4+8-0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}6&-2&6\\0&4&4\\1&1&4\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 44: } \text{If }A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix},\text{ show that }A\text{ is a root of the polynomial }f(x)=x^3-6x^2+7x+2.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3-6x^2+7x+2$
$\displaystyle A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$\displaystyle A^2=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}$
$\displaystyle =\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}$
$\displaystyle A^3=A^2A$
$\displaystyle =\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39\end{bmatrix}$
$\displaystyle =\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}$
$\displaystyle f(A)=A^3-6A^2+7A+2I_3$
$\displaystyle =\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}+\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O.$
$\displaystyle \therefore f(A)=O.$
$\displaystyle \text{Hence, }A\text{ is a root of }f(x)=x^3-6x^2+7x+2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 45: } A=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix},\text{ then prove that }A^2-4A-5I=O. \hspace{2.0cm}\text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$
$\displaystyle A^2=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+4+4&2+2+4&2+4+2\\2+2+4&4+1+4&4+2+2\\2+4+2&4+2+2&4+4+1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}$
$\displaystyle A^2-4A-5I=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-4\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}-5\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-\begin{bmatrix}4&8&8\\8&4&8\\8&8&4\end{bmatrix}-\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}$
$\displaystyle =\begin{bmatrix}9-4-5&8-8-0&8-8-0\\8-8-0&9-4-5&8-8-0\\8-8-0&8-8-0&9-4-5\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O.$
$\displaystyle \therefore A^2-4A-5I=O.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 46: } A=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}\text{ show that }A^2-7A+10I_3=O.$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}$
$\displaystyle A^2=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}$
$\displaystyle =\begin{bmatrix}9+2+0&6+8+0&0+0+0\\3+4+0&2+16+0&0+0+0\\0+0+0&0+0+0&0+0+25\end{bmatrix}$
$\displaystyle =\begin{bmatrix}11&14&0\\7&18&0\\0&0&25\end{bmatrix}$
$\displaystyle A^2-7A+10I_3=\begin{bmatrix}11&14&0\\7&18&0\\0&0&25\end{bmatrix}-7\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}+10\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}11&14&0\\7&18&0\\0&0&25\end{bmatrix}-\begin{bmatrix}21&14&0\\7&28&0\\0&0&35\end{bmatrix}+\begin{bmatrix}10&0&0\\0&10&0\\0&0&10\end{bmatrix}$
$\displaystyle =\begin{bmatrix}11-21+10&14-14+0&0-0+0\\7-7+0&18-28+10&0-0+0\\0-0+0&0-0+0&25-35+10\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O.$
$\displaystyle \therefore A^2-7A+10I_3=O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 47: } \text{Without using the concept of inverse of a matrix, find the matrix }\begin{bmatrix}x&y\\z&u\end{bmatrix}\text{ such that}$
$\displaystyle \begin{bmatrix}5&-7\\-2&3\end{bmatrix}\begin{bmatrix}x&y\\z&u\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{bmatrix}5&-7\\-2&3\end{bmatrix}\begin{bmatrix}x&y\\z&u\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}5x-7z&5y-7u\\-2x+3z&-2y+3u\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
$\displaystyle \text{Equating the corresponding elements,}$
$\displaystyle 5x-7z=-16\qquad ...(1)$
$\displaystyle 5y-7u=-6\qquad ...(2)$
$\displaystyle -2y+3u=2$
$\displaystyle \Rightarrow 3u=2+2y$
$\displaystyle \Rightarrow u=\frac{2+2y}{3}\qquad ...(3)$
$\displaystyle -2x+3z=7$
$\displaystyle \Rightarrow 3z=7+2x$
$\displaystyle \Rightarrow z=\frac{7+2x}{3}\qquad ...(4)$
$\displaystyle \text{Substituting the value of }z\text{ from equation (4) in equation (1),}$
$\displaystyle 5x-7\left(\frac{7+2x}{3}\right)=-16$
$\displaystyle \Rightarrow 15x-49-14x=-48$
$\displaystyle \Rightarrow x-49=-48$
$\displaystyle \Rightarrow x=1$
$\displaystyle \text{Substituting }x=1\text{ in equation (4),}$
$\displaystyle z=\frac{7+2(1)}{3}=\frac{9}{3}=3$
$\displaystyle \text{Substituting the value of }u\text{ from equation (3) in equation (2),}$
$\displaystyle 5y-7\left(\frac{2+2y}{3}\right)=-6$
$\displaystyle \Rightarrow 15y-14-14y=-18$
$\displaystyle \Rightarrow y-14=-18$
$\displaystyle \Rightarrow y=-4$
$\displaystyle \text{Substituting }y=-4\text{ in equation (3),}$
$\displaystyle u=\frac{2+2(-4)}{3}=\frac{-6}{3}=-2$
$\displaystyle \therefore\begin{bmatrix}x&y\\z&u\end{bmatrix}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: } \text{Find the matrix }A\text{ such that}$
$\displaystyle \text{(i) }\begin{bmatrix}1&1\\0&1\end{bmatrix}A=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{Let }A=\begin{bmatrix}x&y&z\\a&b&c\end{bmatrix}.$
$\displaystyle \begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}x&y&z\\a&b&c\end{bmatrix}=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}x+a&y+b&z+c\\a&b&c\end{bmatrix}=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}$
$\displaystyle \text{Equating the corresponding elements,}$
$\displaystyle x+a=3,\qquad y+b=3,\qquad z+c=5$
$\displaystyle a=1,\qquad b=0,\qquad c=1$
$\displaystyle \text{Substituting }a=1\text{ in }x+a=3,$
$\displaystyle x+1=3\Rightarrow x=2$
$\displaystyle \text{Substituting }b=0\text{ in }y+b=3,$
$\displaystyle y+0=3\Rightarrow y=3$
$\displaystyle \text{Substituting }c=1\text{ in }z+c=5,$
$\displaystyle z+1=5\Rightarrow z=4$
$\displaystyle \therefore A=\begin{bmatrix}2&3&4\\1&0&1\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: } \text{Find the matrix }A\text{ such that}$
$\displaystyle \text{(ii) }A\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(ii)}$
$\displaystyle \text{Let }A=\begin{bmatrix}w&x\\y&z\end{bmatrix}.$
$\displaystyle \begin{bmatrix}w&x\\y&z\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}w+4x&2w+5x&3w+6x\\y+4z&2y+5z&3y+6z\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$
$\displaystyle \text{Equating the corresponding elements,}$
$\displaystyle w+4x=-7\qquad ...(1)$
$\displaystyle 3w+6x=-9\qquad ...(2)$
$\displaystyle y+4z=2\qquad ...(3)$
$\displaystyle 2y+5z=4\qquad ...(4)$
$\displaystyle \text{From equation (1),}$
$\displaystyle w=-7-4x\qquad ...(5)$
$\displaystyle \text{Substituting the value of }w\text{ from equation (5) in equation (2),}$
$\displaystyle 3(-7-4x)+6x=-9$
$\displaystyle \Rightarrow -21-12x+6x=-9$
$\displaystyle \Rightarrow -6x=12$
$\displaystyle \Rightarrow x=-2$
$\displaystyle \text{Substituting }x=-2\text{ in equation (5),}$
$\displaystyle w=-7-4(-2)=1$
$\displaystyle \text{From equation (3),}$
$\displaystyle y=2-4z\qquad ...(6)$
$\displaystyle \text{Substituting the value of }y\text{ from equation (6) in equation (4),}$
$\displaystyle 2(2-4z)+5z=4$
$\displaystyle \Rightarrow 4-8z+5z=4$
$\displaystyle \Rightarrow -3z=0$
$\displaystyle \Rightarrow z=0$
$\displaystyle \text{Substituting }z=0\text{ in equation (6),}$
$\displaystyle y=2-4(0)=2$
$\displaystyle \therefore A=\begin{bmatrix}1&-2\\2&0\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: } \text{Find the matrix }A\text{ such that}$
$\displaystyle \text{(iii) }\begin{bmatrix}4\\1\\3\end{bmatrix}A=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iii)}$
$\displaystyle \text{Let }A=\begin{bmatrix}x&y&z\end{bmatrix}.$
$\displaystyle \begin{bmatrix}4\\1\\3\end{bmatrix}\begin{bmatrix}x&y&z\end{bmatrix}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}4x&4y&4z\\x&y&z\\3x&3y&3z\end{bmatrix}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}$
$\displaystyle \text{Equating the corresponding elements,}$
$\displaystyle 4x=-4,\qquad 4y=8,\qquad 4z=4$
$\displaystyle \Rightarrow x=-1,\qquad y=2,\qquad z=1$
$\displaystyle \therefore A=\begin{bmatrix}-1&2&1\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: } \text{Find the matrix }A\text{ such that}$
$\displaystyle \text{(iv) }\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=A.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iv)}$
$\displaystyle A=\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-2-1+0&0+1+3&-2+0+3\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-3&4&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-3+0-1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-4\end{bmatrix}.$
$\displaystyle \therefore A=\begin{bmatrix}-4\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: } \text{Find the matrix }A\text{ such that}$
$\displaystyle \text{(v) }\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}A=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}\hspace{2.0cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(v)}$
$\displaystyle \text{Let }A=\begin{bmatrix}x&y&z\\a&b&c\end{bmatrix}.$
$\displaystyle \begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\begin{bmatrix}x&y&z\\a&b&c\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}2x-a&2y-b&2z-c\\x&y&z\\-3x+4a&-3y+4b&-3z+4c\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}$
$\displaystyle \text{Equating the corresponding elements of the second row,}$
$\displaystyle x=1,\qquad y=-2,\qquad z=-5$
$\displaystyle \text{Equating the corresponding elements of the first row,}$
$\displaystyle 2x-a=-1$
$\displaystyle \Rightarrow 2(1)-a=-1$
$\displaystyle \Rightarrow a=3$
$\displaystyle 2y-b=-8$
$\displaystyle \Rightarrow 2(-2)-b=-8$
$\displaystyle \Rightarrow b=4$
$\displaystyle 2z-c=-10$
$\displaystyle \Rightarrow 2(-5)-c=-10$
$\displaystyle \Rightarrow c=0$
$\displaystyle \therefore A=\begin{bmatrix}1&-2&-5\\3&4&0\end{bmatrix}.$
$\displaystyle \text{Verification:}$
$\displaystyle \begin{bmatrix}-3&4\end{bmatrix}\begin{bmatrix}1&-2&-5\\3&4&0\end{bmatrix}=\begin{bmatrix}-3+12&6+16&15+0\end{bmatrix}=\begin{bmatrix}9&22&15\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: } \text{Find the matrix }A\text{ such that}$
$\displaystyle \text{(vi) }A\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}\hspace{2.0cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(vi)}$
$\displaystyle \text{Let }A=\begin{bmatrix}x&a\\y&b\\z&c\end{bmatrix}.$
$\displaystyle \begin{bmatrix}x&a\\y&b\\z&c\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}$
$\displaystyle \Rightarrow\begin{bmatrix}x+4a&2x+5a&3x+6a\\y+4b&2y+5b&3y+6b\\z+4c&2z+5c&3z+6c\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}$
$\displaystyle \text{Equating the corresponding elements of the first row,}$
$\displaystyle x+4a=-7\qquad ...(1)$
$\displaystyle 2x+5a=-8\qquad ...(2)$
$\displaystyle \text{Multiplying equation (1) by }2\text{ and subtracting equation (2),}$
$\displaystyle 3a=-6\Rightarrow a=-2$
$\displaystyle \text{Substituting }a=-2\text{ in equation (1),}$
$\displaystyle x+4(-2)=-7\Rightarrow x=1$
$\displaystyle \text{Equating the corresponding elements of the second row,}$
$\displaystyle y+4b=2\qquad ...(3)$
$\displaystyle 2y+5b=4\qquad ...(4)$
$\displaystyle \text{Multiplying equation (3) by }2\text{ and subtracting equation (4),}$
$\displaystyle 3b=0\Rightarrow b=0$
$\displaystyle \text{Substituting }b=0\text{ in equation (3),}$
$\displaystyle y=2$
$\displaystyle \text{Equating the corresponding elements of the third row,}$
$\displaystyle z+4c=11\qquad ...(5)$
$\displaystyle 2z+5c=10\qquad ...(6)$
$\displaystyle \text{Multiplying equation (5) by }2\text{ and subtracting equation (6),}$
$\displaystyle 3c=12\Rightarrow c=4$
$\displaystyle \text{Substituting }c=4\text{ in equation (5),}$
$\displaystyle z+16=11\Rightarrow z=-5$
$\displaystyle \therefore A=\begin{bmatrix}1&-2\\2&0\\-5&4\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 49: }\text{Find a }2\times2\text{ matrix }A\text{ such that }A\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6I_2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}w&x\\y&z\end{bmatrix}.$
$\displaystyle \begin{bmatrix}w&x\\y&z\end{bmatrix}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6I_2$
$\displaystyle \Rightarrow\begin{bmatrix}w+x&-2w+4x\\y+z&-2y+4z\end{bmatrix}=\begin{bmatrix}6&0\\0&6\end{bmatrix}$
$\displaystyle \text{Equating the corresponding elements,}$
$\displaystyle w+x=6\qquad ...(1)$
$\displaystyle -2w+4x=0\qquad ...(2)$
$\displaystyle \text{From equation (1),}$
$\displaystyle w=6-x\qquad ...(3)$
$\displaystyle \text{Substituting the value of }w\text{ from equation (3) in equation (2),}$
$\displaystyle -2(6-x)+4x=0$
$\displaystyle \Rightarrow -12+2x+4x=0$
$\displaystyle \Rightarrow 6x=12$
$\displaystyle \Rightarrow x=2$
$\displaystyle \text{Substituting }x=2\text{ in equation (3),}$
$\displaystyle w=6-2=4$
$\displaystyle y+z=0\qquad ...(4)$
$\displaystyle -2y+4z=6\qquad ...(5)$
$\displaystyle \text{From equation (4),}$
$\displaystyle y=-z\qquad ...(6)$
$\displaystyle \text{Substituting the value of }y\text{ from equation (6) in equation (5),}$
$\displaystyle -2(-z)+4z=6$
$\displaystyle \Rightarrow 2z+4z=6$
$\displaystyle \Rightarrow 6z=6$
$\displaystyle \Rightarrow z=1$
$\displaystyle \text{Substituting }z=1\text{ in equation (6),}$
$\displaystyle y=-1$
$\displaystyle \therefore A=\begin{bmatrix}4&2\\-1&1\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 50: }\text{If }A=\begin{bmatrix}0&0\\4&0\end{bmatrix},\text{ find }A^{16}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }A=\begin{bmatrix}0&0\\4&0\end{bmatrix}.$
$\displaystyle A^2=AA$
$\displaystyle \Rightarrow A^2=\begin{bmatrix}0&0\\4&0\end{bmatrix}\begin{bmatrix}0&0\\4&0\end{bmatrix}$
$\displaystyle \Rightarrow A^2=\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$\displaystyle \Rightarrow A^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O$
$\displaystyle \text{Hence, }A^n=O\text{ for all }n\ge2.$
$\displaystyle \therefore A^{16}=O=\begin{bmatrix}0&0\\0&0\end{bmatrix}.$
$\displaystyle \text{Thus, }A^{16}\text{ is a null matrix.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 51: }\text{If }A=\begin{bmatrix}0&-x\\x&0\end{bmatrix},\ B=\begin{bmatrix}0&1\\1&0\end{bmatrix}\text{ and }x^2=-1,\text{ then show that }(A+B)^2=A^2+B^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }A=\begin{bmatrix}0&-x\\x&0\end{bmatrix},\ B=\begin{bmatrix}0&1\\1&0\end{bmatrix}\text{ and }x^2=-1.$
$\displaystyle \text{LHS:}$
$\displaystyle A+B=\begin{bmatrix}0&-x\\x&0\end{bmatrix}+\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}0&1-x\\x+1&0\end{bmatrix}$
$\displaystyle (A+B)^2=\begin{bmatrix}0&1-x\\x+1&0\end{bmatrix}\begin{bmatrix}0&1-x\\x+1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}(1-x)(x+1)&0\\0&(x+1)(1-x)\end{bmatrix}=\begin{bmatrix}1-x^2&0\\0&1-x^2\end{bmatrix}\qquad ...(1)$
$\displaystyle \text{RHS:}$
$\displaystyle A^2=\begin{bmatrix}0&-x\\x&0\end{bmatrix}\begin{bmatrix}0&-x\\x&0\end{bmatrix}=\begin{bmatrix}-x^2&0\\0&-x^2\end{bmatrix}$
$\displaystyle B^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle A^2+B^2=\begin{bmatrix}-x^2+1&0\\0&-x^2+1\end{bmatrix}=\begin{bmatrix}1-x^2&0\\0&1-x^2\end{bmatrix}\qquad ...(2)$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle (A+B)^2=A^2+B^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 52: }\text{If }A=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix},\text{ then verify that }A^2+A=A(A+I),\text{ where }I\text{ is the identity matrix.}$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$\displaystyle A^2=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1+0+0&0+0-3&-3+0-3\\2+2+0&0+1+3&-6+3+3\\0+2+0&0+1+1&0+3+1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&-3&-6\\4&4&0\\2&2&4\end{bmatrix}$
$\displaystyle \text{LHS:}$
$\displaystyle A^2+A=\begin{bmatrix}1&-3&-6\\4&4&0\\2&2&4\end{bmatrix}+\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2&-3&-9\\6&5&3\\2&3&5\end{bmatrix}$
$\displaystyle \text{RHS:}$
$\displaystyle A+I=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}+\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}2&0&-3\\2&2&3\\0&1&2\end{bmatrix}$
$\displaystyle A(A+I)=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}\begin{bmatrix}2&0&-3\\2&2&3\\0&1&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2+0+0&0+0-3&-3+0-6\\4+2+0&0+2+3&-6+3+6\\0+2+0&0+2+1&0+3+2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2&-3&-9\\6&5&3\\2&3&5\end{bmatrix}$
$\displaystyle \therefore \text{LHS}=\text{RHS}.$
$\displaystyle \text{Hence, }A^2+A=A(A+I)\text{ is verified.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 53: }\text{If }A=\begin{bmatrix}3&-5\\-4&2\end{bmatrix},\text{ then find }A^2-5A-14I.\text{ Hence, obtain }A^3.$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
$\displaystyle A^2=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}9+20&-15-10\\-12-8&20+4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}29&-25\\-20&24\end{bmatrix}$
$\displaystyle A^2-5A-14I=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-5\begin{bmatrix}3&-5\\-4&2\end{bmatrix}-14\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}29-15-14&-25+25\\-20+20&24-10-14\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O$
$\displaystyle \therefore A^2-5A-14I=O\qquad ...(1)$
$\displaystyle \text{Premultiplying equation (1) by }A,\text{ we get}$
$\displaystyle A(A^2-5A-14I)=AO$
$\displaystyle \Rightarrow A^3-5A^2-14A=O$
$\displaystyle \Rightarrow A^3=5A^2+14A$
$\displaystyle A^3=5\begin{bmatrix}29&-25\\-20&24\end{bmatrix}+14\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}145+42&-125-70\\-100-56&120+28\end{bmatrix}$
$\displaystyle =\begin{bmatrix}187&-195\\-156&148\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 54:}$
$\displaystyle \text{(i) If }P(x)=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix},\text{ then show that }P(x)P(y)=P(x+y)=P(y)P(x).$
$\displaystyle \text{(ii) If }P=\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}\text{ and }Q=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix},\text{ prove that }PQ=\begin{bmatrix}xa&0&0\\0&yb&0\\0&0&zc\end{bmatrix}=QP.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle P(x)=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix},\qquad P(y)=\begin{bmatrix}\cos y&\sin y\\-\sin y&\cos y\end{bmatrix}$
$\displaystyle P(x)P(y)=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}\begin{bmatrix}\cos y&\sin y\\-\sin y&\cos y\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos x\cos y-\sin x\sin y&\cos x\sin y+\sin x\cos y\\-\sin x\cos y-\cos x\sin y&-\sin x\sin y+\cos x\cos y\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos(x+y)&\sin(x+y)\\-\sin(x+y)&\cos(x+y)\end{bmatrix}$
$\displaystyle =P(x+y)\qquad ...(1)$
$\displaystyle P(y)P(x)=\begin{bmatrix}\cos y&\sin y\\-\sin y&\cos y\end{bmatrix}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos y\cos x-\sin y\sin x&\cos y\sin x+\sin y\cos x\\-\sin y\cos x-\cos y\sin x&-\sin y\sin x+\cos y\cos x\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos(x+y)&\sin(x+y)\\-\sin(x+y)&\cos(x+y)\end{bmatrix}$
$\displaystyle =P(x+y)\qquad ...(2)$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle P(x)P(y)=P(x+y)=P(y)P(x).$
$\displaystyle \\$
$\displaystyle \text{(ii)}$
$\displaystyle PQ=\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}$
$\displaystyle =\begin{bmatrix}xa&0&0\\0&yb&0\\0&0&zc\end{bmatrix}\qquad ...(3)$
$\displaystyle QP=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}$
$\displaystyle =\begin{bmatrix}ax&0&0\\0&by&0\\0&0&cz\end{bmatrix}$
$\displaystyle =\begin{bmatrix}xa&0&0\\0&yb&0\\0&0&zc\end{bmatrix}\qquad ...(4)$
$\displaystyle \text{From (3) and (4),}$
$\displaystyle PQ=\begin{bmatrix}xa&0&0\\0&yb&0\\0&0&zc\end{bmatrix}=QP.$
$\displaystyle \\$

$\displaystyle \textbf{Question 55: }A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix},\text{ find }A^2-5A+4I\text{ and hence find a matrix }X\text{ such that }A^2-5A+4I+X=O.\hspace{2.0cm}\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}$
$\displaystyle A^2=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4+0+1&0+0-1&2+0+0\\4+2+3&0+1-3&2+3+0\\2-2+0&0-1+0&1-3+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}$
$\displaystyle A^2-5A+4I=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}+4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}5-10+4&-1-0+0&2-5+0\\9-10+0&-2-5+4&5-15+0\\0-5+0&-1+5+0&-2-0+4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}$
$\displaystyle \text{Now, }A^2-5A+4I+X=O$
$\displaystyle \Rightarrow X=-(A^2-5A+4I)$
$\displaystyle \therefore X=-\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&1&3\\1&3&10\\5&-4&-2\end{bmatrix}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 56: }\text{If }A=\begin{bmatrix}1&1\\0&1\end{bmatrix},\text{ prove that }A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix}\text{ for all positive integers }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on }n.$
$\displaystyle \text{Step 1: For }n=1,$
$\displaystyle A^1=A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&n\\0&1\end{bmatrix}\text{ for }n=1.$
$\displaystyle \text{Therefore, the result is true for }n=1.$
$\displaystyle \text{Step 2: Assume that the result is true for }n=m.$
$\displaystyle \text{Then, }A^m=\begin{bmatrix}1&m\\0&1\end{bmatrix}\qquad ...(1)$
$\displaystyle \text{We shall now prove that the result is true for }n=m+1.$
$\displaystyle A^{m+1}=A^mA$
$\displaystyle =\begin{bmatrix}1&m\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}\qquad \text{[Using (1)]}$
$\displaystyle =\begin{bmatrix}1+0&1+m\\0+0&0+1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&m+1\\0&1\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=m+1.$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix}\text{ for every positive integer }n.$
$\displaystyle \\$

$\displaystyle \textbf{Question 57: }\text{If }A=\begin{bmatrix}a&b\\0&1\end{bmatrix},\ a\ne1,\text{ prove that }A^n=\begin{bmatrix}a^n&\dfrac{b(a^n-1)}{a-1}\\0&1\end{bmatrix}\text{ for every positive integer }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on }n.$
$\displaystyle \text{Step 1: For }n=1,$
$\displaystyle \begin{bmatrix}a^1&\dfrac{b(a^1-1)}{a-1}\\0&1\end{bmatrix}=\begin{bmatrix}a&b\\0&1\end{bmatrix}=A=A^1$
$\displaystyle \text{Therefore, the result is true for }n=1.$
$\displaystyle \text{Step 2: Assume that the result is true for }n=m.$
$\displaystyle \text{Then, }A^m=\begin{bmatrix}a^m&\dfrac{b(a^m-1)}{a-1}\\0&1\end{bmatrix}\qquad ...(1)$
$\displaystyle \text{We shall now prove that the result is true for }n=m+1.$
$\displaystyle A^{m+1}=A^mA$
$\displaystyle =\begin{bmatrix}a^m&\dfrac{b(a^m-1)}{a-1}\\0&1\end{bmatrix}\begin{bmatrix}a&b\\0&1\end{bmatrix}\qquad \text{[Using (1)]}$
$\displaystyle =\begin{bmatrix}a^{m+1}&a^mb+\dfrac{b(a^m-1)}{a-1}\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^{m+1}&\dfrac{a^mb(a-1)+b(a^m-1)}{a-1}\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^{m+1}&\dfrac{a^{m+1}b-a^mb+a^mb-b}{a-1}\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^{m+1}&\dfrac{b(a^{m+1}-1)}{a-1}\\0&1\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=m+1.$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle A^n=\begin{bmatrix}a^n&\dfrac{b(a^n-1)}{a-1}\\0&1\end{bmatrix}\text{ for every positive integer }n.$
$\displaystyle \\$

$\displaystyle \textbf{Question 58: }\text{If }A=\begin{bmatrix}\cos\theta&i\sin\theta\\i\sin\theta&\cos\theta\end{bmatrix},\text{ then prove by the principle of mathematical induction that}$
$\displaystyle A^n=\begin{bmatrix}\cos n\theta&i\sin n\theta\\i\sin n\theta&\cos n\theta\end{bmatrix}\text{ for all }n\in N.\hspace{2.0cm}\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on }n.$
$\displaystyle \text{Step 1: For }n=1,$
$\displaystyle A^1=A=\begin{bmatrix}\cos\theta&i\sin\theta\\i\sin\theta&\cos\theta\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos1\theta&i\sin1\theta\\i\sin1\theta&\cos1\theta\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=1.$
$\displaystyle \text{Step 2: Assume that the result is true for }n=m.$
$\displaystyle \text{Then, }A^m=\begin{bmatrix}\cos m\theta&i\sin m\theta\\i\sin m\theta&\cos m\theta\end{bmatrix}\qquad ...(1)$
$\displaystyle \text{We shall now prove that the result is true for }n=m+1.$
$\displaystyle A^{m+1}=A^mA$
$\displaystyle =\begin{bmatrix}\cos m\theta&i\sin m\theta\\i\sin m\theta&\cos m\theta\end{bmatrix}\begin{bmatrix}\cos\theta&i\sin\theta\\i\sin\theta&\cos\theta\end{bmatrix}\qquad \text{[Using (1)]}$
$\displaystyle =\begin{bmatrix}\cos m\theta\cos\theta+i^2\sin m\theta\sin\theta&i(\cos m\theta\sin\theta+\sin m\theta\cos\theta)\\i(\sin m\theta\cos\theta+\cos m\theta\sin\theta)&i^2\sin m\theta\sin\theta+\cos m\theta\cos\theta\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos m\theta\cos\theta-\sin m\theta\sin\theta&i(\cos m\theta\sin\theta+\sin m\theta\cos\theta)\\i(\sin m\theta\cos\theta+\cos m\theta\sin\theta)&\cos m\theta\cos\theta-\sin m\theta\sin\theta\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos(m\theta+\theta)&i\sin(m\theta+\theta)\\i\sin(m\theta+\theta)&\cos(m\theta+\theta)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos((m+1)\theta)&i\sin((m+1)\theta)\\i\sin((m+1)\theta)&\cos((m+1)\theta)\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=m+1.$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle A^n=\begin{bmatrix}\cos n\theta&i\sin n\theta\\i\sin n\theta&\cos n\theta\end{bmatrix}\text{ for all }n\in N.$
$\displaystyle \\$

$\displaystyle \textbf{Question 59: }A=\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix}.\text{ Prove that}$
$\displaystyle A^n=\begin{bmatrix}\cos n\alpha+\sin n\alpha&\sqrt{2}\sin n\alpha\\-\sqrt{2}\sin n\alpha&\cos n\alpha-\sin n\alpha\end{bmatrix},\quad \text{for all }n\in\mathbb{N}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on }n.$
$\displaystyle \text{Step 1: For }n=1,$
$\displaystyle A^1=A=\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos1\alpha+\sin1\alpha&\sqrt{2}\sin1\alpha\\-\sqrt{2}\sin1\alpha&\cos1\alpha-\sin1\alpha\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=1.$
$\displaystyle \text{Step 2: Assume that the result is true for }n=m.$
$\displaystyle \text{Then, }A^m=\begin{bmatrix}\cos m\alpha+\sin m\alpha&\sqrt{2}\sin m\alpha\\-\sqrt{2}\sin m\alpha&\cos m\alpha-\sin m\alpha\end{bmatrix}\qquad ...(1)$
$\displaystyle \text{We shall now prove that the result is true for }n=m+1.$
$\displaystyle A^{m+1}=A^mA$
$\displaystyle =\begin{bmatrix}\cos m\alpha+\sin m\alpha&\sqrt{2}\sin m\alpha\\-\sqrt{2}\sin m\alpha&\cos m\alpha-\sin m\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix}\qquad \text{[Using (1)]}$
$\displaystyle =\begin{bmatrix}(\cos m\alpha+\sin m\alpha)(\cos\alpha+\sin\alpha)-2\sin m\alpha\sin\alpha&\sqrt{2}\{(\cos m\alpha+\sin m\alpha)\sin\alpha+\sin m\alpha(\cos\alpha-\sin\alpha)\}\\-\sqrt{2}\{\sin m\alpha(\cos\alpha+\sin\alpha)+(\cos m\alpha-\sin m\alpha)\sin\alpha\}&-2\sin m\alpha\sin\alpha+(\cos m\alpha-\sin m\alpha)(\cos\alpha-\sin\alpha)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos m\alpha\cos\alpha-\sin m\alpha\sin\alpha+\cos m\alpha\sin\alpha+\sin m\alpha\cos\alpha&\sqrt{2}(\cos m\alpha\sin\alpha+\sin m\alpha\cos\alpha)\\-\sqrt{2}(\sin m\alpha\cos\alpha+\cos m\alpha\sin\alpha)&\cos m\alpha\cos\alpha-\sin m\alpha\sin\alpha-\cos m\alpha\sin\alpha-\sin m\alpha\cos\alpha\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos(m\alpha+\alpha)+\sin(m\alpha+\alpha)&\sqrt{2}\sin(m\alpha+\alpha)\\-\sqrt{2}\sin(m\alpha+\alpha)&\cos(m\alpha+\alpha)-\sin(m\alpha+\alpha)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}\cos((m+1)\alpha)+\sin((m+1)\alpha)&\sqrt{2}\sin((m+1)\alpha)\\-\sqrt{2}\sin((m+1)\alpha)&\cos((m+1)\alpha)-\sin((m+1)\alpha)\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=m+1.$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle A^n=\begin{bmatrix}\cos n\alpha+\sin n\alpha&\sqrt{2}\sin n\alpha\\-\sqrt{2}\sin n\alpha&\cos n\alpha-\sin n\alpha\end{bmatrix}\text{ for all }n\in\mathbb{N}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 60:}$
$\displaystyle \text{Let }A=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}.\text{ Use the principle of mathematical induction to show that}$
$\displaystyle A^n=\begin{bmatrix}1&n&\dfrac{n(n+1)}{2}\\0&1&n\\0&0&1\end{bmatrix}\text{ for every positive integer }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on }n.$
$\displaystyle \text{Step 1: For }n=1,$
$\displaystyle A^1=A=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&1&\dfrac{1(1+1)}{2}\\0&1&1\\0&0&1\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=1.$
$\displaystyle \text{Step 2: Assume that the result is true for }n=m.$
$\displaystyle \text{Then, }A^m=\begin{bmatrix}1&m&\dfrac{m(m+1)}{2}\\0&1&m\\0&0&1\end{bmatrix}\qquad ...(1)$
$\displaystyle \text{We shall now prove that the result is true for }n=m+1.$
$\displaystyle A^{m+1}=A^mA$
$\displaystyle =\begin{bmatrix}1&m&\dfrac{m(m+1)}{2}\\0&1&m\\0&0&1\end{bmatrix}\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\qquad \text{[Using (1)]}$
$\displaystyle =\begin{bmatrix}1+0+0&1+m+0&1+m+\dfrac{m(m+1)}{2}\\0+0+0&0+1+0&0+1+m\\0+0+0&0+0+0&0+0+1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&m+1&\dfrac{2+2m+m^2+m}{2}\\0&1&m+1\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&m+1&\dfrac{m^2+3m+2}{2}\\0&1&m+1\\0&0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1&m+1&\dfrac{(m+1)(m+2)}{2}\\0&1&m+1\\0&0&1\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=m+1.$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle A^n=\begin{bmatrix}1&n&\dfrac{n(n+1)}{2}\\0&1&n\\0&0&1\end{bmatrix}\text{ for every positive integer }n.$
$\displaystyle \\$

$\displaystyle \textbf{Question 61: }\text{If }B\text{ and }C\text{ are square matrices of the same order and }A=B+C,\ BC=CB,\ C^2=O,\text{ then show that}$
$\displaystyle A^{n+1}=B^n\bigl(B+(n+1)C\bigr)\text{ for every }n\in\mathbb{N}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(n)\text{ denote the statement}$
$\displaystyle P(n):\ A^{n+1}=B^n\bigl(B+(n+1)C\bigr).$
$\displaystyle \text{For }n=1,$
$\displaystyle A^2=(B+C)(B+C)$
$\displaystyle =B^2+BC+CB+C^2$
$\displaystyle =B^2+2BC\qquad [\because\,BC=CB\text{ and }C^2=O]$
$\displaystyle =B(B+2C).$
$\displaystyle \text{Therefore, }P(1)\text{ is true.}$
$\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in\mathbb{N}.$
$\displaystyle \text{Then, }A^{k+1}=B^k\bigl(B+(k+1)C\bigr).\qquad ...(1)$
$\displaystyle \text{Now,}$
$\displaystyle A^{k+2}=A^{k+1}A$
$\displaystyle =B^k\bigl(B+(k+1)C\bigr)(B+C)\qquad \text{[Using (1)]}$
$\displaystyle =B^k\bigl(B^2+BC+(k+1)CB+(k+1)C^2\bigr)$
$\displaystyle =B^k\bigl(B^2+BC+(k+1)BC\bigr)$
$\displaystyle =B^k\bigl(B^2+(k+2)BC\bigr)$
$\displaystyle =B^{k+1}\bigl(B+(k+2)C\bigr).$
$\displaystyle \text{Therefore, }P(k+1)\text{ is true.}$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle A^{n+1}=B^n\bigl(B+(n+1)C\bigr)\text{ for every }n\in\mathbb{N}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 62: }\text{If }A=\text{diag}(a,b,c),\text{ show that }A^n=\text{diag}(a^n,b^n,c^n)\text{ for all positive integers }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on }n.$
$\displaystyle \text{Step 1: For }n=1,$
$\displaystyle A^1=A=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^1&0&0\\0&b^1&0\\0&0&c^1\end{bmatrix}$
$\displaystyle \text{Therefore, the result is true for }n=1.$
$\displaystyle \text{Step 2: Assume that the result is true for }n=m.$
$\displaystyle \text{Then, }A^m=\begin{bmatrix}a^m&0&0\\0&b^m&0\\0&0&c^m\end{bmatrix}\qquad ...(1)$
$\displaystyle \text{We shall now prove that the result is true for }n=m+1.$
$\displaystyle A^{m+1}=A^mA$
$\displaystyle =\begin{bmatrix}a^m&0&0\\0&b^m&0\\0&0&c^m\end{bmatrix}\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}\qquad \text{[Using (1)]}$
$\displaystyle =\begin{bmatrix}a^ma+0+0&0+0+0&0+0+0\\0+0+0&0+b^mb+0&0+0+0\\0+0+0&0+0+0&0+0+c^mc\end{bmatrix}$
$\displaystyle =\begin{bmatrix}a^{m+1}&0&0\\0&b^{m+1}&0\\0&0&c^{m+1}\end{bmatrix}$
$\displaystyle =\text{diag}(a^{m+1},b^{m+1},c^{m+1}).$
$\displaystyle \text{Therefore, the result is true for }n=m+1.$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle A^n=\text{diag}(a^n,b^n,c^n)\text{ for every positive integer }n.$
$\displaystyle \\$

$\displaystyle \textbf{Question 63: }\text{If }A\text{ is a square matrix, using mathematical induction prove that }(A^T)^n=(A^n)^T\text{ for all }n\in\mathbb{N}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(n)\text{ denote the statement}$
$\displaystyle P(n):\ (A^T)^n=(A^n)^T.$
$\displaystyle \text{For }n=1,$
$\displaystyle (A^T)^1=A^T=(A^1)^T.$
$\displaystyle \text{Therefore, }P(1)\text{ is true.}$
$\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in\mathbb{N}.$
$\displaystyle \text{Then, }(A^T)^k=(A^k)^T.\qquad ...(1)$
$\displaystyle \text{Now,}$
$\displaystyle (A^T)^{k+1}=A^T(A^T)^k$
$\displaystyle =A^T(A^k)^T\qquad \text{[Using (1)]}$
$\displaystyle =(A^kA)^T\qquad [\because\,(XY)^T=Y^TX^T]$
$\displaystyle =(A^{k+1})^T.$
$\displaystyle \text{Therefore, }P(k+1)\text{ is true.}$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle (A^T)^n=(A^n)^T\text{ for all }n\in\mathbb{N}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 64: }\text{A matrix }X\text{ has }(a+b)\text{ rows and }(a+2)\text{ columns, while the matrix } \\ Y\text{ has }(b+1)\text{ rows and }(a+3)\text{ columns. } \text{Both matrices }XY\text{ and }YX \\ \text{ exist. Find }a\text{ and }b.\text{ Can you say }XY\text{ and }YX\text{ are of the same type? Are they equal?}$
$\displaystyle \text{Answer:}$
$\displaystyle [X]_{(a+b)\times(a+2)},\qquad [Y]_{(b+1)\times(a+3)}$
$\displaystyle \text{Since }XY\text{ exists, the number of columns of }X\text{ equals the number of rows of }Y.$
$\displaystyle \Rightarrow a+2=b+1\qquad ...(1)$
$\displaystyle \text{Since }YX\text{ exists, the number of columns of }Y\text{ equals the number of rows of }X.$
$\displaystyle \Rightarrow a+3=a+b$
$\displaystyle \Rightarrow b=3$
$\displaystyle \text{Substituting }b=3\text{ in (1),}$
$\displaystyle a+2=4$
$\displaystyle \Rightarrow a=2$
$\displaystyle \therefore X\text{ is of order }5\times4\text{ and }Y\text{ is of order }4\times5.$
$\displaystyle \therefore XY\text{ is of order }5\times5,\qquad YX\text{ is of order }4\times4.$
$\displaystyle \text{Hence, }XY\text{ and }YX\text{ are not of the same type (order). Therefore, they cannot be equal.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 65:}$
$\displaystyle \text{Give examples of matrices:}$
$\displaystyle \text{(i) }A\text{ and }B\text{ such that }AB\ne BA.$
$\displaystyle \text{(ii) }A\text{ and }B\text{ such that }AB=O\text{ but }A\ne O\text{ and }B\ne O.$
$\displaystyle \text{(iii) }A\text{ and }B\text{ such that }AB=O\text{ but }BA\ne O.$
$\displaystyle \text{(iv) }A,B\text{ and }C\text{ such that }AB=AC\text{ but }B\ne C\text{ and }A\ne O.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{Let }A=\begin{bmatrix}1&-2\\3&2\end{bmatrix}\text{ and }B=\begin{bmatrix}2&3\\-1&2\end{bmatrix}.$
$\displaystyle AB=\begin{bmatrix}1&-2\\3&2\end{bmatrix}\begin{bmatrix}2&3\\-1&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2+2&3-4\\6-2&9+4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4&-1\\4&13\end{bmatrix}$
$\displaystyle BA=\begin{bmatrix}2&3\\-1&2\end{bmatrix}\begin{bmatrix}1&-2\\3&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}2+9&-4+6\\-1+6&2+4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}11&2\\5&6\end{bmatrix}$
$\displaystyle \therefore AB\ne BA.$
$\displaystyle \\$
$\displaystyle \text{(ii)}$
$\displaystyle \text{Let }A=\begin{bmatrix}0&2\\0&0\end{bmatrix}\text{ and }B=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$
$\displaystyle AB=\begin{bmatrix}0&2\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \text{Thus, }AB=O\text{ while }A\ne O\text{ and }B\ne O.$
$\displaystyle \\$
$\displaystyle \text{(iii)}$
$\displaystyle \text{Let }A=\begin{bmatrix}0&1\\0&0\end{bmatrix}\text{ and }B=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$
$\displaystyle AB=\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle BA=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&1\\0&0\end{bmatrix}\ne O.$
$\displaystyle \text{Thus, }AB=O\text{ but }BA\ne O.$
$\displaystyle \\$
$\displaystyle \text{(iv)}$
$\displaystyle \text{Let }A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad B=\begin{bmatrix}0&0\\0&1\end{bmatrix}\text{ and }C=\begin{bmatrix}0&0\\0&2\end{bmatrix}.$
$\displaystyle AB=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle AC=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&2\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$
$\displaystyle \therefore AB=AC.$
$\displaystyle \text{Also, }B\ne C\text{ and }A\ne O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 66: }\text{Let }A\text{ and }B\text{ be square matrices of the same order. Does }(A+B)^2=A^2+2AB+B^2\text{ hold? If not, why?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=(A+B)^2$
$\displaystyle =(A+B)(A+B)$
$\displaystyle =A(A+B)+B(A+B)$
$\displaystyle =A^2+AB+BA+B^2$
$\displaystyle \text{Since matrix multiplication is not commutative in general, }AB\ne BA.$
$\displaystyle \therefore AB+BA\ne2AB\text{ in general.}$
$\displaystyle \therefore (A+B)^2\ne A^2+2AB+B^2\text{ in general.}$
$\displaystyle \text{The equality holds only when }AB=BA.$
$\displaystyle \\$

$\displaystyle \textbf{Question 67:}$
$\displaystyle \text{If }A\text{ and }B\text{ are square matrices of the same order, explain why in general}$
$\displaystyle \text{(i) }(A+B)^2\ne A^2+2AB+B^2.$
$\displaystyle \text{(ii) }(A-B)^2\ne A^2-2AB+B^2.$
$\displaystyle \text{(iii) }(A+B)(A-B)\ne A^2-B^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{LHS}=(A+B)^2$
$\displaystyle =(A+B)(A+B)$
$\displaystyle =A(A+B)+B(A+B)$
$\displaystyle =A^2+AB+BA+B^2$
$\displaystyle \text{Since matrix multiplication is not commutative in general, }AB\ne BA.$
$\displaystyle \therefore AB+BA\ne2AB\text{ in general.}$
$\displaystyle \therefore (A+B)^2\ne A^2+2AB+B^2\text{ in general.}$
$\displaystyle \\$
$\displaystyle \text{(ii)}$
$\displaystyle \text{LHS}=(A-B)^2$
$\displaystyle =(A-B)(A-B)$
$\displaystyle =A(A-B)-B(A-B)$
$\displaystyle =A^2-AB-BA+B^2$
$\displaystyle \text{Since matrix multiplication is not commutative in general, }AB\ne BA.$
$\displaystyle \therefore -AB-BA\ne-2AB\text{ in general.}$
$\displaystyle \therefore (A-B)^2\ne A^2-2AB+B^2\text{ in general.}$
$\displaystyle \\$
$\displaystyle \text{(iii)}$
$\displaystyle \text{LHS}=(A+B)(A-B)$
$\displaystyle =A(A-B)+B(A-B)$
$\displaystyle =A^2-AB+BA-B^2$
$\displaystyle \text{Since matrix multiplication is not commutative in general, }AB\ne BA.$
$\displaystyle \therefore -AB+BA\ne0\text{ in general.}$
$\displaystyle \therefore (A+B)(A-B)\ne A^2-B^2\text{ in general.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 68: }\text{Let }A\text{ and }B\text{ be square matrices of order }3\times3.\text{ Is }(AB)^2=A^2B^2\text{? Give reasons.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{No, }(AB)^2=A^2B^2\text{ does not hold in general.}$
$\displaystyle (AB)^2=(AB)(AB)=A(BA)B$
$\displaystyle \text{Since matrix multiplication is not commutative in general, }BA\ne AB.$
$\displaystyle \therefore A(BA)B\ne A(AB)B=A^2B^2\text{ in general.}$
$\displaystyle \text{For example, let}$
$\displaystyle A=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix},\qquad B=\begin{bmatrix}0&0&0\\1&0&0\\0&0&0\end{bmatrix}.$
$\displaystyle AB=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\displaystyle \Rightarrow (AB)^2=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}.$
$\displaystyle \text{Also, }A^2=O\text{ and }B^2=O.$
$\displaystyle \therefore A^2B^2=O.$
$\displaystyle \therefore (AB)^2\ne A^2B^2.$
$\displaystyle \text{However, if }AB=BA,\text{ then}$
$\displaystyle (AB)^2=A(BA)B=A(AB)B=A^2B^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 69: }\text{If }A\text{ and }B\text{ are square matrices of the same order such that }AB=BA,\text{ then show that }(A+B)^2=A^2+2AB+B^2.$
$\displaystyle \text{Answer:}$
$\displaystyle (A+B)^2=(A+B)(A+B)$
$\displaystyle =A^2+AB+BA+B^2$
$\displaystyle =A^2+AB+AB+B^2\qquad [\because\,AB=BA]$
$\displaystyle =A^2+2AB+B^2.$
$\displaystyle \text{Hence, }(A+B)^2=A^2+2AB+B^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 70: }\text{Let }A=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix},\ B=\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}\text{ and }C=\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}.$
$\displaystyle \text{Verify that }AB=AC\text{ though }B\ne C,\ A\ne O.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }A=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix},\quad B=\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}\text{ and }C=\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}.$
$\displaystyle AB=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix}\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$
$\displaystyle =\begin{bmatrix}3+5-2&1+2+4\\9+15-6&3+6+12\end{bmatrix}$
$\displaystyle =\begin{bmatrix}6&7\\18&21\end{bmatrix}$
$\displaystyle AC=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix}\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4-3+5&2+5+0\\12-9+15&6+15+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}6&7\\18&21\end{bmatrix}$
$\displaystyle \therefore AB=AC.$
$\displaystyle \text{Also, }B\ne C\text{ and }A\ne O.$
$\displaystyle \\$

$\displaystyle \textbf{Question 71: }\text{Three shopkeepers A, B and C go to a store to buy stationery.}$
$\displaystyle \text{Shopkeeper A purchases }12\text{ dozen notebooks, }5\text{ dozen pens and }6\text{ dozen pencils.}$
$\displaystyle \text{Shopkeeper B purchases }10\text{ dozen notebooks, }6\text{ dozen pens and }7\text{ dozen pencils.}$
$\displaystyle \text{Shopkeeper C purchases }11\text{ dozen notebooks, }13\text{ dozen pens and }8\text{ dozen pencils.}$
$\displaystyle \text{The cost of one notebook is }40\text{ paise, one pen costs Rs. }1.25\text{ and one pencil costs}$
$\displaystyle 35\text{ paise. Using matrix multiplication, calculate the total bill amount for each}$
$\displaystyle \text{shopkeeper individually.}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|c|}\hline\text{Shopkeepers}&\text{Notebooks (dozen)}&\text{Pens (dozen)}&\text{Pencils (dozen)}\\ \hline A&12&5&6\\ \hline B&10&6&7\\ \hline C&11&13&8\\ \hline\end{array}$
$\displaystyle \text{Cost of one dozen notebooks}=12\times40\text{ paise}=\text{Rs. }4.80$
$\displaystyle \text{Cost of one dozen pens}=12\times1.25=\text{Rs. }15$
$\displaystyle \text{Cost of one dozen pencils}=12\times35\text{ paise}=\text{Rs. }4.20$
$\displaystyle \therefore\ \begin{bmatrix}12&5&6\\10&6&7\\11&13&8\end{bmatrix}\begin{bmatrix}4.80\\15\\4.20\end{bmatrix}=\begin{bmatrix}12(4.80)+5(15)+6(4.20)\\10(4.80)+6(15)+7(4.20)\\11(4.80)+13(15)+8(4.20)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}57.60+75+25.20\\48+90+29.40\\52.80+195+33.60\end{bmatrix}=\begin{bmatrix}157.80\\167.40\\281.40\end{bmatrix}$
$\displaystyle \text{Hence, the total bills of A, B and C are Rs. }157.80,\ \text{Rs. }167.40\text{ and Rs. }281.40\text{ respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 72: }\text{The cooperative store of a particular school has }10\text{ dozen physics books,}$
$\displaystyle \text{ }8\text{ dozen chemistry books and }5\text{ dozen mathematics books. The selling}$
$\displaystyle \text{price of each physics book is Rs. }8.30,\text{ each chemistry book is Rs. }3.45$
$\displaystyle \text{and each mathematics book is Rs. }4.50.\text{ Find the total amount of money}$
$\displaystyle \text{the store will receive by selling all the books.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Stock of books (in numbers) is}$
$\displaystyle X=\begin{bmatrix}120&96&60\end{bmatrix}$
$\displaystyle \text{where the entries represent Physics, Chemistry and Mathematics books respectively.}$
$\displaystyle \text{Selling prices of the books are}$
$\displaystyle Y=\begin{bmatrix}8.30\\3.45\\4.50\end{bmatrix}\begin{array}{l}\text{Physics}\\\text{Chemistry}\\\text{Mathematics}\end{array}$
$\displaystyle \text{Hence, the total amount received is}$
$\displaystyle XY=\begin{bmatrix}120&96&60\end{bmatrix}\begin{bmatrix}8.30\\3.45\\4.50\end{bmatrix}$
$\displaystyle =\left[(120)(8.30)+(96)(3.45)+(60)(4.50)\right]$
$\displaystyle =\left[996+331.20+270\right]$
$\displaystyle =\left[1597.20\right]$
$\displaystyle \therefore\ \text{The total amount received by the store is Rs. }1597.20.$
$\displaystyle \\$

$\displaystyle \textbf{Question 73:}$
$\displaystyle \text{In a legislative assembly election, a political group hired a public relations}$
$\displaystyle \text{firm to promote its candidates in three ways: telephone, house calls and}$
$\displaystyle \text{letters. The cost per contact (in paise) is given by the matrix }A\text{ as}$
$\displaystyle A=\begin{bmatrix}40\\100\\50\end{bmatrix}\begin{array}{l}\text{Telephone}\\\text{House call}\\\text{Letter}\end{array}$
$\displaystyle \text{The number of contacts of each type made in the two cities }X\text{ and }Y$
$\displaystyle \text{is given by the matrix }B\text{ as}$
$\displaystyle \begin{array}{ccc}\text{Telephone}&\text{House call}&\text{Letter}\end{array}$
$\displaystyle B=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{matrix}\rightarrow X\\\rightarrow Y\end{matrix}$
$\displaystyle \text{Find the total amount spent by the group in the two cities }X\text{ and }Y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The cost per contact (in paise) is given by}$
$\displaystyle A=\begin{bmatrix}40\\100\\50\end{bmatrix}\begin{array}{l}\text{Telephone}\\\text{House call}\\\text{Letter}\end{array}$
$\displaystyle \text{The number of contacts made in the two cities }X\text{ and }Y\text{ is}$
$\displaystyle B=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{array}{l}X\\Y\end{array}$
$\displaystyle \text{The total amount spent by the group in the two cities is}$
$\displaystyle BA=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{bmatrix}40\\100\\50\end{bmatrix}$
$\displaystyle =\begin{bmatrix}40000+50000+250000\\120000+100000+500000\end{bmatrix}$
$\displaystyle =\begin{bmatrix}340000\\720000\end{bmatrix}\text{ paise}$
$\displaystyle =\begin{bmatrix}3400\\7200\end{bmatrix}\text{ rupees}$
$\displaystyle \therefore\ \text{Amount spent in city }X=\text{Rs. }3400$
$\displaystyle \text{Amount spent in city }Y=\text{Rs. }7200$
$\displaystyle \\$

$\displaystyle \textbf{Question 74: }\text{A trust fund has Rs. }30000\text{ that must be invested in two}$
$\displaystyle \text{different types of bonds. The first bond pays }5\%\text{ interest per year and}$
$\displaystyle \text{the second bond pays }7\%\text{ interest per year. Using the matrix method,}$
$\displaystyle \text{determine how the amount of Rs. }30000\text{ should be divided between the two}$
$\displaystyle \text{bonds if the trust fund must earn a total annual interest of}$
$\displaystyle \text{(i) Rs. }1800\qquad \text{(ii) Rs. }2000.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let Rs. }x\text{ be invested in the first bond and Rs. }(30000-x)$
$\displaystyle \text{be invested in the second bond. Then}$
$\displaystyle A=\begin{bmatrix}x&30000-x\end{bmatrix}\text{ represents the investment and}$
$\displaystyle B=\begin{bmatrix}\frac{5}{100}\\[2pt]\frac{7}{100}\end{bmatrix}\text{ represents the rates of interest.}$
$\displaystyle \text{(i)}$
$\displaystyle \begin{bmatrix}x&30000-x\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\[2pt]\frac{7}{100}\end{bmatrix}=\begin{bmatrix}1800\end{bmatrix}$
$\displaystyle \Rightarrow\frac{5x}{100}+\frac{7(30000-x)}{100}=1800$
$\displaystyle \Rightarrow\frac{5x+210000-7x}{100}=1800$
$\displaystyle \Rightarrow210000-2x=180000$
$\displaystyle \Rightarrow2x=30000$
$\displaystyle \Rightarrow x=15000$
$\displaystyle \therefore\ \text{Amount invested in the first bond = Rs. }15000$
$\displaystyle \text{Amount invested in the second bond = Rs. }15000$
$\displaystyle \text{(ii)}$
$\displaystyle \begin{bmatrix}x&30000-x\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\[2pt]\frac{7}{100}\end{bmatrix}=\begin{bmatrix}2000\end{bmatrix}$
$\displaystyle \Rightarrow\frac{5x}{100}+\frac{7(30000-x)}{100}=2000$
$\displaystyle \Rightarrow\frac{5x+210000-7x}{100}=2000$
$\displaystyle \Rightarrow210000-2x=200000$
$\displaystyle \Rightarrow2x=10000$
$\displaystyle \Rightarrow x=5000$
$\displaystyle \therefore\ \text{Amount invested in the first bond = Rs. }5000$
$\displaystyle \text{Amount invested in the second bond = Rs. }25000$
$\displaystyle \\$

$\displaystyle \textbf{Question 75: }\text{To promote the construction of toilets for women, an}$
$\displaystyle \text{organisation tried to generate awareness through (i) house calls, (ii) letters}$
$\displaystyle \text{and (iii) announcements. The cost for each mode per attempt is given as:}$
$\displaystyle \text{(i) Rs. }50\qquad \text{(ii) Rs. }20\qquad \text{(iii) Rs. }40$
$\displaystyle \text{The number of attempts made in three villages }X,\ Y\text{ and }Z\text{ is given below:}$
$\displaystyle \begin{array}{c|ccc}&\text{(i)}&\text{(ii)}&\text{(iii)}\\X&400&300&100\\Y&300&250&75\\Z&500&400&150\end{array}$
$\displaystyle \text{Find the total cost incurred by the organisation for the three villages}$
$\displaystyle \text{separately, using matrices.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A\text{ be the matrix showing the number of attempts made in villages }X,\ Y\text{ and }Z.$
$\displaystyle A=\begin{bmatrix}400&300&100\\300&250&75\\500&400&150\end{bmatrix}$
$\displaystyle \text{Let }B\text{ be the matrix showing the cost of each mode per attempt.}$
$\displaystyle B=\begin{bmatrix}50\\20\\40\end{bmatrix}$
$\displaystyle AB=\begin{bmatrix}400&300&100\\300&250&75\\500&400&150\end{bmatrix}\begin{bmatrix}50\\20\\40\end{bmatrix}$
$\displaystyle =\begin{bmatrix}400(50)+300(20)+100(40)\\300(50)+250(20)+75(40)\\500(50)+400(20)+150(40)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}20000+6000+4000\\15000+5000+3000\\25000+8000+6000\end{bmatrix}$
$\displaystyle =\begin{bmatrix}30000\\23000\\39000\end{bmatrix}$
$\displaystyle \therefore\ \text{The total cost incurred in village }X=\text{Rs. }30000$
$\displaystyle \text{The total cost incurred in village }Y=\text{Rs. }23000$
$\displaystyle \text{The total cost incurred in village }Z=\text{Rs. }39000$
$\displaystyle \\$

$\displaystyle \textbf{Question 76: }\text{There are two families, A and B. Family A has four men,}$
$\displaystyle \text{six women and two children, while family B has two men, two women and four}$
$\displaystyle \text{children. The recommended daily intake is }2400\text{ calories for men, }1900$
$\displaystyle \text{calories for women and }1800\text{ calories for children. The recommended}$
$\displaystyle \text{daily protein intake is }45\text{ grams for men, }55\text{ grams for women and}$
$\displaystyle 33\text{ grams for children. Represent this information using matrices and, by}$
$\displaystyle \text{applying matrix multiplication, calculate the total daily requirements of}$
$\displaystyle \text{calories and proteins for each family. What value is reflected by this problem?}$
$\displaystyle [\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }X\text{ be the matrix showing the number of members in families A and B.}$
$\displaystyle X=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}\begin{array}{l}\text{Family A}\\\text{Family B}\end{array}$
$\displaystyle \text{The columns represent men, women and children respectively.}$
$\displaystyle \text{Let }Y\text{ be the matrix showing the recommended daily calorie intake.}$
$\displaystyle Y=\begin{bmatrix}2400\\1900\\1800\end{bmatrix}$
$\displaystyle \text{Let }Z\text{ be the matrix showing the recommended daily protein intake.}$
$\displaystyle Z=\begin{bmatrix}45\\55\\33\end{bmatrix}$
$\displaystyle \text{The total daily calorie requirements of the two families are given by }XY.$
$\displaystyle XY=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}\begin{bmatrix}2400\\1900\\1800\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4(2400)+6(1900)+2(1800)\\2(2400)+2(1900)+4(1800)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}9600+11400+3600\\4800+3800+7200\end{bmatrix}$
$\displaystyle =\begin{bmatrix}24600\\15800\end{bmatrix}$
$\displaystyle \text{The total daily protein requirements of the two families are given by }XZ.$
$\displaystyle XZ=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}\begin{bmatrix}45\\55\\33\end{bmatrix}$
$\displaystyle =\begin{bmatrix}4(45)+6(55)+2(33)\\2(45)+2(55)+4(33)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}180+330+66\\90+110+132\end{bmatrix}$
$\displaystyle =\begin{bmatrix}576\\332\end{bmatrix}$
$\displaystyle \therefore\ \begin{array}{c|c|c}\text{Family}&\text{Calories}&\text{Protein (g)}\\\hline\text{A}&24600&576\\\text{B}&15800&332\end{array}$
$\displaystyle \text{The problem creates awareness about the importance of a balanced and planned}$
$\displaystyle \text{diet for maintaining good health.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 77: }\text{In a parliament election, a political party hired a}$
$\displaystyle \text{public relations firm to promote its candidates in three ways: telephone,}$
$\displaystyle \text{house calls and letters. The cost per contact (in paise) is given by matrix }A$
$\displaystyle \text{as follows:}$
$\displaystyle A=\begin{bmatrix}140\\200\\150\end{bmatrix}\begin{array}{l}\text{Telephone}\\\text{House calls}\\\text{Letters}\end{array}$
$\displaystyle \text{The number of contacts of each type made in the two cities }X\text{ and }Y$
$\displaystyle \text{is given by matrix }B\text{ as follows:}$
$\displaystyle \begin{array}{ccc}\text{Telephone}&\text{House calls}&\text{Letters}\end{array}$
$\displaystyle B=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{array}{l}\text{City }X\\\text{City }Y\end{array}$
$\displaystyle \text{Find the total amount spent by the party in the two cities.}$
$\displaystyle \text{What should one consider before casting his or her vote: the party's}$
$\displaystyle \text{promotional activities or its social activities?}\qquad[\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A\text{ be the matrix showing the cost per contact in paise.}$
$\displaystyle A=\begin{bmatrix}140\\200\\150\end{bmatrix}\begin{array}{l}\text{Telephone}\\\text{House calls}\\\text{Letters}\end{array}$
$\displaystyle \text{Let }B\text{ show the number of contacts made in cities }X\text{ and }Y.$
$\displaystyle \begin{array}{ccc}\text{Telephone}&\text{House calls}&\text{Letters}\end{array}$
$\displaystyle B=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{array}{l}\text{City }X\\\text{City }Y\end{array}$
$\displaystyle \text{The total amount spent by the party in the two cities is given by }BA.$
$\displaystyle BA=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{bmatrix}140\\200\\150\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1000(140)+500(200)+5000(150)\\3000(140)+1000(200)+10000(150)\end{bmatrix}$
$\displaystyle =\begin{bmatrix}140000+100000+750000\\420000+200000+1500000\end{bmatrix}$
$\displaystyle =\begin{bmatrix}990000\\2120000\end{bmatrix}\text{ paise}$
$\displaystyle =\begin{bmatrix}9900\\21200\end{bmatrix}\text{ rupees}$
$\displaystyle \therefore\ \text{The total amount spent in city }X=\text{Rs. }9900.$
$\displaystyle \text{The total amount spent in city }Y=\text{Rs. }21200.$
$\displaystyle \text{One should consider the social activities and public service record of a}$
$\displaystyle \text{party rather than merely its promotional activities before casting a vote.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 78: }\text{The monthly incomes of Aryan and Babban are in the ratio}$
$\displaystyle 3:4\text{ and their monthly expenditures are in the ratio }5:7.\text{ If each of}$
$\displaystyle \text{them saves Rs. }15000\text{ per month, find their monthly incomes using the}$
$\displaystyle \text{matrix method. This problem reflects the value of saving money for the future}$
$\displaystyle \text{and financial planning.}\qquad[\text{CBSE 2016}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the monthly incomes of Aryan and Babban be Rs. }3x\text{ and Rs. }4x$
$\displaystyle \text{respectively, and their monthly expenditures be Rs. }5y\text{ and Rs. }7y.$
$\displaystyle \text{Since each of them saves Rs. }15000\text{ per month,}$
$\displaystyle 3x-5y=15000$
$\displaystyle 4x-7y=15000$
$\displaystyle \text{The system of equations can be written in matrix form as}$
$\displaystyle \begin{bmatrix}3&-5\\4&-7\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}15000\\15000\end{bmatrix}$
$\displaystyle \text{or, }AX=B,\text{ where}$
$\displaystyle A=\begin{bmatrix}3&-5\\4&-7\end{bmatrix},\quad X=\begin{bmatrix}x\\y\end{bmatrix},\quad B=\begin{bmatrix}15000\\15000\end{bmatrix}$
$\displaystyle |A|=\begin{vmatrix}3&-5\\4&-7\end{vmatrix}=-21+20=-1$
$\displaystyle \text{adj }A=\begin{bmatrix}-7&5\\-4&3\end{bmatrix}$
$\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A$
$\displaystyle =-\begin{bmatrix}-7&5\\-4&3\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}$
$\displaystyle \therefore X=A^{-1}B$
$\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}\begin{bmatrix}15000\\15000\end{bmatrix}$
$\displaystyle =\begin{bmatrix}105000-75000\\60000-45000\end{bmatrix}$
$\displaystyle =\begin{bmatrix}30000\\15000\end{bmatrix}$
$\displaystyle \therefore x=30000\text{ and }y=15000.$
$\displaystyle \text{Monthly income of Aryan}=3x=3(30000)=\text{Rs. }90000$
$\displaystyle \text{Monthly income of Babban}=4x=4(30000)=\text{Rs. }120000$
$\displaystyle \text{Hence, the monthly incomes of Aryan and Babban are Rs. }90000\text{ and}$
$\displaystyle \text{Rs. }120000\text{ respectively.}$
$\displaystyle \text{The problem highlights the importance of regular saving and proper financial}$
$\displaystyle \text{planning for future needs.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 79: }\text{A trust invested some money in two types of bonds. The}$
$\displaystyle \text{first bond pays }10\%\text{ interest and the second bond pays }12\%\text{ interest.}$
$\displaystyle \text{The trust received Rs. }2800\text{ as interest. However, if the amounts invested}$
$\displaystyle \text{in the two bonds had been interchanged, it would have received Rs. }100\text{ less}$
$\displaystyle \text{as interest. Using the matrix method, find the amount invested by the trust.}$
$\displaystyle [\text{CBSE 2016}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let Rs. }x\text{ be invested in the first bond and Rs. }y\text{ in the second bond.}$
$\displaystyle \text{The investment matrix is }A=[\,x\;\;y\,]\text{ and the interest-rate matrix is}$
$\displaystyle B=\begin{bmatrix}\frac{10}{100}\\[2pt]\frac{12}{100}\end{bmatrix}$
$\displaystyle \text{The total annual interest is }AB=[\,x\;\;y\,]\begin{bmatrix}\frac{10}{100}\\[2pt]\frac{12}{100}\end{bmatrix}=\frac{10x}{100}+\frac{12y}{100}$
$\displaystyle \therefore\ \frac{10x}{100}+\frac{12y}{100}=2800$
$\displaystyle \Rightarrow 10x+12y=280000\qquad(1)$
$\displaystyle \text{If the investments are interchanged, the total interest becomes Rs. }2700.$
$\displaystyle \therefore\ \frac{12x}{100}+\frac{10y}{100}=2700$
$\displaystyle \Rightarrow 12x+10y=270000\qquad(2)$
$\displaystyle \text{Equations (1) and (2) can be written as }PX=Q,$
$\displaystyle P=\begin{bmatrix}10&12\\12&10\end{bmatrix},\quad X=\begin{bmatrix}x\\y\end{bmatrix},\quad Q=\begin{bmatrix}280000\\270000\end{bmatrix}$
$\displaystyle |P|=\begin{vmatrix}10&12\\12&10\end{vmatrix}=100-144=-44\ne0$
$\displaystyle \therefore\ P^{-1}=\frac{1}{|P|}\text{adj }P=\frac{1}{-44}\begin{bmatrix}10&-12\\-12&10\end{bmatrix}$
$\displaystyle \therefore\ X=P^{-1}Q$
$\displaystyle \Rightarrow\ \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{-44}\begin{bmatrix}10&-12\\-12&10\end{bmatrix}\begin{bmatrix}280000\\270000\end{bmatrix}$
$\displaystyle =\frac{1}{-44}\begin{bmatrix}-440000\\-660000\end{bmatrix}=\begin{bmatrix}10000\\15000\end{bmatrix}$
$\displaystyle \therefore\ x=10000\text{ and }y=15000.$
$\displaystyle \text{Hence, Rs. }10000\text{ is invested in the first bond and Rs. }15000\text{ in the second bond.}$
$\displaystyle \\$