\displaystyle \textbf{Question 1: }\text{Compute the following sums:}
\displaystyle \text{(i) }\begin{bmatrix}3&-2\\1&4\end{bmatrix}+\begin{bmatrix}-2&4\\1&3\end{bmatrix}
\displaystyle \text{(ii) }\begin{bmatrix}2&1&3\\0&3&5\\-1&2&5\end{bmatrix}+\begin{bmatrix}1&-2&3\\2&6&1\\0&-3&1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \begin{bmatrix}3&-2\\1&4\end{bmatrix}+\begin{bmatrix}-2&4\\1&3\end{bmatrix}=\begin{bmatrix}3+(-2)&-2+4\\1+1&4+3\end{bmatrix}
\displaystyle =\begin{bmatrix}1&2\\2&7\end{bmatrix}
\displaystyle \text{(ii)}
\displaystyle \begin{bmatrix}2&1&3\\0&3&5\\-1&2&5\end{bmatrix}+\begin{bmatrix}1&-2&3\\2&6&1\\0&-3&1\end{bmatrix}
\displaystyle =\begin{bmatrix}2+1&1+(-2)&3+3\\0+2&3+6&5+1\\-1+0&2+(-3)&5+1\end{bmatrix}
\displaystyle =\begin{bmatrix}3&-1&6\\2&9&6\\-1&-1&6\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }A=\begin{bmatrix}2&4\\3&2\end{bmatrix},\ B=\begin{bmatrix}1&3\\-2&5\end{bmatrix}\text{ and }C=\begin{bmatrix}-2&5\\3&4\end{bmatrix}.
\displaystyle \text{Find each of the following:}
\displaystyle \text{(i) }2A-3B\qquad\text{(ii) }B-4C\qquad\text{(iii) }3A-C\qquad\text{(iv) }3A-2B+3C
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle 2A-3B=2\begin{bmatrix}2&4\\3&2\end{bmatrix}-3\begin{bmatrix}1&3\\-2&5\end{bmatrix}
\displaystyle =\begin{bmatrix}4&8\\6&4\end{bmatrix}-\begin{bmatrix}3&9\\-6&15\end{bmatrix}
\displaystyle =\begin{bmatrix}4-3&8-9\\6-(-6)&4-15\end{bmatrix}=\begin{bmatrix}1&-1\\12&-11\end{bmatrix}
\displaystyle \text{(ii)}
\displaystyle B-4C=\begin{bmatrix}1&3\\-2&5\end{bmatrix}-4\begin{bmatrix}-2&5\\3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}1&3\\-2&5\end{bmatrix}-\begin{bmatrix}-8&20\\12&16\end{bmatrix}
\displaystyle =\begin{bmatrix}1-(-8)&3-20\\-2-12&5-16\end{bmatrix}=\begin{bmatrix}9&-17\\-14&-11\end{bmatrix}
\displaystyle \text{(iii)}
\displaystyle 3A-C=3\begin{bmatrix}2&4\\3&2\end{bmatrix}-\begin{bmatrix}-2&5\\3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}6&12\\9&6\end{bmatrix}-\begin{bmatrix}-2&5\\3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}6-(-2)&12-5\\9-3&6-4\end{bmatrix}=\begin{bmatrix}8&7\\6&2\end{bmatrix}
\displaystyle \text{(iv)}
\displaystyle 3A-2B+3C=3\begin{bmatrix}2&4\\3&2\end{bmatrix}-2\begin{bmatrix}1&3\\-2&5\end{bmatrix}+3\begin{bmatrix}-2&5\\3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}6&12\\9&6\end{bmatrix}-\begin{bmatrix}2&6\\-4&10\end{bmatrix}+\begin{bmatrix}-6&15\\9&12\end{bmatrix}
\displaystyle =\begin{bmatrix}6-2-6&12-6+15\\9-(-4)+9&6-10+12\end{bmatrix}=\begin{bmatrix}-2&21\\22&8\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }A=\begin{bmatrix}2&3\\5&7\end{bmatrix},\ B=\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix}\text{ and }C=\begin{bmatrix}-1&2&3\\2&1&0\end{bmatrix}.
\displaystyle \text{Find each of the following:}
\displaystyle \text{(i) }A+B\text{ and }B+C\qquad\text{(ii) }2B+3A\text{ and }3C-4B
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle A+B=\begin{bmatrix}2&3\\5&7\end{bmatrix}+\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix}
\displaystyle \text{Since }A\text{ is of order }2\times2\text{ and }B\text{ is of order }2\times3,\ A+B\text{ is not defined.}
\displaystyle B+C=\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix}+\begin{bmatrix}-1&2&3\\2&1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}-1+(-1)&0+2&2+3\\3+2&4+1&1+0\end{bmatrix}=\begin{bmatrix}-2&2&5\\5&5&1\end{bmatrix}
\displaystyle \text{(ii)}
\displaystyle 2B+3A=2\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix}+3\begin{bmatrix}2&3\\5&7\end{bmatrix}
\displaystyle \text{Since }B\text{ is of order }2\times3\text{ and }A\text{ is of order }2\times2,\ 2B+3A\text{ is not defined.}
\displaystyle 3C-4B=3\begin{bmatrix}-1&2&3\\2&1&0\end{bmatrix}-4\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-3&6&9\\6&3&0\end{bmatrix}-\begin{bmatrix}-4&0&8\\12&16&4\end{bmatrix}
\displaystyle =\begin{bmatrix}-3-(-4)&6-0&9-8\\6-12&3-16&0-4\end{bmatrix}=\begin{bmatrix}1&6&1\\-6&-13&-4\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=\begin{bmatrix}-1&0&2\\3&1&4\end{bmatrix},\quad B=\begin{bmatrix}0&-2&5\\1&-3&1\end{bmatrix}
\displaystyle \text{and }C=\begin{bmatrix}1&-5&2\\6&0&-4\end{bmatrix}.\text{ Compute }2A-3B+4C.
\displaystyle \text{Answer:}
\displaystyle 2A-3B+4C=2\begin{bmatrix}-1&0&2\\3&1&4\end{bmatrix}-3\begin{bmatrix}0&-2&5\\1&-3&1\end{bmatrix}
\displaystyle \qquad\qquad\quad+4\begin{bmatrix}1&-5&2\\6&0&-4\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&0&4\\6&2&8\end{bmatrix}-\begin{bmatrix}0&-6&15\\3&-9&3\end{bmatrix}+\begin{bmatrix}4&-20&8\\24&0&-16\end{bmatrix}
\displaystyle =\begin{bmatrix}-2-0+4&0-(-6)-20&4-15+8\\6-3+24&2-(-9)+0&8-3-16\end{bmatrix}
\displaystyle =\begin{bmatrix}2&-14&-3\\27&11&-11\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A=\text{diag}(2\ -5\ 9),\ B=\text{diag}(1\ 1\ -4)
\displaystyle \text{and }C=\text{diag}(-6\ 3\ 4),\text{ find:}
\displaystyle \text{(i) }A-2B\qquad\text{(ii) }B+C-2A\qquad\text{(iii) }2A+3B-5C
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2&0&0\\0&-5&0\\0&0&9\end{bmatrix},\quad B=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}
\displaystyle \text{and }C=\begin{bmatrix}-6&0&0\\0&3&0\\0&0&4\end{bmatrix}.
\displaystyle \text{(i)}
\displaystyle A-2B=\begin{bmatrix}2&0&0\\0&-5&0\\0&0&9\end{bmatrix}-2\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}
\displaystyle =\begin{bmatrix}2&0&0\\0&-5&0\\0&0&9\end{bmatrix}-\begin{bmatrix}2&0&0\\0&2&0\\0&0&-8\end{bmatrix}
\displaystyle =\begin{bmatrix}2-2&0&0\\0&-5-2&0\\0&0&9-(-8)\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&-7&0\\0&0&17\end{bmatrix}
\displaystyle \text{(ii)}
\displaystyle B+C-2A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}+\begin{bmatrix}-6&0&0\\0&3&0\\0&0&4\end{bmatrix}
\displaystyle \qquad\qquad\quad-2\begin{bmatrix}2&0&0\\0&-5&0\\0&0&9\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}+\begin{bmatrix}-6&0&0\\0&3&0\\0&0&4\end{bmatrix}-\begin{bmatrix}4&0&0\\0&-10&0\\0&0&18\end{bmatrix}
\displaystyle =\begin{bmatrix}1-6-4&0&0\\0&1+3-(-10)&0\\0&0&-4+4-18\end{bmatrix}
\displaystyle =\begin{bmatrix}-9&0&0\\0&14&0\\0&0&-18\end{bmatrix}
\displaystyle \text{(iii)}
\displaystyle 2A+3B-5C=2\begin{bmatrix}2&0&0\\0&-5&0\\0&0&9\end{bmatrix}+3\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}
\displaystyle \qquad\qquad\quad-5\begin{bmatrix}-6&0&0\\0&3&0\\0&0&4\end{bmatrix}
\displaystyle =\begin{bmatrix}4&0&0\\0&-10&0\\0&0&18\end{bmatrix}+\begin{bmatrix}3&0&0\\0&3&0\\0&0&-12\end{bmatrix}-\begin{bmatrix}-30&0&0\\0&15&0\\0&0&20\end{bmatrix}
\displaystyle =\begin{bmatrix}4+3-(-30)&0&0\\0&-10+3-15&0\\0&0&18-12-20\end{bmatrix}
\displaystyle =\begin{bmatrix}37&0&0\\0&-22&0\\0&0&-14\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Given the matrices}
\displaystyle A=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix},\quad B=\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}
\displaystyle \text{and }C=\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}.
\displaystyle \text{Verify that }(A+B)+C=A+(B+C).
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(A+B)+C
\displaystyle =\left(\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}\right)+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}
\displaystyle =\begin{bmatrix}2+9&1+7&1+(-1)\\3+3&-1+5&0+4\\0+2&2+1&4+6\end{bmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}
\displaystyle =\begin{bmatrix}11&8&0\\6&4&4\\2&3&10\end{bmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}
\displaystyle =\begin{bmatrix}11+2&8+(-4)&0+3\\6+1&4+(-1)&4+0\\2+9&3+4&10+5\end{bmatrix}
\displaystyle =\begin{bmatrix}13&4&3\\7&3&4\\11&7&15\end{bmatrix}
\displaystyle \text{RHS}=A+(B+C)
\displaystyle =\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\left(\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}\right)
\displaystyle =\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{bmatrix}9+2&7+(-4)&-1+3\\3+1&5+(-1)&4+0\\2+9&1+4&6+5\end{bmatrix}
\displaystyle =\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{bmatrix}11&3&2\\4&4&4\\11&5&11\end{bmatrix}
\displaystyle =\begin{bmatrix}2+11&1+3&1+2\\3+4&-1+4&0+4\\0+11&2+5&4+11\end{bmatrix}
\displaystyle =\begin{bmatrix}13&4&3\\7&3&4\\11&7&15\end{bmatrix}
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \text{Hence, }(A+B)+C=A+(B+C).
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find matrices }X\text{ and }Y\text{ if}
\displaystyle X+Y=\begin{bmatrix}5&2\\0&9\end{bmatrix}\text{ and }X-Y=\begin{bmatrix}3&6\\0&-1\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }X+Y=\begin{bmatrix}5&2\\0&9\end{bmatrix}\text{ and }X-Y=\begin{bmatrix}3&6\\0&-1\end{bmatrix}.
\displaystyle \text{Adding the two equations,}
\displaystyle 2X=\begin{bmatrix}5&2\\0&9\end{bmatrix}+\begin{bmatrix}3&6\\0&-1\end{bmatrix}=\begin{bmatrix}5+3&2+6\\0+0&9+(-1)\end{bmatrix}
\displaystyle =\begin{bmatrix}8&8\\0&8\end{bmatrix}
\displaystyle \Rightarrow X=\frac12\begin{bmatrix}8&8\\0&8\end{bmatrix}=\begin{bmatrix}4&4\\0&4\end{bmatrix}
\displaystyle \text{Now, }Y=\begin{bmatrix}5&2\\0&9\end{bmatrix}-X
\displaystyle =\begin{bmatrix}5&2\\0&9\end{bmatrix}-\begin{bmatrix}4&4\\0&4\end{bmatrix}=\begin{bmatrix}5-4&2-4\\0-0&9-4\end{bmatrix}
\displaystyle =\begin{bmatrix}1&-2\\0&5\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}4&4\\0&4\end{bmatrix},\qquad Y=\begin{bmatrix}1&-2\\0&5\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find }X\text{ if }Y=\begin{bmatrix}3&2\\1&4\end{bmatrix}\text{ and }2X+Y=\begin{bmatrix}1&0\\-3&2\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }2X+Y=\begin{bmatrix}1&0\\-3&2\end{bmatrix}.
\displaystyle \Rightarrow 2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}=\begin{bmatrix}1-3&0-2\\-3-1&2-4\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}
\displaystyle \Rightarrow X=\frac12\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find matrices }X\text{ and }Y,\text{ if }2X-Y=\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}\text{ and}
\displaystyle X+2Y=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }2X-Y=\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}\text{ and }X+2Y=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}.
\displaystyle \Rightarrow 4X-2Y=2\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}=\begin{bmatrix}12&-12&0\\-8&4&2\end{bmatrix}
\displaystyle \text{Adding the two equations,}
\displaystyle 5X=\begin{bmatrix}12&-12&0\\-8&4&2\end{bmatrix}+\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}
\displaystyle =\begin{bmatrix}12+3&-12+2&0+5\\-8-2&4+1&2-7\end{bmatrix}=\begin{bmatrix}15&-10&5\\-10&5&-5\end{bmatrix}
\displaystyle \Rightarrow X=\frac15\begin{bmatrix}15&-10&5\\-10&5&-5\end{bmatrix}=\begin{bmatrix}3&-2&1\\-2&1&-1\end{bmatrix}
\displaystyle \text{Now, }X+2Y=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}
\displaystyle \Rightarrow 2Y=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}-\begin{bmatrix}3&-2&1\\-2&1&-1\end{bmatrix}=\begin{bmatrix}0&4&4\\0&0&-6\end{bmatrix}
\displaystyle \Rightarrow Y=\frac12\begin{bmatrix}0&4&4\\0&0&-6\end{bmatrix}=\begin{bmatrix}0&2&2\\0&0&-3\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }X-Y=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}\text{ and }X+Y=\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix},
\displaystyle \text{find }X\text{ and }Y.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }X-Y=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}\text{ and }X+Y=\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}.
\displaystyle \text{Adding the two equations,}
\displaystyle 2X=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}+\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}
\displaystyle =\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}
\displaystyle \Rightarrow X=\frac12\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}
\displaystyle \text{Now, }Y=\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}-\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}
\displaystyle =\begin{bmatrix}3-2&5-3&1-1\\-1-0&1-1&4-2\\11-6&8-4&0-0\end{bmatrix}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix},\quad Y=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find matrix }A\text{, if}
\displaystyle \begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}+A=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}+A=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}.
\displaystyle \Rightarrow A=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}-\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}
\displaystyle =\begin{bmatrix}9-1&-1-2&4-(-1)\\-2-0&1-4&3-9\end{bmatrix}
\displaystyle =\begin{bmatrix}8&-3&5\\-2&-3&-6\end{bmatrix}
\displaystyle \therefore A=\begin{bmatrix}8&-3&5\\-2&-3&-6\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }A=\begin{bmatrix}9&1\\7&8\end{bmatrix}\text{ and }B=\begin{bmatrix}1&5\\7&12\end{bmatrix},
\displaystyle \text{find matrix }C\text{ such that }5A+3B+2C\text{ is a null matrix.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }5A+3B+2C=\begin{bmatrix}0&0\\0&0\end{bmatrix}.
\displaystyle \Rightarrow 2C=\begin{bmatrix}0&0\\0&0\end{bmatrix}-5A-3B
\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}-5\begin{bmatrix}9&1\\7&8\end{bmatrix}-3\begin{bmatrix}1&5\\7&12\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}-\begin{bmatrix}45&5\\35&40\end{bmatrix}-\begin{bmatrix}3&15\\21&36\end{bmatrix}
\displaystyle =\begin{bmatrix}0-45-3&0-5-15\\0-35-21&0-40-36\end{bmatrix}
\displaystyle =\begin{bmatrix}-48&-20\\-56&-76\end{bmatrix}
\displaystyle \Rightarrow C=\frac12\begin{bmatrix}-48&-20\\-56&-76\end{bmatrix}=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}
\displaystyle \therefore C=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }A=\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix}\text{ and }B=\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix},
\displaystyle \text{find matrix }X\text{ such that }2A+3X=5B.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }2A+3X=5B.
\displaystyle \Rightarrow 3X=5B-2A
\displaystyle =5\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}-2\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix}
\displaystyle =\begin{bmatrix}40&0\\20&-10\\15&30\end{bmatrix}-\begin{bmatrix}4&-4\\8&4\\-10&2\end{bmatrix}
\displaystyle =\begin{bmatrix}40-4&0-(-4)\\20-8&-10-4\\15-(-10)&30-2\end{bmatrix}
\displaystyle =\begin{bmatrix}36&4\\12&-14\\25&28\end{bmatrix}
\displaystyle \Rightarrow X=\frac13\begin{bmatrix}36&4\\12&-14\\25&28\end{bmatrix}=\begin{bmatrix}12&\frac43\\4&-\frac{14}{3}\\\frac{25}{3}&\frac{28}{3}\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}12&\frac43\\4&-\frac{14}{3}\\\frac{25}{3}&\frac{28}{3}\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }A=\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix}\text{ and }B=\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix},
\displaystyle \text{find matrix }C\text{ such that }A+B+C\text{ is a zero matrix.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A+B+C=\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}.
\displaystyle \Rightarrow C=\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}-\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix}-\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix}
\displaystyle =\begin{bmatrix}0-1-2&0-(-3)-(-1)&0-2-(-1)\\0-2-1&0-0-0&0-2-(-1)\end{bmatrix}
\displaystyle =\begin{bmatrix}-3&4&-1\\-3&0&-1\end{bmatrix}
\displaystyle \therefore C=\begin{bmatrix}-3&4&-1\\-3&0&-1\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find the unknowns satisfying the following matrix equations:}
\displaystyle \text{(i) }\begin{bmatrix}x-y&2&-2\\4&x&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2x+y&5\end{bmatrix}
\displaystyle \text{(ii) }\begin{bmatrix}x&y+2&z-3\end{bmatrix}+\begin{bmatrix}y&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}
\displaystyle \text{(iii) }x\begin{bmatrix}2\\1\end{bmatrix}+y\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \begin{bmatrix}x-y&2&-2\\4&x&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2x+y&5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x-y+3&0&0\\5&x&5\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2x+y&5\end{bmatrix}
\displaystyle \text{By equality of matrices, }x-y+3=6\text{ and }x=2x+y.
\displaystyle \Rightarrow x-y=3\qquad\ldots\ldots(1)
\displaystyle \Rightarrow x=-y\qquad\ldots\ldots(2)
\displaystyle \text{Substituting }x=-y\text{ in (1),}
\displaystyle -y-y=3\Rightarrow -2y=3\Rightarrow y=-\frac32
\displaystyle \Rightarrow x=-y=\frac32
\displaystyle \therefore x=\frac32\text{ and }y=-\frac32.
\displaystyle \text{(ii)}
\displaystyle \begin{bmatrix}x&y+2&z-3\end{bmatrix}+\begin{bmatrix}y&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x+y&y+6&z+2\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}
\displaystyle \text{By equality of matrices, }x+y=4,\quad y+6=9,\quad z+2=12.
\displaystyle y+6=9\Rightarrow y=3
\displaystyle z+2=12\Rightarrow z=10
\displaystyle x+y=4\Rightarrow x+3=4\Rightarrow x=1
\displaystyle \therefore x=1,\quad y=3\text{ and }z=10.
\displaystyle \text{(iii)}
\displaystyle x\begin{bmatrix}2\\1\end{bmatrix}+y\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x\\x\end{bmatrix}+\begin{bmatrix}3y\\5y\end{bmatrix}=\begin{bmatrix}8\\11\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+3y\\x+5y\end{bmatrix}=\begin{bmatrix}8\\11\end{bmatrix}
\displaystyle \text{By equality of matrices, }2x+3y=8\qquad\ldots\ldots(1)
\displaystyle x+5y=11\qquad\ldots\ldots(2)
\displaystyle \text{Multiplying (2) by }2,\quad 2x+10y=22\qquad\ldots\ldots(3)
\displaystyle \text{Subtracting (1) from (3),}\quad 7y=14\Rightarrow y=2
\displaystyle \text{Substituting }y=2\text{ in (2),}\quad x+10=11\Rightarrow x=1
\displaystyle \therefore x=1\text{ and }y=2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }2\begin{bmatrix}3&4\\5&x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix},
\displaystyle \text{find }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }2\begin{bmatrix}3&4\\5&x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}.
\displaystyle \Rightarrow \begin{bmatrix}6&8\\10&2x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}7&8+y\\10&2x+1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}
\displaystyle \text{By equality of matrices, }8+y=0\text{ and }2x+1=5.
\displaystyle \Rightarrow y=-8
\displaystyle \Rightarrow 2x=4\Rightarrow x=2
\displaystyle \therefore x=2\text{ and }y=-8.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the value of the non-zero scalar }\lambda\text{, if}
\displaystyle \lambda\begin{bmatrix}1&0&2\\3&4&5\end{bmatrix}+2\begin{bmatrix}1&2&3\\-1&-3&2\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\lambda\begin{bmatrix}1&0&2\\3&4&5\end{bmatrix}+2\begin{bmatrix}1&2&3\\-1&-3&2\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}.
\displaystyle \Rightarrow \begin{bmatrix}\lambda&0&2\lambda\\3\lambda&4\lambda&5\lambda\end{bmatrix}+\begin{bmatrix}2&4&6\\-2&-6&4\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}\lambda&0&2\lambda\\3\lambda&4\lambda&5\lambda\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}-\begin{bmatrix}2&4&6\\-2&-6&4\end{bmatrix}
\displaystyle =\begin{bmatrix}4-2&4-4&10-6\\4-(-2)&2-(-6)&14-4\end{bmatrix}
\displaystyle =\begin{bmatrix}2&0&4\\6&8&10\end{bmatrix}
\displaystyle \text{By equality of matrices, }\lambda=2.
\displaystyle \therefore \lambda=2.
\displaystyle \\

\displaystyle \textbf{Question 18:}
\displaystyle \text{(i) Find a matrix }X\text{ such that }2A+B+X=O,\text{ where}
\displaystyle A=\begin{bmatrix}-1&2\\3&4\end{bmatrix},\quad B=\begin{bmatrix}3&-2\\1&5\end{bmatrix}.\quad\text{[CBSE 2000]}
\displaystyle \text{(ii) If }A=\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}\text{ and }B=\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix},
\displaystyle \text{find the matrix }X\text{ of order }3\times2\text{ such that }2A+3X=5B.
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{Given, }2A+B+X=O.
\displaystyle \Rightarrow 2\begin{bmatrix}-1&2\\3&4\end{bmatrix}+\begin{bmatrix}3&-2\\1&5\end{bmatrix}+X=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-2&4\\6&8\end{bmatrix}+\begin{bmatrix}3&-2\\1&5\end{bmatrix}+X=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-2+3&4+(-2)\\6+1&8+5\end{bmatrix}+X=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}1&2\\7&13\end{bmatrix}+X=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow X=\begin{bmatrix}0&0\\0&0\end{bmatrix}-\begin{bmatrix}1&2\\7&13\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-1&-2\\-7&-13\end{bmatrix}.
\displaystyle \text{(ii)}
\displaystyle \text{Given, }2A+3X=5B.
\displaystyle \Rightarrow 2\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}+3X=5\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}16&0\\8&-4\\6&12\end{bmatrix}+3X=\begin{bmatrix}10&-10\\20&10\\-25&5\end{bmatrix}
\displaystyle \Rightarrow 3X=\begin{bmatrix}10&-10\\20&10\\-25&5\end{bmatrix}-\begin{bmatrix}16&0\\8&-4\\6&12\end{bmatrix}
\displaystyle =\begin{bmatrix}10-16&-10-0\\20-8&10-(-4)\\-25-6&5-12\end{bmatrix}
\displaystyle =\begin{bmatrix}-6&-10\\12&14\\-31&-7\end{bmatrix}
\displaystyle \Rightarrow X=\frac13\begin{bmatrix}-6&-10\\12&14\\-31&-7\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-2&-\frac{10}{3}\\4&\frac{14}{3}\\-\frac{31}{3}&-\frac{7}{3}\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find }x,y,z\text{ and }t\text{, if}
\displaystyle \text{(i) }3\begin{bmatrix}x&y\\z&t\end{bmatrix}=\begin{bmatrix}x&6\\-1&2t\end{bmatrix}+\begin{bmatrix}4&x+y\\z+t&3\end{bmatrix}
\displaystyle \text{(ii) }2\begin{bmatrix}x&5\\7&y-3\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}
\displaystyle \hspace{6.0cm}\text{[CBSE 2002, 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{Given, }3\begin{bmatrix}x&y\\z&t\end{bmatrix}=\begin{bmatrix}x&6\\-1&2t\end{bmatrix}+\begin{bmatrix}4&x+y\\z+t&3\end{bmatrix}.
\displaystyle \Rightarrow \begin{bmatrix}3x&3y\\3z&3t\end{bmatrix}=\begin{bmatrix}x+4&6+x+y\\-1+z+t&2t+3\end{bmatrix}
\displaystyle \text{By equality of matrices, }3x=x+4,\quad 3y=6+x+y,
\displaystyle 3z=-1+z+t\text{ and }3t=2t+3.
\displaystyle 3t=2t+3\Rightarrow t=3
\displaystyle 3z=-1+z+t\Rightarrow 2z=-1+3\Rightarrow 2z=2\Rightarrow z=1
\displaystyle 3x=x+4\Rightarrow 2x=4\Rightarrow x=2
\displaystyle 3y=6+x+y\Rightarrow 2y=6+2\Rightarrow 2y=8\Rightarrow y=4
\displaystyle \therefore x=2,\quad y=4,\quad z=1\text{ and }t=3.
\displaystyle \text{(ii)}
\displaystyle \text{Given, }2\begin{bmatrix}x&5\\7&y-3\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}.
\displaystyle \Rightarrow \begin{bmatrix}2x&10\\14&2y-6\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}-\begin{bmatrix}3&4\\1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}7-3&14-4\\15-1&14-2\end{bmatrix}=\begin{bmatrix}4&10\\14&12\end{bmatrix}
\displaystyle \text{By equality of matrices, }2x=4\text{ and }2y-6=12.
\displaystyle 2x=4\Rightarrow x=2
\displaystyle 2y-6=12\Rightarrow 2y=18\Rightarrow y=9
\displaystyle \therefore x=2\text{ and }y=9.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }X\text{ and }Y\text{ are }2\times2\text{ matrices, solve the following}
\displaystyle \text{matrix equations for }X\text{ and }Y.
\displaystyle 2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix},\qquad 3X+2Y=\begin{bmatrix}-2&2\\1&-5\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{Given,}
\displaystyle 2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\qquad\ldots\ldots(1)
\displaystyle 3X+2Y=\begin{bmatrix}-2&2\\1&-5\end{bmatrix}\qquad\ldots\ldots(2)
\displaystyle \text{Multiplying equation (1) by }3\text{ and equation (2) by }2,\text{ then subtracting (2) from (1),}
\displaystyle \Rightarrow 9Y-4Y=3\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\begin{bmatrix}-2&2\\1&-5\end{bmatrix}
\displaystyle \Rightarrow 5Y=\begin{bmatrix}6&9\\12&0\end{bmatrix}-\begin{bmatrix}-4&4\\2&-10\end{bmatrix}
\displaystyle \Rightarrow 5Y=\begin{bmatrix}6-(-4)&9-4\\12-2&0-(-10)\end{bmatrix}
\displaystyle \Rightarrow 5Y=\begin{bmatrix}10&5\\10&10\end{bmatrix}
\displaystyle \Rightarrow Y=\frac15\begin{bmatrix}10&5\\10&10\end{bmatrix}=\begin{bmatrix}2&1\\2&2\end{bmatrix}
\displaystyle \text{Substituting }Y=\begin{bmatrix}2&1\\2&2\end{bmatrix}\text{ in equation (1),}
\displaystyle \Rightarrow 2X=\begin{bmatrix}2&3\\4&0\end{bmatrix}-3\begin{bmatrix}2&1\\2&2\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}2&3\\4&0\end{bmatrix}-\begin{bmatrix}6&3\\6&6\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}2-6&3-3\\4-6&0-6\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}-4&0\\-2&-6\end{bmatrix}
\displaystyle \Rightarrow X=\frac12\begin{bmatrix}-4&0\\-2&-6\end{bmatrix}=\begin{bmatrix}-2&0\\-1&-3\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-2&0\\-1&-3\end{bmatrix},\qquad Y=\begin{bmatrix}2&1\\2&2\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In a certain city there are }30\text{ colleges. Each college has }15\text{ peons, }6\text{ clerks,}
\displaystyle 1\text{ typist and }1\text{ section officer. Express the given information as a column matrix. Using scalar}
\displaystyle \text{multiplication, find the total number of posts of each kind in all the colleges.}
\displaystyle \text{Answer:}
\displaystyle \text{The number of posts in one college is represented by the column matrix}
\displaystyle X=\begin{bmatrix}15\\6\\1\\1\end{bmatrix}
\displaystyle \text{where the entries represent peons, clerks, typists and section officers respectively.}
\displaystyle \text{Therefore, the total number of posts of each kind in all }30\text{ colleges is}
\displaystyle 30X=30\begin{bmatrix}15\\6\\1\\1\end{bmatrix}=\begin{bmatrix}450\\180\\30\\30\end{bmatrix}
\displaystyle \therefore \text{The total number of peons, clerks, typists and section officers is }450,\ 180,\ 30\text{ and }30\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The monthly incomes of Aryan and Babban are in the ratio }
\displaystyle 3:4\text{ and their} \ \text{monthly expenditures are in the ratio }5:7.\text{ If each saves Rs. }15{,}000
\displaystyle \text{per month, find their} \ \text{monthly incomes using the matrix method. This problem} \\ \text{reflects which value? }\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the monthly incomes of Aryan and Babban be Rs. }3x\text{ and Rs. }4x\text{ respectively,}
\displaystyle \text{and their monthly expenditures be Rs. }5y\text{ and Rs. }7y\text{ respectively.}
\displaystyle \text{Since each saves Rs. }15{,}000\text{ per month,}
\displaystyle 3x-5y=15{,}000\qquad\ldots\ldots(1)
\displaystyle 4x-7y=15{,}000\qquad\ldots\ldots(2)
\displaystyle \text{The system of equations can be written in matrix form as }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}15{,}000\\15{,}000\end{bmatrix}.
\displaystyle AX=B\Rightarrow X=A^{-1}B
\displaystyle |A|=3(-7)-(-5)(4)=-21+20=-1\ne0
\displaystyle \text{Therefore, }A^{-1}=\frac{1}{|A|}\begin{bmatrix}-7&5\\-4&3\end{bmatrix}
\displaystyle =\frac{1}{-1}\begin{bmatrix}-7&5\\-4&3\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}
\displaystyle X=A^{-1}B=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}\begin{bmatrix}15{,}000\\15{,}000\end{bmatrix}
\displaystyle =\begin{bmatrix}105{,}000-75{,}000\\60{,}000-45{,}000\end{bmatrix}=\begin{bmatrix}30{,}000\\15{,}000\end{bmatrix}
\displaystyle \therefore x=30{,}000\text{ and }y=15{,}000.
\displaystyle \text{Aryan's monthly income}=3x=3(30{,}000)=\text{Rs. }90{,}000.
\displaystyle \text{Babban's monthly income}=4x=4(30{,}000)=\text{Rs. }1{,}20{,}000.
\displaystyle \text{The problem reflects the values of saving, financial discipline and prudent financial planning.}
\displaystyle \\

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