\displaystyle \textbf{Question 1: }\text{Given }A=\begin{bmatrix}3&0\\0&4\end{bmatrix},\ B=\begin{bmatrix}a&b\\0&c\end{bmatrix}\text{ and }AB=A+B,
\displaystyle \text{find the values of }a,\ b\text{ and }c.
\displaystyle \text{Answer:}
\displaystyle AB=A+B
\displaystyle \Rightarrow \begin{bmatrix}3&0\\0&4\end{bmatrix}\begin{bmatrix}a&b\\0&c\end{bmatrix}=\begin{bmatrix}3&0\\0&4\end{bmatrix}+\begin{bmatrix}a&b\\0&c\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}3a&3b\\0&4c\end{bmatrix}=\begin{bmatrix}3+a&b\\0&4+c\end{bmatrix}
\displaystyle 3a=3+a\Rightarrow 2a=3\Rightarrow a=\frac{3}{2}
\displaystyle 3b=b\Rightarrow 2b=0\Rightarrow b=0
\displaystyle 4c=4+c\Rightarrow 3c=4\Rightarrow c=\frac{4}{3}
\displaystyle \therefore a=\frac{3}{2},\ b=0,\ c=\frac{4}{3}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }P=\begin{bmatrix}1&2\\2&-1\end{bmatrix}\text{ and }Q=\begin{bmatrix}1&0\\2&1\end{bmatrix},\text{ find:}
\displaystyle \text{i) }P^2-Q^2\qquad \text{ii) }(P+Q)(P-Q)\qquad \text{Also find if }P^2-Q^2=(P+Q)(P-Q).
\displaystyle \text{Answer:}
\displaystyle \text{i) }P^2-Q^2
\displaystyle =\begin{bmatrix}1&2\\2&-1\end{bmatrix}\begin{bmatrix}1&2\\2&-1\end{bmatrix}-\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}1&0\\2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}5&0\\0&5\end{bmatrix}-\begin{bmatrix}1&0\\4&1\end{bmatrix}=\begin{bmatrix}4&0\\-4&4\end{bmatrix}
\displaystyle \text{ii) }(P+Q)(P-Q)
\displaystyle =\left(\begin{bmatrix}1&2\\2&-1\end{bmatrix}+\begin{bmatrix}1&0\\2&1\end{bmatrix}\right)\left(\begin{bmatrix}1&2\\2&-1\end{bmatrix}-\begin{bmatrix}1&0\\2&1\end{bmatrix}\right)
\displaystyle =\begin{bmatrix}2&2\\4&0\end{bmatrix}\begin{bmatrix}0&2\\0&-2\end{bmatrix}=\begin{bmatrix}0&0\\0&8\end{bmatrix}
\displaystyle \therefore P^2-Q^2\ne(P+Q)(P-Q).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Given }A=\begin{bmatrix}2&1\\4&2\end{bmatrix},\ B=\begin{bmatrix}3&4\\-1&-2\end{bmatrix}\text{ and }C=\begin{bmatrix}-3&1\\0&-2\end{bmatrix},\text{ find:}
\displaystyle \text{i) }ABC\qquad \text{ii) }ACB\qquad \text{Find whether }ABC=ACB.
\displaystyle \text{Answer:}
\displaystyle \text{i) }ABC=\begin{bmatrix}2&1\\4&2\end{bmatrix}\begin{bmatrix}3&4\\-1&-2\end{bmatrix}\begin{bmatrix}-3&1\\0&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}5&6\\10&12\end{bmatrix}\begin{bmatrix}-3&1\\0&-2\end{bmatrix}=\begin{bmatrix}-15&-7\\-30&-14\end{bmatrix}
\displaystyle \text{ii) }ACB=\begin{bmatrix}2&1\\4&2\end{bmatrix}\begin{bmatrix}-3&1\\0&-2\end{bmatrix}\begin{bmatrix}3&4\\-1&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}-6&0\\-12&0\end{bmatrix}\begin{bmatrix}3&4\\-1&-2\end{bmatrix}=\begin{bmatrix}-18&-24\\-36&-48\end{bmatrix}
\displaystyle \therefore ABC\ne ACB.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }A=\begin{bmatrix}1&2\\3&4\end{bmatrix},\ B=\begin{bmatrix}6&1\\1&1\end{bmatrix}\text{ and }C=\begin{bmatrix}-2&-3\\0&1\end{bmatrix},
\displaystyle \text{find i) }CA+B\qquad \text{ii) }A+CB.\text{ Are these equal?}
\displaystyle \text{Answer:}
\displaystyle \text{i) }CA+B=\begin{bmatrix}-2&-3\\0&1\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}6&1\\1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-11&-16\\3&4\end{bmatrix}+\begin{bmatrix}6&1\\1&1\end{bmatrix}=\begin{bmatrix}-5&-15\\4&5\end{bmatrix}
\displaystyle \text{ii) }A+CB=\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}-2&-3\\0&1\end{bmatrix}\begin{bmatrix}6&1\\1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}-15&-5\\1&1\end{bmatrix}=\begin{bmatrix}-14&-3\\4&5\end{bmatrix}
\displaystyle \therefore CA+B\ne A+CB.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A=\begin{bmatrix}2&1\\1&3\end{bmatrix},\ B=\begin{bmatrix}3\\-11\end{bmatrix},\text{ find the matrix }X\text{ such that }AX=B.
\displaystyle \text{Answer:}
\displaystyle A_{2\times2}X_{m\times n}=B_{2\times1}
\displaystyle \Rightarrow m=2,\ n=1
\displaystyle \therefore X\text{ is of order }2\times1.
\displaystyle \text{Let }X=\begin{bmatrix}a\\b\end{bmatrix}
\displaystyle AX=B
\displaystyle \Rightarrow \begin{bmatrix}2&1\\1&3\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}3\\-11\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2a+b\\a+3b\end{bmatrix}=\begin{bmatrix}3\\-11\end{bmatrix}
\displaystyle 2a+b=3
\displaystyle a+3b=-11
\displaystyle \text{Solving the above equations, we get }a=4,\ b=-5.
\displaystyle \therefore X=\begin{bmatrix}4\\-5\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }A=\begin{bmatrix}4&2\\1&1\end{bmatrix},\text{ find }(A-2I)(A-3I).
\displaystyle \text{Answer:}
\displaystyle I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\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 =\begin{bmatrix}2&2\\1&-1\end{bmatrix}\begin{bmatrix}1&2\\1&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}4&0\\0&4\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }A=\begin{bmatrix}2&1&-1\\0&1&-2\end{bmatrix},\text{ find i) }A^tA\qquad \text{ii) }AA^t.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2&1&-1\\0&1&-2\end{bmatrix}
\displaystyle \therefore A^t=\begin{bmatrix}2&0\\1&1\\-1&-2\end{bmatrix}
\displaystyle \text{i) }A^tA=\begin{bmatrix}2&0\\1&1\\-1&-2\end{bmatrix}\begin{bmatrix}2&1&-1\\0&1&-2\end{bmatrix}=\begin{bmatrix}4&2&-2\\2&2&-3\\-2&-3&5\end{bmatrix}
\displaystyle \text{ii) }AA^t=\begin{bmatrix}2&1&-1\\0&1&-2\end{bmatrix}\begin{bmatrix}2&0\\1&1\\-1&-2\end{bmatrix}=\begin{bmatrix}6&3\\3&5\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }M=\begin{bmatrix}4&1\\-1&2\end{bmatrix},\text{ show that }6M-M^2=9I,\text{ where }I\text{ is a }2\times2\text{ unit matrix.}
\displaystyle \text{Answer:}
\displaystyle LHS=6M-M^2
\displaystyle =6\begin{bmatrix}4&1\\-1&2\end{bmatrix}-\begin{bmatrix}4&1\\-1&2\end{bmatrix}\begin{bmatrix}4&1\\-1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}24&6\\-6&12\end{bmatrix}-\begin{bmatrix}15&6\\-6&3\end{bmatrix}
\displaystyle =\begin{bmatrix}9&0\\0&9\end{bmatrix}
\displaystyle =9\begin{bmatrix}1&0\\0&1\end{bmatrix}=9I
\displaystyle \therefore LHS=RHS.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }P=\begin{bmatrix}2&6\\3&9\end{bmatrix},\ Q=\begin{bmatrix}3&x\\y&2\end{bmatrix},\text{ find }x\text{ and }y\text{ such that }PQ=\text{null matrix}.
\displaystyle \text{Answer:}
\displaystyle PQ=\text{null matrix}
\displaystyle \Rightarrow \begin{bmatrix}2&6\\3&9\end{bmatrix}\begin{bmatrix}3&x\\y&2\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6+6y&2x+12\\9+9y&3x+18\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle 6+6y=0\Rightarrow y=-1
\displaystyle 2x+12=0\Rightarrow x=-6
\displaystyle \therefore x=-6,\ y=-1.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Evaluate }\begin{bmatrix}2\cos60^\circ&-2\sin30^\circ\\-\tan45^\circ&\cos0^\circ\end{bmatrix}\begin{bmatrix}\cot45^\circ&\mathrm{cosec}\,30^\circ\\\sec60^\circ&\sin90^\circ\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}2\cos60^\circ&-2\sin30^\circ\\-\tan45^\circ&\cos0^\circ\end{bmatrix}\begin{bmatrix}\cot45^\circ&\mathrm{cosec}\,30^\circ\\\sec60^\circ&\sin90^\circ\end{bmatrix}
\displaystyle =\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&1\\1&-1\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{State True or False with reason.}
\displaystyle \text{Answer:}
\displaystyle \text{i) }A+B=B+A:\text{ True, since addition of matrices is commutative.}
\displaystyle \text{ii) }A-B=B-A:\text{ False, since subtraction of matrices is not commutative.}
\displaystyle \text{iii) }(BC)A=B(CA):\text{ True, since multiplication of matrices is associative.}
\displaystyle \text{iv) }(A+B)C=AC+BC:\text{ True, since multiplication is distributive over addition.}
\displaystyle \text{v) }A(B-C)=AB-AC:\text{ True, since multiplication is distributive over subtraction.}
\displaystyle \text{vi) }(A-B)C=AC-BC:\text{ True, since multiplication is distributive over subtraction.}
\displaystyle \text{vii) }A^2-B^2=(A+B)(A-B):\text{ False, since matrices need not commute.}
\displaystyle \text{In general, }(A+B)(A-B)=A^2-AB+BA-B^2\ne A^2-B^2.
\displaystyle \text{viii) }(A-B)^2=A^2-2AB+B^2:\text{ False, since matrices need not commute.}
\displaystyle \text{In general, }(A-B)^2=A^2-AB-BA+B^2\ne A^2-2AB+B^2.
\displaystyle \\

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