\displaystyle \textbf{Question 1: }\text{Evaluate:}
\displaystyle \text{i) }3\begin{bmatrix}5\\-2\end{bmatrix}\qquad \text{ii) }7\begin{bmatrix}-1&2\\0&1\end{bmatrix}
\displaystyle \text{iii) }2\begin{bmatrix}-1&0\\2&-3\end{bmatrix}+\begin{bmatrix}3&3\\5&0\end{bmatrix}
\displaystyle \text{iv) }6\begin{bmatrix}3\\2\end{bmatrix}-2\begin{bmatrix}-8\\1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{i) }3\begin{bmatrix}5\\-2\end{bmatrix}=\begin{bmatrix}15\\-6\end{bmatrix}
\displaystyle \text{ii) }7\begin{bmatrix}-1&2\\0&1\end{bmatrix}=\begin{bmatrix}-7&14\\0&7\end{bmatrix}
\displaystyle \text{iii) }2\begin{bmatrix}-1&0\\2&-3\end{bmatrix}+\begin{bmatrix}3&3\\5&0\end{bmatrix}=\begin{bmatrix}-2&0\\4&-6\end{bmatrix}+\begin{bmatrix}3&3\\5&0\end{bmatrix}
\displaystyle =\begin{bmatrix}1&3\\9&-6\end{bmatrix}
\displaystyle \text{iv) }6\begin{bmatrix}3\\2\end{bmatrix}-2\begin{bmatrix}-8\\1\end{bmatrix}=\begin{bmatrix}18\\12\end{bmatrix}-\begin{bmatrix}-16\\2\end{bmatrix}
\displaystyle =\begin{bmatrix}18-(-16)\\12-2\end{bmatrix}=\begin{bmatrix}34\\10\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find }x\text{ and }y\text{ if}
\displaystyle \text{i) }3\begin{bmatrix}4&x\end{bmatrix}+2\begin{bmatrix}y&-3\end{bmatrix}=\begin{bmatrix}10&0\end{bmatrix}
\displaystyle \text{ii) }x\begin{bmatrix}-1\\2\end{bmatrix}-4\begin{bmatrix}-2\\y\end{bmatrix}=\begin{bmatrix}7\\-8\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{i) }3\begin{bmatrix}4&x\end{bmatrix}+2\begin{bmatrix}y&-3\end{bmatrix}=\begin{bmatrix}10&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}12&3x\end{bmatrix}+\begin{bmatrix}2y&-6\end{bmatrix}=\begin{bmatrix}10&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}12+2y&3x-6\end{bmatrix}=\begin{bmatrix}10&0\end{bmatrix}
\displaystyle 12+2y=10\Rightarrow y=-1
\displaystyle 3x-6=0\Rightarrow x=2
\displaystyle \text{ii) }x\begin{bmatrix}-1\\2\end{bmatrix}-4\begin{bmatrix}-2\\y\end{bmatrix}=\begin{bmatrix}7\\-8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-x\\2x\end{bmatrix}-\begin{bmatrix}-8\\4y\end{bmatrix}=\begin{bmatrix}7\\-8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-x+8\\2x-4y\end{bmatrix}=\begin{bmatrix}7\\-8\end{bmatrix}
\displaystyle -x+8=7\Rightarrow x=1
\displaystyle 2x-4y=-8
\displaystyle 2(1)-4y=-8\Rightarrow -4y=-10\Rightarrow y=\frac{5}{2}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Given }A=\begin{bmatrix}2&1\\3&0\end{bmatrix},\ B=\begin{bmatrix}1&1\\5&2\end{bmatrix},\ C=\begin{bmatrix}-3&-1\\0&0\end{bmatrix},\text{ find:}
\displaystyle \text{i) }2A-3B+C\qquad\qquad \text{ii) }A+2C-B
\displaystyle \text{Answer:}
\displaystyle \text{i) }2A-3B+C
\displaystyle =2\begin{bmatrix}2&1\\3&0\end{bmatrix}-3\begin{bmatrix}1&1\\5&2\end{bmatrix}+\begin{bmatrix}-3&-1\\0&0\end{bmatrix}
\displaystyle =\begin{bmatrix}4&2\\6&0\end{bmatrix}-\begin{bmatrix}3&3\\15&6\end{bmatrix}+\begin{bmatrix}-3&-1\\0&0\end{bmatrix}
\displaystyle =\begin{bmatrix}1&-1\\-9&-6\end{bmatrix}+\begin{bmatrix}-3&-1\\0&0\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&-2\\-9&-6\end{bmatrix}
\displaystyle \text{ii) }A+2C-B
\displaystyle =\begin{bmatrix}2&1\\3&0\end{bmatrix}+2\begin{bmatrix}-3&-1\\0&0\end{bmatrix}-\begin{bmatrix}1&1\\5&2\end{bmatrix}
\displaystyle =\begin{bmatrix}2&1\\3&0\end{bmatrix}+\begin{bmatrix}-6&-2\\0&0\end{bmatrix}-\begin{bmatrix}1&1\\5&2\end{bmatrix}
\displaystyle =\begin{bmatrix}-4&-1\\3&0\end{bmatrix}-\begin{bmatrix}1&1\\5&2\end{bmatrix}
\displaystyle =\begin{bmatrix}-5&-2\\-2&-2\end{bmatrix}
\displaystyle \\

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

\displaystyle \textbf{Question 5: }\text{Given }A=\begin{bmatrix}1&4\\2&3\end{bmatrix},\ B=\begin{bmatrix}-4&-1\\-3&-2\end{bmatrix},\text{ find:}
\displaystyle \text{i) }2A+B\qquad \text{ii) Matrix }C\text{ such that }C+B=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{i) }2A+B=2\begin{bmatrix}1&4\\2&3\end{bmatrix}+\begin{bmatrix}-4&-1\\-3&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}2&8\\4&6\end{bmatrix}+\begin{bmatrix}-4&-1\\-3&-2\end{bmatrix}=\begin{bmatrix}-2&7\\1&4\end{bmatrix}
\displaystyle \text{ii) }C+B=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow C=\begin{bmatrix}0&0\\0&0\end{bmatrix}-\begin{bmatrix}-4&-1\\-3&-2\end{bmatrix}
\displaystyle \Rightarrow C=\begin{bmatrix}4&1\\3&2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }2\begin{bmatrix}3&x\\0&1\end{bmatrix}+3\begin{bmatrix}1&3\\y&2\end{bmatrix}=\begin{bmatrix}z&-7\\15&8\end{bmatrix},
\displaystyle \text{find the values of }x,\ y\text{ and }z.
\displaystyle \text{Answer:}
\displaystyle 2\begin{bmatrix}3&x\\0&1\end{bmatrix}+3\begin{bmatrix}1&3\\y&2\end{bmatrix}=\begin{bmatrix}z&-7\\15&8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6&2x\\0&2\end{bmatrix}+\begin{bmatrix}3&9\\3y&6\end{bmatrix}=\begin{bmatrix}z&-7\\15&8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}9&2x+9\\3y&8\end{bmatrix}=\begin{bmatrix}z&-7\\15&8\end{bmatrix}
\displaystyle \Rightarrow z=9
\displaystyle 2x+9=-7\Rightarrow 2x=-16\Rightarrow x=-8
\displaystyle 3y=15\Rightarrow y=5
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Given }A=\begin{bmatrix}-3&6\\0&-9\end{bmatrix}\text{ and }A^t\text{ is the transpose matrix. Find:}
\displaystyle \text{i) }2A+3A^t\qquad \text{ii) }2A^t-3A\qquad \text{iii) }\frac{1}{2}A-\frac{1}{3}A^t\qquad \text{iv) }A^t-\frac{1}{3}A
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}-3&6\\0&-9\end{bmatrix}\Rightarrow A^t=\begin{bmatrix}-3&0\\6&-9\end{bmatrix}
\displaystyle \text{i) }2A+3A^t=2\begin{bmatrix}-3&6\\0&-9\end{bmatrix}+3\begin{bmatrix}-3&0\\6&-9\end{bmatrix}
\displaystyle =\begin{bmatrix}-6&12\\0&-18\end{bmatrix}+\begin{bmatrix}-9&0\\18&-27\end{bmatrix}=\begin{bmatrix}-15&12\\18&-45\end{bmatrix}
\displaystyle \text{ii) }2A^t-3A=2\begin{bmatrix}-3&0\\6&-9\end{bmatrix}-3\begin{bmatrix}-3&6\\0&-9\end{bmatrix}
\displaystyle =\begin{bmatrix}-6&0\\12&-18\end{bmatrix}-\begin{bmatrix}-9&18\\0&-27\end{bmatrix}=\begin{bmatrix}3&-18\\12&9\end{bmatrix}
\displaystyle \text{iii) }\frac{1}{2}A-\frac{1}{3}A^t=\frac{1}{2}\begin{bmatrix}-3&6\\0&-9\end{bmatrix}-\frac{1}{3}\begin{bmatrix}-3&0\\6&-9\end{bmatrix}
\displaystyle =\begin{bmatrix}-\frac{3}{2}&3\\0&-\frac{9}{2}\end{bmatrix}-\begin{bmatrix}-1&0\\2&-3\end{bmatrix}=\begin{bmatrix}-\frac{1}{2}&3\\-2&-\frac{3}{2}\end{bmatrix}
\displaystyle \text{iv) }A^t-\frac{1}{3}A=\begin{bmatrix}-3&0\\6&-9\end{bmatrix}-\frac{1}{3}\begin{bmatrix}-3&6\\0&-9\end{bmatrix}
\displaystyle =\begin{bmatrix}-3&0\\6&-9\end{bmatrix}-\begin{bmatrix}-1&2\\0&-3\end{bmatrix}=\begin{bmatrix}-2&-2\\6&-6\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Given }A=\begin{bmatrix}1&1\\-2&0\end{bmatrix}\text{ and }B=\begin{bmatrix}2&-1\\1&1\end{bmatrix},\text{ solve for:}
\displaystyle \text{i) }X+2A=B\qquad \text{ii) }3X+B+2A=0\qquad \text{iii) }3A-2X=X-2B
\displaystyle \text{Answer:}
\displaystyle \text{i) }X+2A=B
\displaystyle \Rightarrow X=B-2A
\displaystyle X=\begin{bmatrix}2&-1\\1&1\end{bmatrix}-2\begin{bmatrix}1&1\\-2&0\end{bmatrix}
\displaystyle =\begin{bmatrix}2&-1\\1&1\end{bmatrix}-\begin{bmatrix}2&2\\-4&0\end{bmatrix}=\begin{bmatrix}0&-3\\5&1\end{bmatrix}
\displaystyle \text{ii) }3X+B+2A=0
\displaystyle \Rightarrow 3X=-(B+2A)
\displaystyle \Rightarrow X=-\frac{1}{3}(B+2A)
\displaystyle X=-\frac{1}{3}\left(\begin{bmatrix}2&-1\\1&1\end{bmatrix}+2\begin{bmatrix}1&1\\-2&0\end{bmatrix}\right)
\displaystyle X=-\frac{1}{3}\left(\begin{bmatrix}2&-1\\1&1\end{bmatrix}+\begin{bmatrix}2&2\\-4&0\end{bmatrix}\right)
\displaystyle X=-\frac{1}{3}\begin{bmatrix}4&1\\-3&1\end{bmatrix}=\begin{bmatrix}-\frac{4}{3}&-\frac{1}{3}\\1&-\frac{1}{3}\end{bmatrix}
\displaystyle \text{iii) }3A-2X=X-2B
\displaystyle \Rightarrow 3A+2B=3X
\displaystyle \Rightarrow X=\frac{1}{3}(3A+2B)
\displaystyle X=\frac{1}{3}\left(3\begin{bmatrix}1&1\\-2&0\end{bmatrix}+2\begin{bmatrix}2&-1\\1&1\end{bmatrix}\right)
\displaystyle X=\frac{1}{3}\left(\begin{bmatrix}3&3\\-6&0\end{bmatrix}+\begin{bmatrix}4&-2\\2&2\end{bmatrix}\right)
\displaystyle X=\frac{1}{3}\begin{bmatrix}7&1\\-4&2\end{bmatrix}=\begin{bmatrix}\frac{7}{3}&\frac{1}{3}\\-\frac{4}{3}&\frac{2}{3}\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }M=\begin{bmatrix}0\\1\end{bmatrix}\text{ and }N=\begin{bmatrix}1\\0\end{bmatrix},\text{ show that }3M+5N=\begin{bmatrix}5\\3\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle 3M+5N=3\begin{bmatrix}0\\1\end{bmatrix}+5\begin{bmatrix}1\\0\end{bmatrix}
\displaystyle =\begin{bmatrix}0\\3\end{bmatrix}+\begin{bmatrix}5\\0\end{bmatrix}=\begin{bmatrix}5\\3\end{bmatrix}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }I\text{ is the unit matrix of order }2\times2,\text{ find the matrix }M\text{ such that:}
\displaystyle \text{i) }M-2I=3\begin{bmatrix}-1&0\\4&1\end{bmatrix}\qquad \text{ii) }5M+3I=4\begin{bmatrix}2&-5\\0&-3\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{Since }I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.
\displaystyle \text{i) }M-2I=3\begin{bmatrix}-1&0\\4&1\end{bmatrix}
\displaystyle \Rightarrow M=2I+3\begin{bmatrix}-1&0\\4&1\end{bmatrix}
\displaystyle =2\begin{bmatrix}1&0\\0&1\end{bmatrix}+3\begin{bmatrix}-1&0\\4&1\end{bmatrix}
\displaystyle =\begin{bmatrix}2&0\\0&2\end{bmatrix}+\begin{bmatrix}-3&0\\12&3\end{bmatrix}=\begin{bmatrix}-1&0\\12&5\end{bmatrix}
\displaystyle \text{ii) }5M+3I=4\begin{bmatrix}2&-5\\0&-3\end{bmatrix}
\displaystyle \Rightarrow 5M=4\begin{bmatrix}2&-5\\0&-3\end{bmatrix}-3I
\displaystyle \Rightarrow M=\frac{1}{5}\left(4\begin{bmatrix}2&-5\\0&-3\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}\right)
\displaystyle =\frac{1}{5}\left(\begin{bmatrix}8&-20\\0&-12\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}\right)
\displaystyle =\frac{1}{5}\begin{bmatrix}5&-20\\0&-15\end{bmatrix}=\begin{bmatrix}1&-4\\0&-3\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\begin{bmatrix}1&4\\-2&3\end{bmatrix}+2M=3\begin{bmatrix}3&2\\0&-3\end{bmatrix},\text{ find the matrix }M.
\displaystyle \text{Answer:}
\displaystyle 2M=3\begin{bmatrix}3&2\\0&-3\end{bmatrix}-\begin{bmatrix}1&4\\-2&3\end{bmatrix}
\displaystyle =\begin{bmatrix}9&6\\0&-9\end{bmatrix}-\begin{bmatrix}1&4\\-2&3\end{bmatrix}=\begin{bmatrix}8&2\\2&-12\end{bmatrix}
\displaystyle \Rightarrow M=\frac{1}{2}\begin{bmatrix}8&2\\2&-12\end{bmatrix}=\begin{bmatrix}4&1\\1&-6\end{bmatrix}
\displaystyle \\

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