\displaystyle \textbf{Question 1: }\text{State True or False. If False, state the reason.}
\displaystyle \text{1) If }A\text{ and }B\text{ are two matrices of order }3\times2\text{ and }2\times3\text{ respectively,}
\displaystyle \text{then their sum }A+B\text{ is possible.}
\displaystyle \text{2) The matrices }A_{2\times3}\text{ and }B_{2\times3}\text{ are conformable for subtraction.}
\displaystyle \text{3) Transpose of a }2\times1\text{ matrix is a }2\times1\text{ matrix.}
\displaystyle \text{4) Transpose of a square matrix is a square matrix.}
\displaystyle \text{5) A column matrix has many columns and only one row.}
\displaystyle \text{Answer:}
\displaystyle \text{1) False: Two matrices can be added only if they are of the same order.}
\displaystyle \text{Here }A\text{ is of order }3\times2\text{ and }B\text{ is of order }2\times3.
\displaystyle \therefore \text{The sum }A+B\text{ is not possible.}
\displaystyle \text{2) True: Two matrices can be subtracted only if they are of the same order.}
\displaystyle \text{Here both }A\text{ and }B\text{ are of order }2\times3.
\displaystyle \therefore \text{They are conformable for subtraction.}
\displaystyle \text{3) False: The transpose of a matrix is obtained by interchanging rows and columns.}
\displaystyle \therefore \text{The transpose of a }2\times1\text{ matrix is a }1\times2\text{ matrix.}
\displaystyle \text{4) True: The transpose of a square matrix is also a square matrix.}
\displaystyle \text{The number of rows and columns remains the same after transposition.}
\displaystyle \text{5) False: A column matrix has one column and many rows.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Given }\begin{bmatrix}x&y+2\\3&z-1\end{bmatrix}=\begin{bmatrix}3&1\\3&2\end{bmatrix},\text{ find }x,\ y\text{ and }z.
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}x&y+2\\3&z-1\end{bmatrix}=\begin{bmatrix}3&1\\3&2\end{bmatrix}
\displaystyle \Rightarrow x=3
\displaystyle y+2=1\Rightarrow y=-1
\displaystyle z-1=2\Rightarrow z=3
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve for }a,\ b\text{ and }c\text{ if:}
\displaystyle \text{1) }\begin{bmatrix}-4&a+5\\3&2\end{bmatrix}=\begin{bmatrix}b+4&2\\3&c-1\end{bmatrix}
\displaystyle \text{2) }\begin{bmatrix}a&a-b\\b+c&0\end{bmatrix}=\begin{bmatrix}3&-1\\2&0\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{1) }\begin{bmatrix}-4&a+5\\3&2\end{bmatrix}=\begin{bmatrix}b+4&2\\3&c-1\end{bmatrix}
\displaystyle -4=b+4\Rightarrow b=-8
\displaystyle a+5=2\Rightarrow a=-3
\displaystyle 2=c-1\Rightarrow c=3
\displaystyle \text{2) }\begin{bmatrix}a&a-b\\b+c&0\end{bmatrix}=\begin{bmatrix}3&-1\\2&0\end{bmatrix}
\displaystyle a=3
\displaystyle a-b=-1\Rightarrow b=a+1=3+1=4
\displaystyle b+c=2\Rightarrow c=2-b=2-4=-2
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }A=\begin{bmatrix}8&-3\end{bmatrix}\text{ and }B=\begin{bmatrix}4&-5\end{bmatrix},\text{ find:}
\displaystyle \text{1) }A+B\qquad\qquad \text{2) }B-A
\displaystyle \text{Answer:}
\displaystyle \text{1) }A+B=\begin{bmatrix}8&-3\end{bmatrix}+\begin{bmatrix}4&-5\end{bmatrix}
\displaystyle \phantom{\text{1) }A+B}=\begin{bmatrix}8+4&-3-5\end{bmatrix}=\begin{bmatrix}12&-8\end{bmatrix}
\displaystyle \text{2) }B-A=\begin{bmatrix}4&-5\end{bmatrix}-\begin{bmatrix}8&-3\end{bmatrix}
\displaystyle \phantom{\text{2) }B-A}=\begin{bmatrix}4-8&-5-(-3)\end{bmatrix}=\begin{bmatrix}-4&-2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A=\begin{bmatrix}2\\5\end{bmatrix},\ B=\begin{bmatrix}1\\4\end{bmatrix}\text{ and }C=\begin{bmatrix}6\\-2\end{bmatrix},\text{ find:}
\displaystyle \text{1) }B+C\qquad \text{2) }A-C\qquad \text{3) }A+B-C\qquad \text{4) }A-B+C
\displaystyle \text{Answer:}
\displaystyle \text{1) }B+C=\begin{bmatrix}1\\4\end{bmatrix}+\begin{bmatrix}6\\-2\end{bmatrix}=\begin{bmatrix}1+6\\4-2\end{bmatrix}=\begin{bmatrix}7\\2\end{bmatrix}
\displaystyle \text{2) }A-C=\begin{bmatrix}2\\5\end{bmatrix}-\begin{bmatrix}6\\-2\end{bmatrix}=\begin{bmatrix}2-6\\5-(-2)\end{bmatrix}=\begin{bmatrix}-4\\7\end{bmatrix}
\displaystyle \text{3) }A+B-C=\begin{bmatrix}2\\5\end{bmatrix}+\begin{bmatrix}1\\4\end{bmatrix}-\begin{bmatrix}6\\-2\end{bmatrix}=\begin{bmatrix}2+1-6\\5+4-(-2)\end{bmatrix}=\begin{bmatrix}-3\\11\end{bmatrix}
\displaystyle \text{4) }A-B+C=\begin{bmatrix}2\\5\end{bmatrix}-\begin{bmatrix}1\\4\end{bmatrix}+\begin{bmatrix}6\\-2\end{bmatrix}=\begin{bmatrix}2-1+6\\5-4+(-2)\end{bmatrix}=\begin{bmatrix}7\\-1\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Wherever possible, write each of the following in a single matrix:}
\displaystyle \text{1) }\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}-1&-2\\1&-7\end{bmatrix}
\displaystyle \text{2) }\begin{bmatrix}2&3&4\\5&6&7\end{bmatrix}-\begin{bmatrix}0&2&3\\6&-1&0\end{bmatrix}
\displaystyle \text{3) }\begin{bmatrix}0&1&2\\4&6&7\end{bmatrix}+\begin{bmatrix}3&4\\6&8\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{1) }\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}-1&-2\\1&-7\end{bmatrix}=\begin{bmatrix}1-1&2-2\\3+1&4-7\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0\\4&-3\end{bmatrix}
\displaystyle \text{2) }\begin{bmatrix}2&3&4\\5&6&7\end{bmatrix}-\begin{bmatrix}0&2&3\\6&-1&0\end{bmatrix}=\begin{bmatrix}2-0&3-2&4-3\\5-6&6-(-1)&7-0\end{bmatrix}
\displaystyle =\begin{bmatrix}2&1&1\\-1&7&7\end{bmatrix}
\displaystyle \text{3) }\text{The first matrix is of order }2\times3\text{ and the second matrix is of order }2\times2.
\displaystyle \therefore \text{The addition is not possible, as the matrices are not of the same order.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find }x\text{ and }y\text{ from the following equations:}
\displaystyle \text{1) }\begin{bmatrix}5&2\\-1&y-1\end{bmatrix}-\begin{bmatrix}1&x-1\\2&-3\end{bmatrix}=\begin{bmatrix}4&7\\-3&2\end{bmatrix}
\displaystyle \text{2) }\begin{bmatrix}-8&x\end{bmatrix}+\begin{bmatrix}y&-2\end{bmatrix}=\begin{bmatrix}-3&2\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{1) }\begin{bmatrix}5&2\\-1&y-1\end{bmatrix}-\begin{bmatrix}1&x-1\\2&-3\end{bmatrix}=\begin{bmatrix}4&7\\-3&2\end{bmatrix}
\displaystyle \begin{bmatrix}5-1&2-(x-1)\\-1-2&y-1-(-3)\end{bmatrix}=\begin{bmatrix}4&7\\-3&2\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4&3-x\\-3&y+2\end{bmatrix}=\begin{bmatrix}4&7\\-3&2\end{bmatrix}
\displaystyle 3-x=7\Rightarrow x=-4
\displaystyle y+2=2\Rightarrow y=0
\displaystyle \text{2) }\begin{bmatrix}-8&x\end{bmatrix}+\begin{bmatrix}y&-2\end{bmatrix}=\begin{bmatrix}-3&2\end{bmatrix}
\displaystyle \begin{bmatrix}-8+y&x-2\end{bmatrix}=\begin{bmatrix}-3&2\end{bmatrix}
\displaystyle -8+y=-3\Rightarrow y=5
\displaystyle x-2=2\Rightarrow x=4
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Given }M=\begin{bmatrix}5&-3\\-2&4\end{bmatrix},\text{ find its transpose }M^{t}.\text{ If possible, find:}
\displaystyle \text{1) }M+M^{t}\qquad\qquad \text{2) }M^{t}-M
\displaystyle \text{Answer:}
\displaystyle M=\begin{bmatrix}5&-3\\-2&4\end{bmatrix},\qquad M^{t}=\begin{bmatrix}5&-2\\-3&4\end{bmatrix}
\displaystyle \text{1) }M+M^{t}=\begin{bmatrix}5&-3\\-2&4\end{bmatrix}+\begin{bmatrix}5&-2\\-3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}10&-5\\-5&8\end{bmatrix}
\displaystyle \text{2) }M^{t}-M=\begin{bmatrix}5&-2\\-3&4\end{bmatrix}-\begin{bmatrix}5&-3\\-2&4\end{bmatrix}
\displaystyle =\begin{bmatrix}0&1\\-1&0\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write the additive inverse of matrices }A,\ B\text{ and }C\text{ where }A=\begin{bmatrix}6&-5\end{bmatrix},
\displaystyle B=\begin{bmatrix}-2&0\\4&-1\end{bmatrix}\text{ and }C=\begin{bmatrix}-2\\4\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Additive inverse of }A=\begin{bmatrix}6&-5\end{bmatrix}\text{ is }-\!A=\begin{bmatrix}-6&5\end{bmatrix}
\displaystyle \text{Additive inverse of }B=\begin{bmatrix}-2&0\\4&-1\end{bmatrix}\text{ is }-\!B=\begin{bmatrix}2&0\\-4&1\end{bmatrix}
\displaystyle \text{Additive inverse of }C=\begin{bmatrix}-2\\4\end{bmatrix}\text{ is }-\!C=\begin{bmatrix}2\\-4\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Given }A=\begin{bmatrix}2&-3\end{bmatrix},\ B=\begin{bmatrix}0&2\end{bmatrix},\ C=\begin{bmatrix}-1&4\end{bmatrix}.
\displaystyle \text{Find matrix }X\text{ in each of the following:}
\displaystyle \text{1) }X+B=C-A\qquad\qquad \text{2) }A-X=B+C
\displaystyle \text{Answer:}
\displaystyle \text{Let }X=\begin{bmatrix}a&b\end{bmatrix}.
\displaystyle \text{1) }X+B=C-A
\displaystyle \begin{bmatrix}a&b\end{bmatrix}+\begin{bmatrix}0&2\end{bmatrix}=\begin{bmatrix}-1&4\end{bmatrix}-\begin{bmatrix}2&-3\end{bmatrix}
\displaystyle \begin{bmatrix}a&b+2\end{bmatrix}=\begin{bmatrix}-3&7\end{bmatrix}
\displaystyle \Rightarrow a=-3,\ b=5
\displaystyle \therefore X=\begin{bmatrix}-3&5\end{bmatrix}
\displaystyle \text{2) }A-X=B+C
\displaystyle \begin{bmatrix}2&-3\end{bmatrix}-\begin{bmatrix}a&b\end{bmatrix}=\begin{bmatrix}0&2\end{bmatrix}+\begin{bmatrix}-1&4\end{bmatrix}
\displaystyle \begin{bmatrix}2-a&-3-b\end{bmatrix}=\begin{bmatrix}-1&6\end{bmatrix}
\displaystyle \Rightarrow a=3,\ b=-9
\displaystyle \therefore X=\begin{bmatrix}3&-9\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Given }A=\begin{bmatrix}-1&0\\2&-4\end{bmatrix}\text{ and }B=\begin{bmatrix}3&-3\\-2&0\end{bmatrix}.
\displaystyle \text{Find the matrix }X\text{ in each of the following:}
\displaystyle \text{1) }A+X=B\qquad \text{2) }A-X=B\qquad \text{3) }X-B=A
\displaystyle \text{Answer:}
\displaystyle \text{Let }X=\begin{bmatrix}a&b\\c&d\end{bmatrix}.
\displaystyle \text{1) }A+X=B
\displaystyle \begin{bmatrix}-1&0\\2&-4\end{bmatrix}+\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}3&-3\\-2&0\end{bmatrix}
\displaystyle \begin{bmatrix}-1+a&b\\2+c&-4+d\end{bmatrix}=\begin{bmatrix}3&-3\\-2&0\end{bmatrix}
\displaystyle \Rightarrow a=4,\ b=-3,\ c=-4,\ d=4
\displaystyle \therefore X=\begin{bmatrix}4&-3\\-4&4\end{bmatrix}
\displaystyle \text{2) }A-X=B
\displaystyle \begin{bmatrix}-1&0\\2&-4\end{bmatrix}-\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}3&-3\\-2&0\end{bmatrix}
\displaystyle \begin{bmatrix}-1-a&-b\\2-c&-4-d\end{bmatrix}=\begin{bmatrix}3&-3\\-2&0\end{bmatrix}
\displaystyle \Rightarrow a=-4,\ b=3,\ c=4,\ d=-4
\displaystyle \therefore X=\begin{bmatrix}-4&3\\4&-4\end{bmatrix}
\displaystyle \text{3) }X-B=A
\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix}-\begin{bmatrix}3&-3\\-2&0\end{bmatrix}=\begin{bmatrix}-1&0\\2&-4\end{bmatrix}
\displaystyle \begin{bmatrix}a-3&b+3\\c+2&d\end{bmatrix}=\begin{bmatrix}-1&0\\2&-4\end{bmatrix}
\displaystyle \Rightarrow a=2,\ b=-3,\ c=0,\ d=-4
\displaystyle \therefore X=\begin{bmatrix}2&-3\\0&-4\end{bmatrix}
\displaystyle \\

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