\displaystyle \textbf{Question 1: }\text{Evaluate:}
\displaystyle \text{i) }\begin{bmatrix}3&2\end{bmatrix}\begin{bmatrix}2\\0\end{bmatrix}\qquad \text{ii) }\begin{bmatrix}1&-2\end{bmatrix}\begin{bmatrix}-2&3\\-1&4\end{bmatrix}
\displaystyle \text{iii) }\begin{bmatrix}6&4\\3&-1\end{bmatrix}\begin{bmatrix}-1\\3\end{bmatrix}\qquad \text{iv) }\begin{bmatrix}6&4\\3&-1\end{bmatrix}\begin{bmatrix}-1&3\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{i) }\begin{bmatrix}3&2\end{bmatrix}\begin{bmatrix}2\\0\end{bmatrix}=\begin{bmatrix}3(2)+2(0)\end{bmatrix}=\begin{bmatrix}6\end{bmatrix}
\displaystyle \text{ii) }\begin{bmatrix}1&-2\end{bmatrix}\begin{bmatrix}-2&3\\-1&4\end{bmatrix}=\begin{bmatrix}1(-2)+(-2)(-1)&1(3)+(-2)(4)\end{bmatrix}
\displaystyle =\begin{bmatrix}0&-5\end{bmatrix}
\displaystyle \text{iii) }\begin{bmatrix}6&4\\3&-1\end{bmatrix}\begin{bmatrix}-1\\3\end{bmatrix}=\begin{bmatrix}6(-1)+4(3)\\3(-1)+(-1)(3)\end{bmatrix}
\displaystyle =\begin{bmatrix}6\\-6\end{bmatrix}
\displaystyle \text{iv) }\begin{bmatrix}6&4\\3&-1\end{bmatrix}\begin{bmatrix}-1&3\end{bmatrix}
\displaystyle \text{This multiplication is not possible, as the number of columns of the first matrix}
\displaystyle \text{is not equal to the number of rows of the second matrix.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }A=\begin{bmatrix}0&2\\5&-2\end{bmatrix},\ B=\begin{bmatrix}1&-1\\3&2\end{bmatrix}\text{ and }I
\displaystyle \text{is the unit matrix of order }2\times2,\text{ find:}
\displaystyle \text{i) }AB\qquad \text{ii) }BA\qquad \text{iii) }AI\qquad \text{iv) }IB\qquad \text{v) }A^2\qquad \text{vi) }B^2A
\displaystyle \text{Answer:}
\displaystyle I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \text{i) }AB=\begin{bmatrix}0&2\\5&-2\end{bmatrix}\begin{bmatrix}1&-1\\3&2\end{bmatrix}=\begin{bmatrix}6&4\\-1&-9\end{bmatrix}
\displaystyle \text{ii) }BA=\begin{bmatrix}1&-1\\3&2\end{bmatrix}\begin{bmatrix}0&2\\5&-2\end{bmatrix}=\begin{bmatrix}-5&4\\10&2\end{bmatrix}
\displaystyle \text{iii) }AI=\begin{bmatrix}0&2\\5&-2\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}0&2\\5&-2\end{bmatrix}
\displaystyle \text{iv) }IB=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}1&-1\\3&2\end{bmatrix}=\begin{bmatrix}1&-1\\3&2\end{bmatrix}
\displaystyle \text{v) }A^2=\begin{bmatrix}0&2\\5&-2\end{bmatrix}\begin{bmatrix}0&2\\5&-2\end{bmatrix}=\begin{bmatrix}10&-4\\-10&14\end{bmatrix}
\displaystyle \text{vi) }B^2A=\begin{bmatrix}1&-1\\3&2\end{bmatrix}\begin{bmatrix}1&-1\\3&2\end{bmatrix}\begin{bmatrix}0&2\\5&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&-3\\9&1\end{bmatrix}\begin{bmatrix}0&2\\5&-2\end{bmatrix}=\begin{bmatrix}-15&2\\5&16\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }M=\begin{bmatrix}2&1\\1&-2\end{bmatrix},\text{ find }M^2,\ M^3\text{ and }M^5.
\displaystyle \text{Answer:}
\displaystyle M^2=\begin{bmatrix}2&1\\1&-2\end{bmatrix}\begin{bmatrix}2&1\\1&-2\end{bmatrix}=\begin{bmatrix}5&0\\0&5\end{bmatrix}
\displaystyle M^3=M^2M=\begin{bmatrix}5&0\\0&5\end{bmatrix}\begin{bmatrix}2&1\\1&-2\end{bmatrix}=\begin{bmatrix}10&5\\5&-10\end{bmatrix}
\displaystyle M^5=M^2M^3=\begin{bmatrix}5&0\\0&5\end{bmatrix}\begin{bmatrix}10&5\\5&-10\end{bmatrix}=\begin{bmatrix}50&25\\25&-50\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find }x\text{ and }y\text{ if:}
\displaystyle \text{i) }\begin{bmatrix}4&3x\\x&-2\end{bmatrix}\begin{bmatrix}5\\1\end{bmatrix}=\begin{bmatrix}y\\8\end{bmatrix}\qquad \text{ii) }\begin{bmatrix}x&0\\-3&1\end{bmatrix}\begin{bmatrix}1&1\\0&y\end{bmatrix}=\begin{bmatrix}2&2\\-3&-2\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{i) }\begin{bmatrix}4&3x\\x&-2\end{bmatrix}\begin{bmatrix}5\\1\end{bmatrix}=\begin{bmatrix}y\\8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}20+3x\\5x-2\end{bmatrix}=\begin{bmatrix}y\\8\end{bmatrix}
\displaystyle 5x-2=8\Rightarrow 5x=10\Rightarrow x=2
\displaystyle y=20+3x=20+3(2)=26
\displaystyle \text{ii) }\begin{bmatrix}x&0\\-3&1\end{bmatrix}\begin{bmatrix}1&1\\0&y\end{bmatrix}=\begin{bmatrix}2&2\\-3&-2\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x&x\\-3&-3+y\end{bmatrix}=\begin{bmatrix}2&2\\-3&-2\end{bmatrix}
\displaystyle x=2
\displaystyle -3+y=-2\Rightarrow y=1
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\ B=\begin{bmatrix}1&2\\4&3\end{bmatrix}\text{ and }C=\begin{bmatrix}4&3\\1&2\end{bmatrix},\text{ find:}
\displaystyle \text{i) }(AB)C\qquad \text{ii) }A(BC)\qquad \text{Is }A(BC)=(AB)C?
\displaystyle \text{Answer:}
\displaystyle \text{i) }(AB)C=\left(\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}1&2\\4&3\end{bmatrix}\right)\begin{bmatrix}4&3\\1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}13&11\\18&16\end{bmatrix}\begin{bmatrix}4&3\\1&2\end{bmatrix}=\begin{bmatrix}63&61\\88&86\end{bmatrix}
\displaystyle \text{ii) }A(BC)=\begin{bmatrix}1&3\\2&4\end{bmatrix}\left(\begin{bmatrix}1&2\\4&3\end{bmatrix}\begin{bmatrix}4&3\\1&2\end{bmatrix}\right)
\displaystyle =\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}6&7\\19&18\end{bmatrix}=\begin{bmatrix}63&61\\88&86\end{bmatrix}
\displaystyle \therefore A(BC)=(AB)C.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }A=\begin{bmatrix}0&4&6\\3&0&-1\end{bmatrix},\ B=\begin{bmatrix}0&-1\\-1&2\\-5&-6\end{bmatrix},\text{ calculate:}
\displaystyle \text{i) }AB\qquad \text{ii) }BA\qquad \text{iii) }A^2
\displaystyle \text{Answer:}
\displaystyle \text{i) }AB=\begin{bmatrix}0&4&6\\3&0&-1\end{bmatrix}\begin{bmatrix}0&-1\\-1&2\\-5&-6\end{bmatrix}=\begin{bmatrix}-34&-28\\5&3\end{bmatrix}
\displaystyle \text{ii) }BA=\begin{bmatrix}0&-1\\-1&2\\-5&-6\end{bmatrix}\begin{bmatrix}0&4&6\\3&0&-1\end{bmatrix}
\displaystyle =\begin{bmatrix}-3&0&1\\6&-4&-8\\-18&-20&-24\end{bmatrix}
\displaystyle \text{iii) }A^2\text{ is not possible, since }A\text{ is of order }2\times3.
\displaystyle \text{The number of columns of }A\text{ is not equal to the number of rows of }A.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }A=\begin{bmatrix}2&1\\0&-2\end{bmatrix},\ B=\begin{bmatrix}4&1\\-3&-2\end{bmatrix}\text{ and }C=\begin{bmatrix}-3&2\\-1&4\end{bmatrix},
\displaystyle \text{find }A^2+AC-5B.\hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle A^2+AC-5B
\displaystyle =\begin{bmatrix}2&1\\0&-2\end{bmatrix}\begin{bmatrix}2&1\\0&-2\end{bmatrix}+\begin{bmatrix}2&1\\0&-2\end{bmatrix}\begin{bmatrix}-3&2\\-1&4\end{bmatrix}-5\begin{bmatrix}4&1\\-3&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}4&0\\0&4\end{bmatrix}+\begin{bmatrix}-7&8\\2&-8\end{bmatrix}-\begin{bmatrix}20&5\\-15&-10\end{bmatrix}
\displaystyle =\begin{bmatrix}-23&3\\17&6\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }M=\begin{bmatrix}1&2\\1&1\end{bmatrix}\text{ and }I\text{ is the unit matrix of order }2\times2,
\displaystyle \text{verify whether }M^2=2M+3I.
\displaystyle \text{Answer:}
\displaystyle I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle LHS=M^2=\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}1&2\\1&1\end{bmatrix}=\begin{bmatrix}3&4\\2&3\end{bmatrix}
\displaystyle RHS=2M+3I=2\begin{bmatrix}1&2\\1&1\end{bmatrix}+3\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}2&4\\2&2\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}=\begin{bmatrix}5&4\\2&5\end{bmatrix}
\displaystyle \therefore LHS\ne RHS
\displaystyle \text{Hence, the given statement }M^2=2M+3I\text{ is false for the given matrix }M.
\displaystyle \text{In fact, }M^2=2M+I.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }A=\begin{bmatrix}a&0\\0&2\end{bmatrix},\ B=\begin{bmatrix}0&-b\\1&0\end{bmatrix}\text{ and }M=\begin{bmatrix}1&-1\\1&1\end{bmatrix},
\displaystyle \text{and }BA=M^2,\text{ find the values of }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle BA=M^2
\displaystyle \begin{bmatrix}0&-b\\1&0\end{bmatrix}\begin{bmatrix}a&0\\0&2\end{bmatrix}=\begin{bmatrix}1&-1\\1&1\end{bmatrix}\begin{bmatrix}1&-1\\1&1\end{bmatrix}
\displaystyle \begin{bmatrix}0&-2b\\a&0\end{bmatrix}=\begin{bmatrix}0&-2\\2&0\end{bmatrix}
\displaystyle \Rightarrow -2b=-2\text{ and }a=2
\displaystyle \therefore b=1\text{ and }a=2
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }A=\begin{bmatrix}4&3\\2&1\end{bmatrix},\ B=\begin{bmatrix}1&0\\-2&1\end{bmatrix},\text{ find:}
\displaystyle \text{i) }A-B\qquad \text{ii) }A^2\qquad \text{iii) }AB\qquad \text{iv) }A^2-AB+2B
\displaystyle \text{Answer:}
\displaystyle \text{i) }A-B=\begin{bmatrix}4&3\\2&1\end{bmatrix}-\begin{bmatrix}1&0\\-2&1\end{bmatrix}=\begin{bmatrix}3&3\\4&0\end{bmatrix}
\displaystyle \text{ii) }A^2=\begin{bmatrix}4&3\\2&1\end{bmatrix}\begin{bmatrix}4&3\\2&1\end{bmatrix}=\begin{bmatrix}22&15\\10&7\end{bmatrix}
\displaystyle \text{iii) }AB=\begin{bmatrix}4&3\\2&1\end{bmatrix}\begin{bmatrix}1&0\\-2&1\end{bmatrix}=\begin{bmatrix}-2&3\\0&1\end{bmatrix}
\displaystyle \text{iv) }A^2-AB+2B=\begin{bmatrix}22&15\\10&7\end{bmatrix}-\begin{bmatrix}-2&3\\0&1\end{bmatrix}+2\begin{bmatrix}1&0\\-2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}22&15\\10&7\end{bmatrix}-\begin{bmatrix}-2&3\\0&1\end{bmatrix}+\begin{bmatrix}2&0\\-4&2\end{bmatrix}
\displaystyle =\begin{bmatrix}26&12\\6&8\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }A=\begin{bmatrix}1&4\\1&-1\end{bmatrix}\text{ and }B=\begin{bmatrix}1&2\\-1&-1\end{bmatrix},\text{ find:}
\displaystyle \text{i) }(A+B)^2\qquad \text{ii) }A^2+B^2\qquad \text{iii) Is }(A+B)^2=A^2+B^2?
\displaystyle \text{Answer:}
\displaystyle \text{i) }(A+B)^2
\displaystyle =\left(\begin{bmatrix}1&4\\1&-1\end{bmatrix}+\begin{bmatrix}1&2\\-1&-1\end{bmatrix}\right)^2
\displaystyle =\left(\begin{bmatrix}2&6\\0&-2\end{bmatrix}\right)^2
\displaystyle =\begin{bmatrix}2&6\\0&-2\end{bmatrix}\begin{bmatrix}2&6\\0&-2\end{bmatrix}=\begin{bmatrix}4&0\\0&4\end{bmatrix}
\displaystyle \text{ii) }A^2+B^2
\displaystyle =\begin{bmatrix}1&4\\1&-1\end{bmatrix}\begin{bmatrix}1&4\\1&-1\end{bmatrix}+\begin{bmatrix}1&2\\-1&-1\end{bmatrix}\begin{bmatrix}1&2\\-1&-1\end{bmatrix}
\displaystyle =\begin{bmatrix}5&0\\0&5\end{bmatrix}+\begin{bmatrix}-1&0\\0&-1\end{bmatrix}=\begin{bmatrix}4&0\\0&4\end{bmatrix}
\displaystyle \text{iii) Since }(A+B)^2=A^2+B^2=\begin{bmatrix}4&0\\0&4\end{bmatrix},
\displaystyle \therefore (A+B)^2=A^2+B^2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }A=\begin{bmatrix}-1&1\\a&b\end{bmatrix}\text{ and }A^2=I,\text{ find }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{Since }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},
\displaystyle \begin{bmatrix}-1&1\\a&b\end{bmatrix}\begin{bmatrix}-1&1\\a&b\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}1+a&-1+b\\-a+ab&a+b^2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle 1+a=1\Rightarrow a=0
\displaystyle -1+b=0\Rightarrow b=1
\displaystyle \therefore a=0,\ b=1.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }A=\begin{bmatrix}2&1\\0&0\end{bmatrix},\ B=\begin{bmatrix}2&3\\4&1\end{bmatrix}\text{ and }C=\begin{bmatrix}-1&4\\0&2\end{bmatrix},
\displaystyle \text{then show that:}
\displaystyle \text{i) }A(B+C)=AB+AC\qquad \text{ii) }(B-A)C=BC-AC
\displaystyle \text{Answer:}
\displaystyle \text{i) }A(B+C)=AB+AC
\displaystyle LHS=A(B+C)=\begin{bmatrix}2&1\\0&0\end{bmatrix}\left(\begin{bmatrix}2&3\\4&1\end{bmatrix}+\begin{bmatrix}-1&4\\0&2\end{bmatrix}\right)
\displaystyle =\begin{bmatrix}2&1\\0&0\end{bmatrix}\begin{bmatrix}1&7\\4&3\end{bmatrix}=\begin{bmatrix}6&17\\0&0\end{bmatrix}
\displaystyle RHS=AB+AC
\displaystyle =\begin{bmatrix}2&1\\0&0\end{bmatrix}\begin{bmatrix}2&3\\4&1\end{bmatrix}+\begin{bmatrix}2&1\\0&0\end{bmatrix}\begin{bmatrix}-1&4\\0&2\end{bmatrix}
\displaystyle =\begin{bmatrix}8&7\\0&0\end{bmatrix}+\begin{bmatrix}-2&10\\0&0\end{bmatrix}=\begin{bmatrix}6&17\\0&0\end{bmatrix}
\displaystyle \therefore LHS=RHS
\displaystyle \text{Hence, }A(B+C)=AB+AC.
\displaystyle \text{ii) }(B-A)C=BC-AC
\displaystyle LHS=(B-A)C=\left(\begin{bmatrix}2&3\\4&1\end{bmatrix}-\begin{bmatrix}2&1\\0&0\end{bmatrix}\right)\begin{bmatrix}-1&4\\0&2\end{bmatrix}
\displaystyle =\begin{bmatrix}0&2\\4&1\end{bmatrix}\begin{bmatrix}-1&4\\0&2\end{bmatrix}=\begin{bmatrix}0&4\\-4&18\end{bmatrix}
\displaystyle RHS=BC-AC
\displaystyle =\begin{bmatrix}2&3\\4&1\end{bmatrix}\begin{bmatrix}-1&4\\0&2\end{bmatrix}-\begin{bmatrix}2&1\\0&0\end{bmatrix}\begin{bmatrix}-1&4\\0&2\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&14\\-4&18\end{bmatrix}-\begin{bmatrix}-2&10\\0&0\end{bmatrix}=\begin{bmatrix}0&4\\-4&18\end{bmatrix}
\displaystyle \therefore LHS=RHS
\displaystyle \text{Hence, }(B-A)C=BC-AC.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }A=\begin{bmatrix}1&4\\2&1\end{bmatrix},\ B=\begin{bmatrix}-3&2\\4&0\end{bmatrix}\text{ and }C=\begin{bmatrix}1&0\\0&2\end{bmatrix},
\displaystyle \text{find the value of }A^2+BC.
\displaystyle \text{Answer:}
\displaystyle A^2+BC
\displaystyle =\begin{bmatrix}1&4\\2&1\end{bmatrix}\begin{bmatrix}1&4\\2&1\end{bmatrix}+\begin{bmatrix}-3&2\\4&0\end{bmatrix}\begin{bmatrix}1&0\\0&2\end{bmatrix}
\displaystyle =\begin{bmatrix}9&8\\4&9\end{bmatrix}+\begin{bmatrix}-3&4\\4&0\end{bmatrix}
\displaystyle =\begin{bmatrix}6&12\\8&9\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Solve for }x\text{ and }y.
\displaystyle \text{i) }\begin{bmatrix}2&5\\5&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-7\\14\end{bmatrix}\qquad \text{ii) }\begin{bmatrix}x+y&x-4\end{bmatrix}\begin{bmatrix}-1&-2\\2&2\end{bmatrix}=\begin{bmatrix}-7&14\end{bmatrix}
\displaystyle \text{iii) }\begin{bmatrix}-2&0\\3&1\end{bmatrix}\begin{bmatrix}-1\\2x\end{bmatrix}+3\begin{bmatrix}-2\\1\end{bmatrix}=2\begin{bmatrix}y\\3\end{bmatrix}\hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{i) }\begin{bmatrix}2&5\\5&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-7\\14\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+5y\\5x+2y\end{bmatrix}=\begin{bmatrix}-7\\14\end{bmatrix}
\displaystyle 2x+5y=-7
\displaystyle 5x+2y=14
\displaystyle \text{Solving the above equations, we get }x=4,\ y=-3.
\displaystyle \text{ii) }\begin{bmatrix}x+y&x-4\end{bmatrix}\begin{bmatrix}-1&-2\\2&2\end{bmatrix}=\begin{bmatrix}-7&14\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x-y-8&-2y-8\end{bmatrix}=\begin{bmatrix}-7&14\end{bmatrix}
\displaystyle x-y-8=-7\Rightarrow x-y=1
\displaystyle -2y-8=14\Rightarrow -2y=22\Rightarrow y=-11
\displaystyle x-y=1\Rightarrow x-(-11)=1\Rightarrow x=-10
\displaystyle \text{iii) }\begin{bmatrix}-2&0\\3&1\end{bmatrix}\begin{bmatrix}-1\\2x\end{bmatrix}+3\begin{bmatrix}-2\\1\end{bmatrix}=2\begin{bmatrix}y\\3\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2\\-3+2x\end{bmatrix}+\begin{bmatrix}-6\\3\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-4\\2x\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle 2y=-4\Rightarrow y=-2
\displaystyle 2x=6\Rightarrow x=3
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find i) the order of matrix }M\text{ ii) and find }M.
\displaystyle \text{i) }M\begin{bmatrix}1&1\\0&2\end{bmatrix}=\begin{bmatrix}1&2\end{bmatrix}\qquad \text{ii) }\begin{bmatrix}1&1\\0&2\end{bmatrix}M=\begin{bmatrix}13\\5\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{i) }M\begin{bmatrix}1&1\\0&2\end{bmatrix}=\begin{bmatrix}1&2\end{bmatrix}
\displaystyle \text{Since }M\text{ multiplied by a }2\times2\text{ matrix gives a }1\times2\text{ matrix,}
\displaystyle \therefore \text{ order of }M\text{ is }1\times2.
\displaystyle \text{Let }M=\begin{bmatrix}a&b\end{bmatrix}.
\displaystyle \begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}1&1\\0&2\end{bmatrix}=\begin{bmatrix}1&2\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}a&a+2b\end{bmatrix}=\begin{bmatrix}1&2\end{bmatrix}
\displaystyle a=1
\displaystyle a+2b=2\Rightarrow 1+2b=2\Rightarrow b=\frac{1}{2}
\displaystyle \therefore M=\begin{bmatrix}1&\frac{1}{2}\end{bmatrix}
\displaystyle \text{ii) }\begin{bmatrix}1&1\\0&2\end{bmatrix}M=\begin{bmatrix}13\\5\end{bmatrix}
\displaystyle \text{Since a }2\times2\text{ matrix multiplied by }M\text{ gives a }2\times1\text{ matrix,}
\displaystyle \therefore \text{ order of }M\text{ is }2\times1.
\displaystyle \text{Let }M=\begin{bmatrix}a\\b\end{bmatrix}.
\displaystyle \begin{bmatrix}1&1\\0&2\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}13\\5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}a+b\\2b\end{bmatrix}=\begin{bmatrix}13\\5\end{bmatrix}
\displaystyle a+b=13
\displaystyle 2b=5\Rightarrow b=\frac{5}{2}
\displaystyle a+\frac{5}{2}=13\Rightarrow a=\frac{21}{2}
\displaystyle \therefore M=\begin{bmatrix}\frac{21}{2}\\\frac{5}{2}\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }A=\begin{bmatrix}2&x\\0&1\end{bmatrix}\text{ and }B=\begin{bmatrix}4&36\\0&1\end{bmatrix},\text{ find }x\text{ given }A^2=B.
\displaystyle \text{Answer:}
\displaystyle A^2=B
\displaystyle \Rightarrow \begin{bmatrix}2&x\\0&1\end{bmatrix}\begin{bmatrix}2&x\\0&1\end{bmatrix}=\begin{bmatrix}4&36\\0&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4&3x\\0&1\end{bmatrix}=\begin{bmatrix}4&36\\0&1\end{bmatrix}
\displaystyle 3x=36\Rightarrow x=12
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find positive integers }p\text{ and }q\text{ such that }
\displaystyle \begin{bmatrix}p&q\end{bmatrix}\begin{bmatrix}p\\q\end{bmatrix}=\begin{bmatrix}25\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}p&q\end{bmatrix}\begin{bmatrix}p\\q\end{bmatrix}=\begin{bmatrix}25\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}p^2+q^2\end{bmatrix}=\begin{bmatrix}25\end{bmatrix}
\displaystyle \Rightarrow p^2+q^2=25
\displaystyle \text{The positive integer solutions are }(p,q)=(3,4)\text{ or }(4,3).
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }A\text{ and }B\text{ are any two }2\times2\text{ matrices such that }
\displaystyle AB=BA=B \ \text{ and }B\text{ is not a zero matrix, what can you say about matrix }A?
\displaystyle \text{Answer:}
\displaystyle \text{The given information does not imply that }A\text{ is the identity matrix.}
\displaystyle \text{For example, let }A=\begin{bmatrix}1&0\\0&2\end{bmatrix}\text{ and }B=\begin{bmatrix}1&0\\0&0\end{bmatrix}.
\displaystyle \text{Then }AB=\begin{bmatrix}1&0\\0&0\end{bmatrix}=B\text{ and }BA=\begin{bmatrix}1&0\\0&0\end{bmatrix}=B.
\displaystyle \text{However, }A=\begin{bmatrix}1&0\\0&2\end{bmatrix}\ne\begin{bmatrix}1&0\\0&1\end{bmatrix}.
\displaystyle \therefore \text{The matrix }A\text{ need not be the identity matrix.}
\displaystyle \text{The only conclusion is that }A\text{ acts as the identity on the image of }B.
\displaystyle \\

\displaystyle \textbf{Question 21: }\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 22: }\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 23: }\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 24: }\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 25: }\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 26: }\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 27: }\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 28: }\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 29: }\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 30: }\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 31: }\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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