\displaystyle \textbf{Question 1: } \text{Find} x \text{ and } y \text{, if} \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}.\begin{bmatrix} 2x \\ 1 \end{bmatrix} +2 \begin{bmatrix} -4 \\ 5 \end{bmatrix}=4 \begin{bmatrix} 2 \\ y \end{bmatrix} \hspace{1.0cm} \text{          [ICSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}.\begin{bmatrix} 2x \\ 1 \end{bmatrix} +2 \begin{bmatrix} -4 \\ 5 \end{bmatrix}=4 \begin{bmatrix} 2 \\ y \end{bmatrix}
\displaystyle \begin{bmatrix} 6x-2 \\ -2x+4 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix}= \begin{bmatrix} 8 \\ 4y \end{bmatrix}
Therefore
\displaystyle 6x-10=8 \Rightarrow x = 3
and \displaystyle -2x+14=4y \Rightarrow y = 2
\displaystyle \text{Hence } x = 3 \text{ and } y = 2.

\displaystyle \textbf{Question 2: } \text{Given } \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}
\displaystyle \text{Find: i) the order of matrix }  X \ \ \ \ \ \ \ \ \ \text{ii) the matrix } X  \hspace{1.0cm}  \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}
\displaystyle A_{2 \times 2} . X_{p \times q} = B_{2 \times 1}
Therefore \displaystyle p = 2 \text{ and } q = 1 . Hence the order of Matrix is \displaystyle 2 \times 1
Let \displaystyle X = \begin{bmatrix} x \\ y \end{bmatrix}
Therefore
\displaystyle \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix}
\displaystyle \begin{bmatrix} 2x+y \\ -3x+4y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix}
Therefore
\displaystyle 2x+y = 7
and \displaystyle -3x+4y = 6
Solving we get \displaystyle x = 2 \text{ and } y = 3
\displaystyle \text{Hence } X = \begin{bmatrix} 2 \\ 3 \end{bmatrix}

\displaystyle \textbf{Question 3: } \text{If } \begin{bmatrix} a & 3 \\ 4 & 1 \end{bmatrix}+\begin{bmatrix} 2 & b \\ 1 & -2 \end{bmatrix}-\begin{bmatrix} 1 & 1 \\ -2 & c \end{bmatrix}=\begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}
\displaystyle \text{Find the value of } a, b \text{ and } c \hspace{1.0cm} \text{[ICSE 1981]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix} a & 3 \\ 4 & 1 \end{bmatrix}+\begin{bmatrix} 2 & b \\ 1 & -2 \end{bmatrix}-\begin{bmatrix} 1 & 1 \\ -2 & c \end{bmatrix}=\begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}
\displaystyle \begin{bmatrix} a+2-1 & 3+b-1 \\ 4+1+2 & 1-2-c \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}
\displaystyle \begin{bmatrix} a+1 & b+2 \\ 7 & -c-1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}
Therefore
\displaystyle a+1 = 5 \Rightarrow 4
\displaystyle b+2 = 0 \Rightarrow b = -2
\displaystyle -c-1=3 \Rightarrow c = -4

\displaystyle \textbf{Question 4: } \text{If } A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}  \text{, and } B = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}
\displaystyle \text{Find: i) } A(BA) \ \ \ \ \ \ \ \ \           \text{ii) } (AB) B. \hspace{1.0cm} \text{[ICSE 1991] }
\displaystyle \text{Answer:}
i) \displaystyle A(BA)
\displaystyle = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} [ \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}]
\displaystyle =\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}. \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}
\displaystyle = \begin{bmatrix} 14 & 13 \\ 13 & 14 \end{bmatrix}
ii) \displaystyle (AB) B
\displaystyle = (\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}) . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}
\displaystyle = \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}
\displaystyle = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}

\displaystyle \textbf{Question 5: } \text{Find } x \text{ and } y \text{, if} \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix}. \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \hspace{1.0cm} \text{[ICSE 1992, 2013] }
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix}. \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix}
\displaystyle \begin{bmatrix} 2x + 3x \\ 2y+4y \end{bmatrix}. = \begin{bmatrix} 5 \\ 12 \end{bmatrix}
Therefore
\displaystyle 5x=5 \Rightarrow x = 1
and \displaystyle 6y = 12 \Rightarrow y = 2

\displaystyle \textbf{Question 6: } \text{Given } A = \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \text{, } B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}
\displaystyle \text{Find } X  \text{ such that } A+X=2B+C \hspace{1.0cm} \text{[ICSE 2005]}
\displaystyle \text{Answer:}
\displaystyle A+X=2B+C
\displaystyle \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}+X=2\begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}+ \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}
\displaystyle \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}+X = \begin{bmatrix} -5 & 4 \\ 8 & 2 \end{bmatrix}
\displaystyle X = \begin{bmatrix} -5 & 4 \\ 8 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}
\displaystyle X = \begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix}

\displaystyle \textbf{Question 7: } \text{Find the value of  x  given that } A^2=B \text{, } A = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \text{, and } B = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} \hspace{1.0cm} \text{[ICSE 2005]}
\displaystyle \text{Answer:}
\displaystyle A^2=B
\displaystyle \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} . \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}
\displaystyle \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}
Therefore \displaystyle x = 36

\displaystyle \textbf{Question 8: } \text{If } A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \text{ , and } B = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \text{ and } I \text{matrix of the same order and } A^t \\ \text{is the transpose of the matrix, find } A^t.B+BI \hspace{1.0cm} \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}
\displaystyle A^t = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}
\displaystyle A^t.B+BI
\displaystyle = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} . \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}+ \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} . \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\displaystyle = \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}
\displaystyle = \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}

\displaystyle \textbf{Question 9: } \text{Given } A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \text{, } B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}
\displaystyle \text{Find the matrix } X \text{ such that } A+2X=2B+C \hspace{1.0cm} \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle A+2X=2B+C
\displaystyle \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}+2X=2\begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}+\begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}
\displaystyle \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}+2X= \begin{bmatrix} -6 & 4 \\ 8 & 0 \end{bmatrix}+\begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}
\displaystyle \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}+2X= \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix}
\displaystyle 2X= \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix} -\begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}
\displaystyle X= \frac{1}{2} \begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix}
\displaystyle X= \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix}

\displaystyle \textbf{Question 10: } \text{Let } A = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \text{, } B = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix} \text{ and } C = \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix}
\displaystyle \text{Find } A^2-A+BC \hspace{1.0cm} \text{[ICSE 2006] }
\displaystyle \text{Answer:}
\displaystyle A^2-A+BC
\displaystyle = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}.\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}-\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}+\begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}. \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix}
\displaystyle = \begin{bmatrix} 4 & -2 \\ 6 & 3 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ -4 & 4 \end{bmatrix}
\displaystyle = \begin{bmatrix} 2 & -2 \\ -1 & 11 \end{bmatrix}

\displaystyle \textbf{Question 11: } \text{Let } A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \text{, } B = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}
\displaystyle \text{Find } A^2+AB+B^2 \hspace{1.0cm} \text{[ICSE 2007] }
\displaystyle \text{Answer:}
\displaystyle A^2+AB+B^2
\displaystyle =\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}.\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}+\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}.\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}+\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}.\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}
\displaystyle =\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}+\begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix}+\begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}
\displaystyle =\begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}

\displaystyle \textbf{Question 12: } \text{Given } A = \begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix} \text{, } B = \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \text{ and } \\ BA=C^2 \text{ Find the value of } p \text{ and } q \hspace{1.0cm} \text{[ICSE 2008] }
\displaystyle \text{Answer:}
\displaystyle BA=C^2
\displaystyle \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix}.\begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} .\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix}
\displaystyle \begin{bmatrix} 0 & -2q \\ p & 0 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 8 & 0 \end{bmatrix}
Therefore
\displaystyle -2q = -8 \Rightarrow q = 4
and \displaystyle p = 8

\displaystyle \textbf{Question 13: } \text{Given } A = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix} \text{, } B = \begin{bmatrix} 6 \\ 1 \end{bmatrix} \text{, } C = \begin{bmatrix} -4 \\ 5 \end{bmatrix} \text{ and } D = \begin{bmatrix} 2 \\ 2 \end{bmatrix}
\displaystyle \text{Find } AB+2C-4D \hspace{1.0cm} \text{[ICSE 2010] }
\displaystyle \text{Answer:}
\displaystyle AB+2C-4D
\displaystyle = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}. \begin{bmatrix} 6 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} -4 \\ 5 \end{bmatrix} - 4 \begin{bmatrix} 2 \\ 2 \end{bmatrix}
\displaystyle = \begin{bmatrix} 16 \\ -2 \end{bmatrix}+\begin{bmatrix} -8 \\ 10 \end{bmatrix}- \begin{bmatrix} 8 \\ 8 \end{bmatrix}
\displaystyle = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

\displaystyle \textbf{Question 14: } \text{Evaluate } \begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \hspace{1.0cm} \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}
\displaystyle = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}
\displaystyle = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}

\displaystyle \textbf{Question 15: } \text{If } A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \text{, } I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\displaystyle \text{Find } A^2-5A+7I \hspace{1.0cm} \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle A^2-5A+7I
\displaystyle = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}.\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}-5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}+7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\displaystyle = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}
\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

\displaystyle \textbf{Question 16: } \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 \hspace{1.0cm} \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} = \begin{bmatrix} -23 & 3 \\ 17 & 14 \end{bmatrix}

\displaystyle \textbf{Question 17: } \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}
\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 \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} \hspace{1.0cm} \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}
Therefore
\displaystyle 2x+5y = -7
\displaystyle 5x+2y=14
Solving the above two equations we get
\displaystyle x = 4 \text{ and } 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}
Therefore
\displaystyle x-y-8=-7 \Rightarrow x-y=1
Also \displaystyle -2y-8=-11 \Rightarrow y = \frac{3}{2}
Substituting we get \displaystyle x = \frac{3}{2}+1 = \frac{5}{2}
\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 \begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}
Therefore \displaystyle y = -2 and x = \frac{6}{2}=3

\displaystyle \textbf{Question 18: } \text{If } A = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 \\ 4 \end{bmatrix}, \text{is the product of } AB \text{ possible } \text{[ICSE 2011]}
\displaystyle \text{Answer:}
The order of matrix \displaystyle A = 2 \times 2 and the order of matrix \displaystyle B \ is \ 2 \times 1
Since the number of columns in \displaystyle A is equal to the number of rows in \displaystyle B , the product \displaystyle AB is possible.
\displaystyle AB = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} . \begin{bmatrix} 4 \\ 2 \end{bmatrix} = \begin{bmatrix} 6+20 \\ 8-8 \end{bmatrix} = \begin{bmatrix} 26 \\ 0 \end{bmatrix}

\displaystyle \textbf{Question 19: }\text{Find }x\text{ and }y\text{ if }\begin{bmatrix}-3&2\\0&-5\end{bmatrix}\begin{bmatrix}x\\2\end{bmatrix}=\begin{bmatrix}-5\\y\end{bmatrix}.\ \text{[ICSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}-3&2\\0&-5\end{bmatrix}\begin{bmatrix}x\\2\end{bmatrix}=\begin{bmatrix}-5\\y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-3x+2\times2\\0\times x-5\times2\end{bmatrix}=\begin{bmatrix}-5\\y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-3x+4\\-10\end{bmatrix}=\begin{bmatrix}-5\\y\end{bmatrix}
\displaystyle \text{Equating corresponding elements, }-3x+4=-5\text{ and }-10=y.
\displaystyle -3x=-9\Rightarrow x=3\text{ and }y=-10.

\displaystyle \textbf{Question 20: }\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},\text{ then find }x\text{ and }y.\ \text{[ICSE 2007]}
\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}2\times3&2\times4\\2\times5&2\times 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}6+1&8+y\\10+0&2x+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{As above two matrices are equal, their corresponding elements are also equal.}
\displaystyle 8+y=0\text{ and }2x+1=5
\displaystyle \Rightarrow y=-8\text{ and }2x=5-1
\displaystyle \Rightarrow 2x=4\Rightarrow x=2
\displaystyle \therefore y=-8\text{ and }x=2

\displaystyle \textbf{Question 21: }\text{If }\begin{bmatrix}1&4\\-2&3\end{bmatrix}+2M=3\begin{bmatrix}3&2\\0&-3\end{bmatrix},\text{ then find matrix }M.\ \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}1&4\\-2&3\end{bmatrix}+2M=3\begin{bmatrix}3&2\\0&-3\end{bmatrix}
\displaystyle \therefore 2M=3\begin{bmatrix}3&2\\0&-3\end{bmatrix}-\begin{bmatrix}1&4\\-2&3\end{bmatrix}
\displaystyle =\begin{bmatrix}3\times3&3\times2\\3\times0&3\times(-3)\end{bmatrix}-\begin{bmatrix}1&4\\-2&3\end{bmatrix}
\displaystyle \left[\text{in first matrix, each element of a matrix multiply by }3\right]
\displaystyle =\begin{bmatrix}9&6\\0&-9\end{bmatrix}-\begin{bmatrix}1&4\\-2&3\end{bmatrix}=\begin{bmatrix}9-1&6-4\\0-(-2)&-9-3\end{bmatrix}
\displaystyle \Rightarrow 2M=\begin{bmatrix}8&2\\2&-12\end{bmatrix}
\displaystyle \Rightarrow M=\frac{1}{2}\begin{bmatrix}8&2\\2&-12\end{bmatrix}\qquad[\text{dividing both sides by }2]
\displaystyle =\begin{bmatrix}\frac{8}{2}&\frac{2}{2}\\\frac{2}{2}&\frac{-12}{2}\end{bmatrix}\qquad\left[\text{multiplying each element of matrix by }\frac{1}{2}\right]
\displaystyle \therefore M=\begin{bmatrix}4&1\\1&-6\end{bmatrix}

\displaystyle \textbf{Question 22: }\text{Given }A=\begin{bmatrix}p&0\\0&2\end{bmatrix},\ B=\begin{bmatrix}0&-q\\1&0\end{bmatrix},\ C=\begin{bmatrix}2&-2\\2&2\end{bmatrix}\text{ and } \\ BA=C^2,  \text{ find the values of }p\text{ and }q.\ \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}p&0\\0&2\end{bmatrix},\ B=\begin{bmatrix}0&-q\\1&0\end{bmatrix}\text{ and }C=\begin{bmatrix}2&-2\\2&2\end{bmatrix}
\displaystyle \text{Also, given }BA=C^2\qquad\cdots\text{(i)}
\displaystyle \text{Here, }BA=\begin{bmatrix}0&-q\\1&0\end{bmatrix}\begin{bmatrix}p&0\\0&2\end{bmatrix}
\displaystyle =\begin{bmatrix}0\times p+(-q)\times0&0\times0+(-q)\times2\\1\times p+0\times0&1\times0+0\times2\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}0&-2q\\p&0\end{bmatrix}
\displaystyle \text{and }C^2=\begin{bmatrix}2&-2\\2&2\end{bmatrix}\begin{bmatrix}2&-2\\2&2\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times2+(-2)\times2&2\times(-2)+(-2)\times2\\2\times2+2\times2&2\times(-2)+2\times2\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}4-4&-4-4\\4+4&-4+4\end{bmatrix}=\begin{bmatrix}0&-8\\8&0\end{bmatrix}
\displaystyle \text{According to the question, }BA=C^2
\displaystyle \Rightarrow \begin{bmatrix}0&-2q\\p&0\end{bmatrix}=\begin{bmatrix}0&-8\\8&0\end{bmatrix}
\displaystyle \text{As above two matrices are equal, their corresponding elements are also equal.}
\displaystyle -2q=-8\Rightarrow q=\frac{-8}{-2}\Rightarrow q=4
\displaystyle \text{and }p=8
\displaystyle \text{Hence, the values of }p\text{ and }q\text{ are }8\text{ and }4,\text{ respectively.}

\displaystyle \textbf{Question 23: }\text{Find }x\text{ and }y,\text{ if }\begin{bmatrix}2x&x\\y&3y\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}=\begin{bmatrix}16\\9\end{bmatrix}.\ \text{[ICSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}2x&x\\y&3y\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}=\begin{bmatrix}16\\9\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6x+2x\\3y+6y\end{bmatrix}=\begin{bmatrix}16\\9\end{bmatrix}\Rightarrow \begin{bmatrix}8x\\9y\end{bmatrix}=\begin{bmatrix}16\\9\end{bmatrix}
\displaystyle \text{As above two matrices are equal, their corresponding elements are also equal.}
\displaystyle 8x=16\Rightarrow x=\frac{16}{8}=2\text{ and }9y=9\Rightarrow y=\frac{9}{9}=1
\displaystyle \therefore x=2\text{ and }y=1

\displaystyle \textbf{Question 24: }\text{Given }A=\begin{bmatrix}3&-2\\-1&4\end{bmatrix},\ B=\begin{bmatrix}6\\1\end{bmatrix},\ C=\begin{bmatrix}-4\\5\end{bmatrix}\text{ and }D=\begin{bmatrix}2\\2\end{bmatrix}.
\displaystyle \text{Find }AB+2C-4D.\ \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&-2\\-1&4\end{bmatrix},\ B=\begin{bmatrix}6\\1\end{bmatrix},\ C=\begin{bmatrix}-4\\5\end{bmatrix},\ D=\begin{bmatrix}2\\2\end{bmatrix}
\displaystyle \text{We have to find }AB+2C-4D
\displaystyle \text{Here, }AB=\begin{bmatrix}3&-2\\-1&4\end{bmatrix}\begin{bmatrix}6\\1\end{bmatrix}=\begin{bmatrix}3\times6-2\times1\\-1\times6+4\times1\end{bmatrix}
\displaystyle \qquad\qquad\qquad\qquad\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}18-2\\-6+4\end{bmatrix}=\begin{bmatrix}16\\-2\end{bmatrix}
\displaystyle AB+2C-4D=\begin{bmatrix}16\\-2\end{bmatrix}+2\begin{bmatrix}-4\\5\end{bmatrix}-4\begin{bmatrix}2\\2\end{bmatrix}
\displaystyle =\begin{bmatrix}16\\-2\end{bmatrix}+\begin{bmatrix}-8\\10\end{bmatrix}-\begin{bmatrix}8\\8\end{bmatrix}
\displaystyle =\begin{bmatrix}16-8-8\\-2+10-8\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}

\displaystyle \textbf{Question 25: }\text{Evaluate }\begin{bmatrix}4\sin30^\circ&2\cos60^\circ\\\sin90^\circ&2\cos0^\circ\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}.\ \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Evaluate }\begin{bmatrix}4\sin30^\circ&2\cos60^\circ\\\sin90^\circ&2\cos0^\circ\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}
\displaystyle =\begin{bmatrix}4\times\frac{1}{2}&2\times\frac{1}{2}\\1&2\times1\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}
\displaystyle \left[\because\ \sin30^\circ=\frac{1}{2},\ \cos60^\circ=\frac{1}{2},\ \sin90^\circ=1,\ \cos0^\circ=1\right]
\displaystyle =\begin{bmatrix}2&1\\1&2\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times4+1\times5&2\times5+1\times4\\1\times4+2\times5&1\times5+2\times4\end{bmatrix}
\displaystyle \qquad\qquad\qquad\qquad\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}8+5&10+4\\4+10&5+8\end{bmatrix}=\begin{bmatrix}13&14\\14&13\end{bmatrix}

\displaystyle \textbf{Question 26: }\text{If }A=\begin{bmatrix}3&5\\4&-2\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\4\end{bmatrix},\text{ is the product }AB\text{ possible?}
\displaystyle \text{Give a reason. If yes, find it. [ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&5\\4&-2\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\4\end{bmatrix}
\displaystyle \text{Since, matrix }A\text{ has two rows and two columns, therefore order of }A\text{ is }2\times2
\displaystyle \text{and matrix }B\text{ has two rows and one column, therefore order of }B\text{ is }2\times1
\displaystyle \text{Now, }A_{2\times2}\cdot B_{2\times1}\text{ is possible because number of columns in }A
\displaystyle \text{is equal to the number of rows in }B.\text{ Hence, product of }A\text{ and }B\text{ is possible.}
\displaystyle \text{Now, }AB=\begin{bmatrix}3&5\\4&-2\end{bmatrix}\begin{bmatrix}2\\4\end{bmatrix}=\begin{bmatrix}3\times2+5\times4\\4\times2+(-2)\times4\end{bmatrix}
\displaystyle \qquad\qquad\qquad\qquad\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}6+20\\8-8\end{bmatrix}=\begin{bmatrix}26\\0\end{bmatrix}

\displaystyle \textbf{Question 27: }\text{If }A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ then find }A^2-5A+7I.\ \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ we have to find }A^2-5A+7I
\displaystyle \text{Here, }A^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}3\times3+1\times(-1)&3\times1+1\times2\\(-1)\times3+2\times(-1)&(-1)\times1+2\times2\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}
\displaystyle \therefore A^2-5A+7I=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}8&5\\-5&3\end{bmatrix}+\begin{bmatrix}(-5)\times3&(-5)\times1\\(-5)\times(-1)&(-5)\times2\end{bmatrix}+\begin{bmatrix}7\times1&7\times0\\7\times0&7\times1\end{bmatrix}
\displaystyle =\begin{bmatrix}8&5\\-5&3\end{bmatrix}+\begin{bmatrix}-15&-5\\5&-10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}
\displaystyle =\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}

\displaystyle \textbf{Question 28: }\text{Given }\begin{bmatrix}2&1\\-3&4\end{bmatrix}X=\begin{bmatrix}7\\6\end{bmatrix},\text{ write (i) the order of }X \\ \text{ (ii) the matrix }X.\ \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }\begin{bmatrix}2&1\\-3&4\end{bmatrix}X=\begin{bmatrix}7\\6\end{bmatrix}\qquad\cdots\text{(A)}
\displaystyle \text{The right side above matrix is of order }2\times1
\displaystyle \text{Therefore, the matrix on left side should be of order }2\times1
\displaystyle \text{Left side is the product of two matrices. One is of order }2\times2,\text{ therefore the other matrix }X
\displaystyle \text{should be of order }2\times1
\displaystyle \text{Hence, order of matrix }X\text{ is }2\times1
\displaystyle \text{(ii) Let }X=\begin{bmatrix}x_1\\x_2\end{bmatrix}
\displaystyle \text{From Eq. (A), we get }\begin{bmatrix}2&1\\-3&4\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x_1+1x_2\\-3x_1+4x_2\end{bmatrix}=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{cases}2x_1+x_2=7&\cdots\text{(B)}\\-3x_1+4x_2=6&\cdots\text{(C)}\end{cases}
\displaystyle \text{On multiplying Eq. (B) by }4\text{ and then subtracting Eq. (C) from Eq. (B), we get}
\displaystyle 8x_1+4x_2=28
\displaystyle -(-3x_1+4x_2)= -6
\displaystyle 11x_1=22\Rightarrow x_1=\frac{22}{11}\Rightarrow x_1=2
\displaystyle \text{On putting }x_1=2\text{ in Eq. (B), we get }2(2)+x_2=7\Rightarrow x_2=7-4\Rightarrow x_2=3
\displaystyle \therefore X=\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}

\displaystyle \textbf{Question 29: }\text{Given }A=\begin{bmatrix}2&-6\\2&0\end{bmatrix},\ B=\begin{bmatrix}-3&2\\4&0\end{bmatrix},\ C=\begin{bmatrix}4&0\\0&2\end{bmatrix}.
\displaystyle \text{Find matrix }X\text{ such that }A+2X=2B+C.\ \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&-6\\2&0\end{bmatrix},\ B=\begin{bmatrix}-3&2\\4&0\end{bmatrix}\text{ and }C=\begin{bmatrix}4&0\\0&2\end{bmatrix}
\displaystyle \text{Also, given }A+2X=2B+C
\displaystyle \therefore \begin{bmatrix}2&-6\\2&0\end{bmatrix}+2X=2\begin{bmatrix}-3&2\\4&0\end{bmatrix}+\begin{bmatrix}4&0\\0&2\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}-3\times2&2\times2\\4\times2&0\times2\end{bmatrix}+\begin{bmatrix}4&0\\0&2\end{bmatrix}-\begin{bmatrix}2&-6\\2&0\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}-6&4\\8&0\end{bmatrix}+\begin{bmatrix}4&0\\0&2\end{bmatrix}-\begin{bmatrix}2&-6\\2&0\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}-6+4-2&4+0+6\\8+0-2&0+2-0\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}-4&10\\6&2\end{bmatrix}
\displaystyle \text{On dividing both sides by }2,\text{ we get}
\displaystyle X=\frac{1}{2}\begin{bmatrix}-4&10\\6&2\end{bmatrix}=\begin{bmatrix}\frac{-4}{2}&\frac{10}{2}\\\frac{6}{2}&\frac{2}{2}\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-2&5\\3&1\end{bmatrix}\qquad\left[\text{multiplying each element of matrix by }\frac{1}{2}\right]

\displaystyle \textbf{Question 30: }\text{Find }x\text{ and }y,\text{ if }\begin{bmatrix}x&3x\\y&4y\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}.\ \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}x&3x\\y&4y\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle \Rightarrow \begin{bmatrix}x\times2+3x\times1\\y\times2+4y\times1\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+3x\\2y+4y\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}\Rightarrow \begin{bmatrix}5x\\6y\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle \text{As two matrices are equal, their corresponding elements are also equal.}
\displaystyle 5x=5\text{ and }6y=12\Rightarrow x=1\text{ and }y=2

\displaystyle \textbf{Question 31: }\text{Find }x\text{ and }y,\text{ if }\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}.\ \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \textbf{Given, }\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\times(-1)+0\times2x\\3\times(-1)+1\times2x\end{bmatrix}+\begin{bmatrix}-6\\3\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2\\-3+2x\end{bmatrix}+\begin{bmatrix}-6\\3\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}\qquad[\text{adding corresponding elements}]
\displaystyle \Rightarrow \begin{bmatrix}-4\\2x\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle \text{As two matrices are equal, equating corresponding elements are also equal}
\displaystyle -4=2y\text{ and }2x=6
\displaystyle \Rightarrow y=\frac{-4}{2}\text{ and }x=\frac{6}{2}
\displaystyle \therefore y=-2\text{ and }x=3

\displaystyle \textbf{Question 32: }\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{ then find }A^2+AC-5B.\ \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }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{We have to find }A^2+AC-5B
\displaystyle \text{Here, }A^2=\begin{bmatrix}2&1\\0&-2\end{bmatrix}\begin{bmatrix}2&1\\0&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times2+1\times0&2\times1+1\times(-2)\\0\times2+(-2)\times0&0\times1+(-2)\times(-2)\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}4+0&2-2\\0-0&0+4\end{bmatrix}=\begin{bmatrix}4&0\\0&4\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle AC=\begin{bmatrix}2&1\\0&-2\end{bmatrix}\begin{bmatrix}-3&2\\-1&4\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times(-3)+1\times(-1)&2\times2+1\times4\\0\times(-3)+(-2)\times(-1)&0\times2+(-2)\times4\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}-6-1&4+4\\0+2&0-8\end{bmatrix}=\begin{bmatrix}-7&8\\2&-8\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle 5B=5\begin{bmatrix}4&1\\-3&-2\end{bmatrix}=\begin{bmatrix}5\times4&5\times1\\5\times(-3)&5\times(-2)\end{bmatrix}\qquad[\text{multiplying each element of matrix by }5]
\displaystyle =\begin{bmatrix}20&5\\-15&-10\end{bmatrix}\qquad\cdots\text{(iii)}
\displaystyle \text{Now, }A^2+AC-5B=\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}4-7-20&0+8-5\\0+2-(-15)&4-8-(-10)\end{bmatrix}=\begin{bmatrix}-23&3\\17&6\end{bmatrix}

\displaystyle \textbf{Question 33: }\text{If }A=\begin{bmatrix}3&x\\0&1\end{bmatrix}\text{ and }B=\begin{bmatrix}9&16\\0&-y\end{bmatrix},\text{ find }x\text{ and }y\text{ when } \\ A^2=B.\ \text{[ICSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&x\\0&1\end{bmatrix}\text{ and }B=\begin{bmatrix}9&16\\0&-y\end{bmatrix}
\displaystyle \text{Also, it is given }A^2=B\qquad\cdots\text{(i)}
\displaystyle \text{Here, }A^2=A\cdot A=\begin{bmatrix}3&x\\0&1\end{bmatrix}\begin{bmatrix}3&x\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}9+0&3x+x\\0+0&0+1\end{bmatrix}=\begin{bmatrix}9&4x\\0&1\end{bmatrix}
\displaystyle \text{From Eq. (i), }A^2=B
\displaystyle \Rightarrow \begin{bmatrix}9&4x\\0&1\end{bmatrix}=\begin{bmatrix}9&16\\0&-y\end{bmatrix}
\displaystyle \text{By comparing the corresponding elements, we get }4x=16\text{ and }1=-y
\displaystyle \Rightarrow x=\frac{16}{4}\text{ and }y=-1
\displaystyle \therefore x=4\text{ and }y=-1

\displaystyle \textbf{Question 34: }\text{If }A=\begin{bmatrix}3&7\\2&4\end{bmatrix},\ B=\begin{bmatrix}0&2\\5&3\end{bmatrix}\text{ and }C=\begin{bmatrix}1&-5\\-4&6\end{bmatrix},\text{ find } \\ AB-5C.\ \text{[ICSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&7\\2&4\end{bmatrix},\ B=\begin{bmatrix}0&2\\5&3\end{bmatrix}\text{ and }C=\begin{bmatrix}1&-5\\-4&6\end{bmatrix}
\displaystyle \text{Here, }AB=\begin{bmatrix}3&7\\2&4\end{bmatrix}\begin{bmatrix}0&2\\5&3\end{bmatrix}=\begin{bmatrix}0+35&6+21\\0+20&4+12\end{bmatrix}=\begin{bmatrix}35&27\\20&16\end{bmatrix}
\displaystyle \text{and }5C=5\begin{bmatrix}1&-5\\-4&6\end{bmatrix}=\begin{bmatrix}5&-25\\-20&30\end{bmatrix}
\displaystyle \therefore AB-5C=\begin{bmatrix}35&27\\20&16\end{bmatrix}-\begin{bmatrix}5&-25\\-20&30\end{bmatrix}=\begin{bmatrix}30&52\\40&-14\end{bmatrix}

\displaystyle \textbf{Question 35: }\text{Given }A=\begin{bmatrix}4\sin30^\circ&\cos0^\circ\\\cos0^\circ&4\sin30^\circ\end{bmatrix}\text{ and }B=\begin{bmatrix}4\\5\end{bmatrix}.
\displaystyle \text{If }AX=B,\text{ then (i) write the order of matrix }X\text{ (ii) find matrix }X.\ \text{[ICSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }A=\begin{bmatrix}4\sin30^\circ&\cos0^\circ\\\cos0^\circ&4\sin30^\circ\end{bmatrix}_{2\times2}
\displaystyle \text{and }B=\begin{bmatrix}4\\5\end{bmatrix}_{2\times1}
\displaystyle \text{It is given that, }AX=B
\displaystyle \text{Since, product of matrices }A\text{ and }X\text{ are possible,}
\displaystyle \text{Number of columns of }A=\text{Number of rows of }X
\displaystyle \Rightarrow \text{Rows in matrix }X\text{ is }2
\displaystyle \text{Since, }AX\text{ results a matrix }B\text{ which have order }2\times1
\displaystyle \text{Matrix }X\text{ will have same number of columns as matrix }B
\displaystyle \Rightarrow \text{Number of columns in matrix }X=1
\displaystyle \therefore \text{Order of matrix }X\text{ is }2\times1
\displaystyle \text{(ii) We have, }A=\begin{bmatrix}4\sin30^\circ&\cos0^\circ\\\cos0^\circ&4\sin30^\circ\end{bmatrix}
\displaystyle =\begin{bmatrix}2&1\\1&2\end{bmatrix}\qquad\left[\because\ \sin30^\circ=\frac{1}{2}\text{ and }\cos0^\circ=1\right]
\displaystyle \text{Let }X=\begin{bmatrix}x_1\\x_2\end{bmatrix}_{2\times1}
\displaystyle \text{Now, }AX=B
\displaystyle \Rightarrow \begin{bmatrix}2&1\\1&2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}4\\5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x_1+x_2\\x_1+2x_2\end{bmatrix}=\begin{bmatrix}4\\5\end{bmatrix}
\displaystyle \text{On equating the corresponding elements both sides, we get}
\displaystyle 2x_1+x_2=4\qquad\cdots\text{(i)}
\displaystyle x_1+2x_2=5\qquad\cdots\text{(ii)}
\displaystyle \text{On solving Eqs. (i) and (ii), we get }x_1=1\text{ and }x_2=2
\displaystyle \therefore X=\begin{bmatrix}1\\2\end{bmatrix}

\displaystyle \textbf{Question 36: }\text{Given }A=\begin{bmatrix}2&0\\-1&7\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\text{ and }A^2=9A+mI,
\displaystyle \text{ find }m.\ \text{[ICSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&0\\-1&7\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \text{Also, it is given that }A^2=9A+mI
\displaystyle \text{Here, }A^2=A\cdot A=\begin{bmatrix}2&0\\-1&7\end{bmatrix}\begin{bmatrix}2&0\\-1&7\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times2+0\times(-1)&2\times0+0\times7\\(-1)\times2+7\times(-1)&(-1)\times0+7\times7\end{bmatrix}
\displaystyle =\begin{bmatrix}4&0\\-9&49\end{bmatrix}
\displaystyle A^2=9A+mI\Rightarrow \begin{bmatrix}4&0\\-9&49\end{bmatrix}=9\begin{bmatrix}2&0\\-1&7\end{bmatrix}+m\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4&0\\-9&49\end{bmatrix}=\begin{bmatrix}18&0\\-9&63\end{bmatrix}+\begin{bmatrix}m&0\\0&m\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4&0\\-9&49\end{bmatrix}=\begin{bmatrix}18+m&0\\-9&63+m\end{bmatrix}
\displaystyle \Rightarrow 18+m=4\qquad\cdots\text{(i)}
\displaystyle \text{and }63+m=49\qquad\cdots\text{(ii)}
\displaystyle \text{From Eq. (i), }m=4-18=-14
\displaystyle \text{From Eq. (ii), }m=49-63=-14
\displaystyle \therefore m=-14

\displaystyle \textbf{Question 37: }\text{If }A=\begin{bmatrix}1&3\\3&4\end{bmatrix}\text{ and }B=\begin{bmatrix}-2&1\\-3&2\end{bmatrix}\text{ and }A^2-5B^2=5C, \\ \text{ find matrix }C,\text{ where }C\text{ is }2\times2.\ \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}1&3\\3&4\end{bmatrix}\text{ and }B=\begin{bmatrix}-2&1\\-3&2\end{bmatrix}
\displaystyle \text{Also, we have }A^2-5B^2=5C
\displaystyle \therefore C=\frac{1}{5}A^2-B^2
\displaystyle \text{Now, }A^2=AA=\begin{bmatrix}1&3\\3&4\end{bmatrix}\begin{bmatrix}1&3\\3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}1+9&3+12\\3+12&9+16\end{bmatrix}=\begin{bmatrix}10&15\\15&25\end{bmatrix}
\displaystyle \therefore \frac{1}{5}A^2=\frac{1}{5}\begin{bmatrix}10&15\\15&25\end{bmatrix}=\begin{bmatrix}2&3\\3&5\end{bmatrix}
\displaystyle \text{and }B^2=BB=\begin{bmatrix}-2&1\\-3&2\end{bmatrix}\begin{bmatrix}-2&1\\-3&2\end{bmatrix}
\displaystyle =\begin{bmatrix}4-3&-2+2\\6-6&-3+4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \therefore C=\frac{1}{5}A^2-B^2=\begin{bmatrix}2&3\\3&5\end{bmatrix}-\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}2-1&3-0\\3-0&5-1\end{bmatrix}=\begin{bmatrix}1&3\\3&4\end{bmatrix}

\displaystyle \textbf{Question 38: }\text{Given }B=\begin{bmatrix}1&1\\8&3\end{bmatrix},\text{ find matrix }X\text{ if }X=B^2-4B.
\displaystyle \text{Hence, solve for }a\text{ and }b\text{ given }X\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}5\\50\end{bmatrix}.\ \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \textbf{Given, }B=\begin{bmatrix}1&1\\8&3\end{bmatrix}
\displaystyle \text{Also, we have }X=B^2-4B
\displaystyle \text{Here, }B^2=BB=\begin{bmatrix}1&1\\8&3\end{bmatrix}\begin{bmatrix}1&1\\8&3\end{bmatrix}
\displaystyle =\begin{bmatrix}1+8&1+3\\8+24&8+9\end{bmatrix}=\begin{bmatrix}9&4\\32&17\end{bmatrix}
\displaystyle \text{and }4B=\begin{bmatrix}4\times1&4\times1\\4\times8&4\times3\end{bmatrix}=\begin{bmatrix}4&4\\32&12\end{bmatrix}
\displaystyle \therefore X=B^2-4B=\begin{bmatrix}9&4\\32&17\end{bmatrix}-\begin{bmatrix}4&4\\32&12\end{bmatrix}
\displaystyle =\begin{bmatrix}9-4&4-4\\32-32&17-12\end{bmatrix}=\begin{bmatrix}5&0\\0&5\end{bmatrix}
\displaystyle \text{Given, }X\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}5\\50\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}5&0\\0&5\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}5\\50\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}5a+0\\0+5b\end{bmatrix}=\begin{bmatrix}5\\50\end{bmatrix}
\displaystyle \Rightarrow 5a=5\Rightarrow a=1\text{ and }5b=50\Rightarrow b=10

\displaystyle \text{Question 39: }\text{If }A=\begin{bmatrix}4&4\\-2&6\end{bmatrix}\text{ and }B=\begin{bmatrix}2&1\\3&-2\end{bmatrix},\text{ find matrix }D\text{ such that } \\ 3A-2B+2D=0.\ \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \because 3A-2B+2D=0
\displaystyle \Rightarrow 2D=2B-3A\Rightarrow D=B-\frac{3}{2}A\qquad\cdots\text{(i)}
\displaystyle \text{Also, given }A=\begin{bmatrix}4&4\\-2&6\end{bmatrix}\text{ and }B=\begin{bmatrix}2&1\\3&-2\end{bmatrix}
\displaystyle \Rightarrow -\frac{3}{2}A=\begin{bmatrix}-6&-6\\3&-9\end{bmatrix}
\displaystyle \text{Now, from Eq.(i), }D=\begin{bmatrix}2&1\\3&-2\end{bmatrix}+\begin{bmatrix}-6&-6\\3&-9\end{bmatrix}
\displaystyle \Rightarrow D=\begin{bmatrix}-4&-5\\6&-11\end{bmatrix}

\displaystyle \textbf{Question 40: }\text{If }\begin{bmatrix}2&4\\6&2\end{bmatrix}\begin{bmatrix}3x\\2\end{bmatrix}+2\begin{bmatrix}3\\4\end{bmatrix}=5\begin{bmatrix}4\\y\end{bmatrix},\text{ find }x\text{ and }y.\ \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}2&4\\6&2\end{bmatrix}\begin{bmatrix}3x\\2\end{bmatrix}+2\begin{bmatrix}3\\4\end{bmatrix}=5\begin{bmatrix}4\\y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6x+8\\18x+4\end{bmatrix}+\begin{bmatrix}6\\8\end{bmatrix}=\begin{bmatrix}20\\5y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6x+14\\18x+12\end{bmatrix}=\begin{bmatrix}20\\5y\end{bmatrix}
\displaystyle \text{On comparing both sides, we get }6x+14=20\qquad\cdots\text{(i)}
\displaystyle \text{and }18x+12=5y\qquad\cdots\text{(ii)}
\displaystyle \text{From Eq.(i), }6x=20-14\Rightarrow x=1
\displaystyle \text{From Eq.(ii), }18\times1+12=5y\Rightarrow 5y=30\Rightarrow y=6

\displaystyle \textbf{Question 41: }\text{Find the values of }x\text{ and }y,\text{ if } \\ 2\begin{bmatrix}x&7\\9&y-5\end{bmatrix}+\begin{bmatrix}6&-7\\4&5\end{bmatrix}=\begin{bmatrix}10&7\\22&15\end{bmatrix}.\ \text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Given that, }2\begin{bmatrix}x&7\\9&y-5\end{bmatrix}+\begin{bmatrix}6&-7\\4&5\end{bmatrix}=\begin{bmatrix}10&7\\22&15\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x&14\\18&2y-10\end{bmatrix}+\begin{bmatrix}6&-7\\4&5\end{bmatrix}=\begin{bmatrix}10&7\\22&15\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+6&7\\22&2y-5\end{bmatrix}=\begin{bmatrix}10&7\\22&15\end{bmatrix}
\displaystyle \text{Here, both the matrices are equal, so on equating corresponding elements,}
\displaystyle 2x+6=10\Rightarrow 2x=10-6\Rightarrow 2x=4\Rightarrow x=2
\displaystyle \text{and }2y-5=15\Rightarrow 2y=15+5\Rightarrow 2y=20\Rightarrow y=10
\displaystyle \text{Hence, }x=2\text{ and }y=10.

\displaystyle \textbf{Question 42: }\text{If }A=\begin{bmatrix}2&3\\5&7\end{bmatrix},\ B=\begin{bmatrix}0&4\\-1&7\end{bmatrix},\ C=\begin{bmatrix}1&0\\-1&4\end{bmatrix},\text{ find } \\ AC+B^2-10C.\ \text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&3\\5&7\end{bmatrix},\ B=\begin{bmatrix}0&4\\-1&7\end{bmatrix}\text{ and }C=\begin{bmatrix}1&0\\-1&4\end{bmatrix}
\displaystyle AC=\begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&0\\-1&4\end{bmatrix}=\begin{bmatrix}2-3&0+12\\5-7&0+28\end{bmatrix}=\begin{bmatrix}-1&12\\-2&28\end{bmatrix}
\displaystyle B^2=\begin{bmatrix}0&4\\-1&7\end{bmatrix}\begin{bmatrix}0&4\\-1&7\end{bmatrix}=\begin{bmatrix}-4&28\\-7&45\end{bmatrix}
\displaystyle \text{Now, }AC+B^2-10C=\begin{bmatrix}-1&12\\-2&28\end{bmatrix}+\begin{bmatrix}-4&28\\-7&45\end{bmatrix}-10\begin{bmatrix}1&0\\-1&4\end{bmatrix}
\displaystyle =\begin{bmatrix}-5&40\\-9&73\end{bmatrix}-\begin{bmatrix}10&0\\-10&40\end{bmatrix}=\begin{bmatrix}-15&40\\1&33\end{bmatrix}

\displaystyle \textbf{Question 43: }\text{Simplify }\sin A\begin{bmatrix}\sin A&-\cos A\\\cos A&\sin A\end{bmatrix}+\cos A\begin{bmatrix}\cos A&\sin A\\-\sin A&\cos A\end{bmatrix}.\ \text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \sin A\begin{bmatrix}\sin A&-\cos A\\\cos A&\sin A\end{bmatrix}+\cos A\begin{bmatrix}\cos A&\sin A\\-\sin A&\cos A\end{bmatrix}
\displaystyle =\begin{bmatrix}\sin^2A&-\sin A\cos A\\\sin A\cos A&\sin^2A\end{bmatrix}+\begin{bmatrix}\cos^2A&\cos A\sin A\\-\sin A\cos A&\cos^2A\end{bmatrix}
\displaystyle =\begin{bmatrix}\sin^2A+\cos^2A&-\sin A\cos A+\cos A\sin A\\\sin A\cos A-\sin A\cos A&\sin^2A+\cos^2A\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0\\0&1\end{bmatrix}\qquad\left[\because\ \cos^2\theta+\sin^2\theta=1\right]

\displaystyle \textbf{Question 44: }\text{Given }\begin{bmatrix}4&2\\-1&1\end{bmatrix}M=6I,\text{ where }M\text{ is a matrix and }I\text{ is the unit}
\displaystyle \text{matrix of order }2\times2,\text{ (i) state the order of matrix }M\text{ (ii) find matrix }M.\ \text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\begin{bmatrix}4&2\\-1&1\end{bmatrix}M=6I\qquad\cdots\text{(A)}
\displaystyle \text{(i) Since, }\begin{bmatrix}4&2\\-1&1\end{bmatrix}\text{ is of order }2\times2\text{ and right side of the equation is}
\displaystyle \text{also of order }2\times2,\text{ so the matrix }M\text{ should be of order }2\times2
\displaystyle \text{(ii) Let }M=\begin{bmatrix}a&b\\c&d\end{bmatrix}
\displaystyle \text{From Eq. (A), we get }\begin{bmatrix}4&2\\-1&1\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=6\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4a+2c&4b+2d\\-a+c&-b+d\end{bmatrix}=\begin{bmatrix}6&0\\0&6\end{bmatrix}
\displaystyle \text{On equating both sides, we get }4a+2c=6\qquad\cdots\text{(B)}
\displaystyle 4b+2d=0\qquad\cdots\text{(C)}
\displaystyle -a+c=0\qquad\cdots\text{(D)}
\displaystyle -b+d=6\qquad\cdots\text{(E)}
\displaystyle \text{On solving Eqs. (B) and (D), we get }a=1\text{ and }c=1
\displaystyle \text{Now, on solving (C) and (E), we get }b=-2\text{ and }d=4
\displaystyle \therefore \text{Matrix }M=\begin{bmatrix}1&-2\\1&4\end{bmatrix}

\displaystyle \textbf{Question 45: }\text{If }A=\begin{bmatrix}3&0\\5&1\end{bmatrix}\text{ and }B=\begin{bmatrix}-4&2\\1&0\end{bmatrix},\text{ find }A^2-2AB+B^2.\ \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that, }A^2-2AB+B^2=(A-B)^2
\displaystyle A=\begin{bmatrix}3&0\\5&1\end{bmatrix},\ B=\begin{bmatrix}-4&2\\1&0\end{bmatrix}
\displaystyle A-B=\begin{bmatrix}3-(-4)&0-2\\5-1&1-0\end{bmatrix}=\begin{bmatrix}7&-2\\4&1\end{bmatrix}
\displaystyle (A-B)^2=\begin{bmatrix}7&-2\\4&1\end{bmatrix}\begin{bmatrix}7&-2\\4&1\end{bmatrix}
\displaystyle =\begin{bmatrix}7\times7+(-2)\times4&7\times(-2)+(-2)\times1\\4\times7+1\times4&4\times(-2)+1\times1\end{bmatrix}
\displaystyle =\begin{bmatrix}41&-16\\32&-7\end{bmatrix}

\displaystyle \textbf{Question 46: }\text{Given }A=\begin{bmatrix}x&3\\y&3\end{bmatrix},\text{ if }A^2=3I,\text{ where }I \\ \text{ is identity of order }2,\text{ find }x\text{ and }y.\ \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}x&3\\y&3\end{bmatrix}
\displaystyle A^2=A\cdot A=\begin{bmatrix}x^2+3y&3x+9\\xy+3y&3y+9\end{bmatrix}
\displaystyle 3I=\begin{bmatrix}3&0\\0&3\end{bmatrix}
\displaystyle \text{Since }A^2=3I,\text{ comparing corresponding elements,}
\displaystyle 3x+9=0\Rightarrow 3x=-9\Rightarrow x=-3
\displaystyle 3y+9=3\Rightarrow 3y=-6\Rightarrow y=-2
\displaystyle \text{Thus, }x=-3\text{ and }y=-2.

\displaystyle \textbf{Question 47: }\text{If }A=\begin{bmatrix}3&-1\\0&2\end{bmatrix},\text{ find matrix }B\text{ such that }A^2-2B=3A-5I,\text{ where }I\text{ is identity of order }2.\ \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}3&-1\\0&2\end{bmatrix},\ I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle A^2=A\cdot A
\displaystyle =\begin{bmatrix}3&-1\\0&2\end{bmatrix}\begin{bmatrix}3&-1\\0&2\end{bmatrix}
\displaystyle =\begin{bmatrix}3\times3+(-1)\times0&3\times(-1)+(-1)\times2\\0\times3+2\times0&0\times(-1)+2\times2\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}9&-5\\0&4\end{bmatrix}
\displaystyle \text{Now, we have }A^2-2B=3A-5I
\displaystyle \Rightarrow 2B=A^2-3A+5I
\displaystyle \Rightarrow 2B=\begin{bmatrix}9&-5\\0&4\end{bmatrix}-3\begin{bmatrix}3&-1\\0&2\end{bmatrix}+5\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \Rightarrow 2B=\begin{bmatrix}9&-5\\0&4\end{bmatrix}+\begin{bmatrix}-9&3\\0&-6\end{bmatrix}+\begin{bmatrix}5&0\\0&5\end{bmatrix}
\displaystyle \Rightarrow 2B=\begin{bmatrix}9-9+5&-5+3+0\\0+0+0&4-6+5\end{bmatrix}
\displaystyle \Rightarrow 2B=\begin{bmatrix}5&-2\\0&3\end{bmatrix}
\displaystyle \therefore B=\frac{1}{2}\begin{bmatrix}5&-2\\0&3\end{bmatrix}=\begin{bmatrix}\frac{5}{2}&-1\\0&\frac{3}{2}\end{bmatrix}

\displaystyle \textbf{Question 48: }\text{Find }x\text{ and }y,\text{ if }3\begin{bmatrix}5&-6\\4&x\end{bmatrix}-\begin{bmatrix}6&y\\0&6\end{bmatrix}=3\begin{bmatrix}3&-2\\4&0\end{bmatrix}.\ \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }3\begin{bmatrix}5&-6\\4&x\end{bmatrix}-\begin{bmatrix}6&y\\0&6\end{bmatrix}=3\begin{bmatrix}3&-2\\4&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}15&-18\\12&3x\end{bmatrix}+\begin{bmatrix}-6&-y\\0&-6\end{bmatrix}=\begin{bmatrix}9&-6\\12&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}15-6&-18-y\\12+0&3x-6\end{bmatrix}=\begin{bmatrix}9&-6\\12&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}9&-18-y\\12&3x-6\end{bmatrix}=\begin{bmatrix}9&-6\\12&0\end{bmatrix}
\displaystyle \Rightarrow -18-y=-6\text{ and }3x-6=0\qquad\text{(by equality of matrix)}
\displaystyle \Rightarrow y=-12\text{ and }x=2

\displaystyle \text{Question 49: }\text{The product }AB\text{ of two matrices }A\text{ and }B\text{ is possible if }\ \text{[ICSE 2021]}
\displaystyle \text{(a) Same number of rows}\quad\text{(b) Columns of }A=\text{ rows of }B\quad \\ \text{(c) Rows of }A=\text{ columns of }B\quad\text{(d) Same number of columns}
\displaystyle \text{Answer:}
\displaystyle \textbf{(b) }\text{The product of two matrices }A\text{ and }B\text{ is possible, if}
\displaystyle \text{the number of columns of }A\text{ is equal to the number of rows of }B.

\displaystyle \textbf{Question 50: }\text{If matrix }A\text{ is of order }3\times2\text{ and }B\text{ is }2\times2,\text{ then }AB\text{ is of order }
\displaystyle \text{(a) }3\times2\qquad\text{(b) }3\times1\qquad\text{(c) }2\times3\qquad\text{(d) }1\times3
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, order of matrix }A\text{ is }3\times2,\text{ order of matrix }B\text{ is }2\times2.
\displaystyle \text{We know that, order of matrix is number of rows}\times\text{number of columns.}
\displaystyle \Rightarrow \text{Order of matrix }AB=\text{Number of rows of }A\times\text{Number of columns of }B
\displaystyle =3\times2
\displaystyle \text{Hence, matrix }AB\text{ is of order }3\times2.

\displaystyle \text{Question 51: }\text{Which of the following statement is not true? [ICSE 2022]}
\displaystyle \text{(a) All identity matrices are square.}\quad\text{(b) All null matrices are square.}
\displaystyle \text{(c) For a square matrix, rows equal columns.}\quad \\ \text{(d) Diagonal matrix has zeros except leading diagonal.}
\displaystyle \text{Answer:}
\displaystyle \text{(b) }\text{Null matrices are matrices of order }m\times n\text{ in which all elements are zero.}
\displaystyle \text{It is square if }m=n\text{ otherwise it is not square.}
\displaystyle \text{Hence, the statement ``all null matrices are square matrices'' is not true.}

\displaystyle \textbf{Question 52: }\text{If }A=\begin{bmatrix}2&0\\-1&7\end{bmatrix},\text{ then }A^2\text{ is }\ \text{[ICSE 2022]}
\displaystyle \text{(a) }\begin{bmatrix}4&0\\1&49\end{bmatrix}\quad\text{(b) }\begin{bmatrix}4&0\\-9&49\end{bmatrix}\quad\text{(c) }\begin{bmatrix}4&0\\9&49\end{bmatrix}\quad\text{(d) }\begin{bmatrix}1&9\\-9&48\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }A=\begin{bmatrix}2&0\\-1&7\end{bmatrix}\Rightarrow A^2=\begin{bmatrix}2&0\\-1&7\end{bmatrix}\begin{bmatrix}2&0\\-1&7\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times2+0\times(-1)&2\times0+0\times7\\(-1)\times2+7\times(-1)&(-1)\times0+7\times7\end{bmatrix}
\displaystyle =\begin{bmatrix}4&0\\-9&49\end{bmatrix}

\displaystyle \textbf{Question 53: }\text{If }A=\begin{bmatrix}3&5\\1&4\end{bmatrix},\ B=\begin{bmatrix}2&4\\0&3\end{bmatrix},\ C=\begin{bmatrix}1&-1\\2&1\end{bmatrix},\text{ then } \\ 5A-BC\text{ is equal to }\ \text{[ICSE 2022]}
\displaystyle \text{(a) }\begin{bmatrix}-5&-23\\1&17\end{bmatrix}\quad\text{(b) }\begin{bmatrix}5&23\\1&17\end{bmatrix}\quad\text{(c) }\begin{bmatrix}-2&8\\-3&3\end{bmatrix}\quad\text{(d) }\begin{bmatrix}5&23\\-1&17\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }A=\begin{bmatrix}3&5\\1&4\end{bmatrix},\ B=\begin{bmatrix}2&4\\0&3\end{bmatrix},\ C=\begin{bmatrix}1&-1\\2&1\end{bmatrix}
\displaystyle \therefore 5A-BC=5\begin{bmatrix}3&5\\1&4\end{bmatrix}-\begin{bmatrix}2&4\\0&3\end{bmatrix}\begin{bmatrix}1&-1\\2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}15&25\\5&20\end{bmatrix}-\begin{bmatrix}10&2\\6&3\end{bmatrix}=\begin{bmatrix}5&23\\-1&17\end{bmatrix}

\displaystyle \textbf{Question 54: }\text{If }A=\begin{bmatrix}5&10\\3&-4\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ find }AI.\ \text{[ICSE 2022]}
\displaystyle \text{(a) }\begin{bmatrix}1&0\\0&1\end{bmatrix}\quad\text{(b) }\begin{bmatrix}5&10\\-3&4\end{bmatrix}\quad\text{(c) }\begin{bmatrix}5&10\\3&-4\end{bmatrix}\quad\text{(d) }\begin{bmatrix}15&15\\-1&-1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }A=\begin{bmatrix}5&10\\3&-4\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \therefore AI=\begin{bmatrix}5&10\\3&-4\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}5&10\\3&-4\end{bmatrix}

\displaystyle \textbf{Question 55: }\text{If }\begin{bmatrix}2&0\\0&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\-8\end{bmatrix},\text{ find }x\text{ and }y.\ \text{[ICSE 2023]}
\displaystyle \text{(a) }1,-2\qquad\text{(b) }-2,1\qquad\text{(c) }1,2\qquad\text{(d) }-2,-1
\displaystyle \text{Answer:}
\displaystyle \text{(a) }2x+0y=2\Rightarrow 2x=2\Rightarrow x=1\qquad\cdots\text{(i)}
\displaystyle 0x+4y=-8\Rightarrow 4y=-8\Rightarrow y=-2
\displaystyle \text{Hence, }x=1,\ y=-2.

\displaystyle \textbf{Question 56: }\text{Given }\begin{bmatrix}a&b\\c&d\end{bmatrix}\times X=\begin{bmatrix}p\\q\end{bmatrix},\text{ find order of }X.\ \text{[ICSE 2023]}
\displaystyle \text{(a) }2\times2\qquad\text{(b) }1\times2\qquad\text{(c) }2\times1\qquad\text{(d) }1\times1
\displaystyle \text{Answer:}
\displaystyle \text{(c) Let }A=\begin{bmatrix}a&b\\c&d\end{bmatrix}_{2\times2}\text{ and }B=\begin{bmatrix}p\\q\end{bmatrix}_{2\times1}
\displaystyle \text{Also, we have }AX=B.
\displaystyle \text{Since product of matrices }A\text{ and }X\text{ is possible,}
\displaystyle \text{Number of columns in }A=\text{Number of rows in }X.
\displaystyle \therefore \text{Number of rows in }X\text{ is }2.
\displaystyle \text{Since }AX\text{ results in a matrix }B\text{ of order }2\times1,
\displaystyle \text{Number of columns in }X=\text{Number of columns in }B.
\displaystyle \therefore \text{Number of columns in }X=1.
\displaystyle \text{Thus, order of matrix }X\text{ is }2\times1.

\displaystyle \textbf{Question 57: }\text{If }\begin{bmatrix}2&x\\0&1\end{bmatrix}+3\begin{bmatrix}2&1\\4&0\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix},\text{ find }x.\ \text{[ICSE 2023]}
\displaystyle \text{(a) }2\qquad\text{(b) }3\qquad\text{(c) }4\qquad\text{(d) }5
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }\begin{bmatrix}2&x\\0&1\end{bmatrix}+3\begin{bmatrix}2&1\\4&0\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2&x\\0&1\end{bmatrix}+\begin{bmatrix}6&3\\12&0\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2+6&x+3\\0+12&1+0\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}8&x+3\\12&1\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix}
\displaystyle \text{Here, both matrices are equal, so equating corresponding elements,}
\displaystyle x+3=8\Rightarrow x=8-3\Rightarrow x=5.

\displaystyle \textbf{Question 58: }\text{If }A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\ B=\begin{bmatrix}1&2\\2&4\end{bmatrix},\ C=\begin{bmatrix}4&1\\1&5\end{bmatrix},\ I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ find } \\ A(B+C)-14I.\ \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\ B=\begin{bmatrix}1&2\\2&4\end{bmatrix},\ C=\begin{bmatrix}4&1\\1&5\end{bmatrix}
\displaystyle \therefore B+C=\begin{bmatrix}1+4&2+1\\2+1&4+5\end{bmatrix}=\begin{bmatrix}5&3\\3&9\end{bmatrix}
\displaystyle A(B+C)=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}5&3\\3&9\end{bmatrix}
\displaystyle =\begin{bmatrix}1\times5+3\times3&1\times3+3\times9\\2\times5+4\times3&2\times3+4\times9\end{bmatrix}
\displaystyle =\begin{bmatrix}14&30\\22&42\end{bmatrix}
\displaystyle 14I=14\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}14&0\\0&14\end{bmatrix}
\displaystyle \therefore A(B+C)-14I=\begin{bmatrix}14-14&30-0\\22-0&42-14\end{bmatrix}=\begin{bmatrix}0&30\\22&28\end{bmatrix}

\displaystyle \textbf{Question 59: }\text{If }A=\begin{bmatrix}3&-2\\-1&4\end{bmatrix},\ B=\begin{bmatrix}6\\1\end{bmatrix},\ C=\begin{bmatrix}-4\\5\end{bmatrix},\text{ evaluate } \\ AB-5C.\ \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&-2\\-1&4\end{bmatrix},\ B=\begin{bmatrix}6\\1\end{bmatrix},\ C=\begin{bmatrix}-4\\5\end{bmatrix}
\displaystyle AB=\begin{bmatrix}3&-2\\-1&4\end{bmatrix}\begin{bmatrix}6\\1\end{bmatrix}=\begin{bmatrix}3\times6-2\times1\\-1\times6+4\times1\end{bmatrix}=\begin{bmatrix}16\\-2\end{bmatrix}
\displaystyle 5C=5\begin{bmatrix}-4\\5\end{bmatrix}=\begin{bmatrix}-20\\25\end{bmatrix}
\displaystyle \therefore AB-5C=\begin{bmatrix}16-(-20)\\-2-25\end{bmatrix}=\begin{bmatrix}36\\-27\end{bmatrix}

\displaystyle \textbf{Question 60: }\text{If matrix }A=\begin{bmatrix}2&2\\0&2\end{bmatrix}\text{ and }A^2=\begin{bmatrix}4&x\\0&4\end{bmatrix},\text{ then find }x.\ \text{[ICSE 2024]}
\displaystyle \text{(a) }2\qquad\text{(b) }4\qquad\text{(c) }8\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }A=\begin{bmatrix}2&2\\0&2\end{bmatrix}\text{ and }A^2=\begin{bmatrix}4&x\\0&4\end{bmatrix}
\displaystyle A^2=A\times A
\displaystyle \Rightarrow \begin{bmatrix}4&x\\0&4\end{bmatrix}=\begin{bmatrix}2&2\\0&2\end{bmatrix}\begin{bmatrix}2&2\\0&2\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4&x\\0&4\end{bmatrix}=\begin{bmatrix}4&8\\0&4\end{bmatrix}
\displaystyle \text{By the definition of equality of two matrices, we get }x=8.

\displaystyle \textbf{Question 61: }\text{If }A=\begin{bmatrix}x&0\\1&1\end{bmatrix},\ B=\begin{bmatrix}4&0\\y&1\end{bmatrix}\text{ and }C=\begin{bmatrix}4&0\\x&1\end{bmatrix},\text{ then}
\displaystyle \text{find the values of }x\text{ and }y,\text{ if }AB=C.\ \text{[ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}x&0\\1&1\end{bmatrix},\ B=\begin{bmatrix}4&0\\y&1\end{bmatrix}\text{ and }C=\begin{bmatrix}4&0\\x&1\end{bmatrix}
\displaystyle AB=C
\displaystyle \Rightarrow \begin{bmatrix}x&0\\1&1\end{bmatrix}\begin{bmatrix}4&0\\y&1\end{bmatrix}=\begin{bmatrix}4&0\\x&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4x&0\\4+y&1\end{bmatrix}=\begin{bmatrix}4&0\\x&1\end{bmatrix}
\displaystyle \text{By the definition of equality of matrices, we get }4x=4\Rightarrow x=1
\displaystyle \text{and }4+y=x\Rightarrow 4+y=1\Rightarrow y=1-4=-3
\displaystyle \text{Hence, }x=1\text{ and }y=-3

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.