\displaystyle \textbf{Question 1. }\text{P and Q are two matrices such that }[P+Q]^T=\begin{pmatrix}7 & -8\\3 & 0\end{pmatrix}. \ \
\displaystyle \text{If }P=\begin{pmatrix}-1 & 1\\3 & 2\end{pmatrix},\text{ which one of the following is }Q?\text{ \hspace{0.2cm} ISC Specimen 2024} \ \
\displaystyle \text{(a) }\begin{pmatrix}8 & -9\\0 & 2\end{pmatrix} \ \
\displaystyle \text{(b) }\begin{pmatrix}-8 & -2\\11 & 2\end{pmatrix} \ \
\displaystyle \text{(c) }\begin{pmatrix}8 & 2\\-11 & -2\end{pmatrix} \ \
\displaystyle \text{(d) }\begin{pmatrix}8 & -11\\2 & -2\end{pmatrix} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, } (P+Q)^{T}=\begin{bmatrix}7&-8\\3&0\end{bmatrix} \text{ and } P=\begin{bmatrix}-1&1\\3&2\end{bmatrix}
\displaystyle \therefore P^{T}+Q^{T}=\begin{bmatrix}7&-8\\3&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-1&1\\3&2\end{bmatrix}^{T}+Q^{T}=\begin{bmatrix}7&-8\\3&0\end{bmatrix}
\displaystyle \Rightarrow Q^{T}=\begin{bmatrix}7&-8\\3&0\end{bmatrix}-\begin{bmatrix}-1&3\\1&2\end{bmatrix}
\displaystyle \Rightarrow Q^{T}=\begin{bmatrix}8&-11\\2&-2\end{bmatrix}
\displaystyle \therefore Q=\begin{bmatrix}8&2\\-11&-2\end{bmatrix}
\\

\displaystyle \textbf{Question 2. }\text{Matrix }P\text{ is a skew-symmetric matrix of odd order.} \ \
\displaystyle \text{Assertion: The inverse of matrix }P\text{ does not exist.} \ \
\displaystyle \text{Reason: The determinant of matrix }P\text{ is zero. \hspace{0.2cm} ISC Specimen 2024} \ \
\displaystyle \text{(a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.} \ \
\displaystyle \text{(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.} \ \
\displaystyle \text{(c) Assertion is true but Reason is false.} \ \
\displaystyle \text{(d) Assertion is false but Reason is true.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let } P \text{ be any skew-symmetric matrix of odd order given by}
\displaystyle P=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}
\displaystyle \therefore |P|=0(0+c^{2})-a(0+bc)+b(ac-0)
\displaystyle =-abc+abc=0
\displaystyle \therefore |P|=0
\displaystyle \therefore P^{-1} \text{ does not exist.}
\\

\displaystyle \textbf{Question 3. }\text{If }A=\begin{pmatrix}5 & x\\y & 0\end{pmatrix}\text{ and }A\text{ is symmetric, then write the} \ \
\displaystyle \text{relation between }x\text{ and }y.\text{ \hspace{0.2cm} ISC Specimen 2024} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } A=\begin{bmatrix}5&x\\y&0\end{bmatrix}
\displaystyle \therefore A \text{ is symmetric.}
\displaystyle \therefore A^{T}=A \Rightarrow \begin{bmatrix}5&y\\x&0\end{bmatrix}=\begin{bmatrix}5&x\\y&0\end{bmatrix} \Rightarrow x=y
\\

\displaystyle \textbf{Question 4. }\text{If }\begin{pmatrix}3 & 5\\7 & 9\end{pmatrix}=X+Y,\text{ where }X\text{ is skew-symmetric} \ \
\displaystyle \text{matrix and }Y\text{ is symmetric matrix. Find }|X|.\text{ \hspace{0.2cm} ISC Specimen 2023} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=\begin{bmatrix}3&5\\7&9\end{bmatrix}=X+Y
\displaystyle \text{where, } X \text{ is skew-symmetric matrix.}
\displaystyle \text{i.e. } X=\frac{1}{2}(A-A^{T})
\displaystyle \text{and } Y \text{ is symmetric matrix}
\displaystyle \text{i.e. } Y=\frac{1}{2}(A+A^{T})
\displaystyle \therefore A=\begin{bmatrix}3&5\\7&9\end{bmatrix} \text{ and } A^{T}=\begin{bmatrix}3&7\\5&9\end{bmatrix}
\displaystyle \Rightarrow X=\frac{1}{2}(A-A^{T})=\frac{1}{2}\left[\begin{bmatrix}3&5\\7&9\end{bmatrix}-\begin{bmatrix}3&7\\5&9\end{bmatrix}\right]
\displaystyle \Rightarrow X=\frac{1}{2}\begin{bmatrix}0&-2\\2&0\end{bmatrix}
\displaystyle \Rightarrow X=\begin{bmatrix}0&-1\\1&0\end{bmatrix}
\displaystyle \therefore |X|=0-(-1)=1
\\

\displaystyle \textbf{Question 5. }\text{If }A=\begin{pmatrix}1 & b\\0 & 1\end{pmatrix},\text{ then }A^{n}\ (n\in N)\text{ is equal to \hspace{0.2cm} ISC Specimen 2023} \ \
\displaystyle \text{(a) }\begin{pmatrix}1 & nb\\0 & 1\end{pmatrix} \ \
\displaystyle \text{(b) }\begin{pmatrix}1 & b^{n}\\0 & 1\end{pmatrix} \ \
\displaystyle \text{(c) }\begin{pmatrix}1 & n^{b}\\0 & 1\end{pmatrix} \ \
\displaystyle \text{(d) }\begin{pmatrix}1 & nb\\0 & 0\end{pmatrix} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(a) Here, } A=\begin{bmatrix}1&b\\0&1\end{bmatrix}
\displaystyle A^{2}=A\cdot A=\begin{bmatrix}1&b\\0&1\end{bmatrix}\begin{bmatrix}1&b\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1+0&b+b\\0+0&0+1\end{bmatrix}=\begin{bmatrix}1&2b\\0&1\end{bmatrix}
\displaystyle A^{3}=A^{2}\cdot A=\begin{bmatrix}1&2b\\0&1\end{bmatrix}\begin{bmatrix}1&b\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1+0&b+2b\\0+0&0+1\end{bmatrix}=\begin{bmatrix}1&3b\\0&1\end{bmatrix}
\displaystyle \text{Similarly, } A^{n}=\begin{bmatrix}1&nb\\0&1\end{bmatrix}
\\

\displaystyle \textbf{Question 6. }\text{What is the transpose of a column matrix? \hspace{0.2cm} ISC Specimen Sem-I 2022} \ \
\displaystyle \text{(a) Zero matrix} \ \
\displaystyle \text{(b) Diagonal matrix} \ \
\displaystyle \text{(c) Column matrix} \ \
\displaystyle \text{(d) Row matrix} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(d) The transpose of a column matrix is a row matrix.}
\\

\displaystyle \textbf{Question 7. }\text{If }A\text{ and }B\text{ are two non-singular matrices and }AB\text{ exists, then }(AB)^{-1}\text{ is \hspace{0.2cm} ISC Specimen Sem-I 2022} \ \
\displaystyle \text{(a) }A^{-1}B^{-1} \ \
\displaystyle \text{(b) }B^{-1}A^{-1} \ \
\displaystyle \text{(c) }AB \ \
\displaystyle \text{(d) None of these} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(b) We know that if } A^{-1}, B^{-1} \text{ and } (AB)^{-1} \text{ exists,}
\displaystyle \text{then } (AB)^{-1}=B^{-1}A^{-1}.
\\

\displaystyle \textbf{Question 8. }\text{From the matrix equation }AB=AC,\text{ it can be} \ \
\displaystyle \text{concluded that }B=C\text{ provided \hspace{0.2cm} ISC Specimen Sem-I 2022} \ \
\displaystyle \text{(a) }A\text{ is singular matrix} \ \
\displaystyle \text{(b) }A\text{ is a non-singular matrix} \ \
\displaystyle \text{(c) }A\text{ is a symmetric matrix} \ \
\displaystyle \text{(d) }A\text{ is a skew-symmetric matrix} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, } AB=AC
\displaystyle \Rightarrow B=C
\displaystyle \text{On pre-multiplying by } A^{-1} \text{ on both sides of the equation } AB=AC, \text{ we get}
\displaystyle A^{-1}AB=A^{-1}AC
\displaystyle \Rightarrow IB=IC \Rightarrow B=C
\displaystyle \text{So, } A \text{ is non-singular matrix.}
\\

\displaystyle \textbf{Question 9. }\text{If }A=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix},\text{ then }A^{2}\text{ is equal to \hspace{0.2cm} ISC Specimen 2021} \ \
\displaystyle \text{(a) }\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix} \ \
\displaystyle \text{(b) }\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix} \ \
\displaystyle \text{(c) }\begin{pmatrix}1 & 0\\1 & 0\end{pmatrix} \ \
\displaystyle \text{(d) }\begin{pmatrix}0 & 0\\1 & 1\end{pmatrix} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, } A=\begin{bmatrix}0&1\\1&0\end{bmatrix}
\displaystyle \text{Now, } A^{2}=A\cdot A
\displaystyle =\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0+1&0+0\\0+0&1+0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\\

\displaystyle \textbf{Question 10. }\text{Solve the matrix equation }A\begin{pmatrix}3 & 4\\-1 & 2\\2 & 1\end{pmatrix}=\begin{pmatrix}0 & 15\\1 & -2\end{pmatrix}.\text{ \hspace{0.2cm} ISC Specimen 2021} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A\begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix}=\begin{bmatrix}0&15\\1&-2\end{bmatrix}
\displaystyle \text{The right side of given matrix equation is of order } 2\times 2, \text{ so left side of the given matrix equation should be of } 2\times 2.
\displaystyle \text{It is clear that product of } A \text{ with } 3\times 2 \text{ matrix is } 2\times 2 \text{ matrix.}
\displaystyle \text{Therefore, } A \text{ is of order } 2\times 3.
\displaystyle \text{Let } A=\begin{bmatrix}u&w&y\\v&x&z\end{bmatrix}
\displaystyle \therefore \text{ From Eq. (i), we have}
\displaystyle \begin{bmatrix}u&w&y\\v&x&z\end{bmatrix}\begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix}=\begin{bmatrix}0&15\\1&-2\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}3u-w+2y&4u+2w+y\\3v-x+2z&4v+2x+z\end{bmatrix}=\begin{bmatrix}0&15\\1&-2\end{bmatrix}
\displaystyle \text{On comparing the elements from both the side, we get}
\displaystyle 3u-w+2y=0,\ 4u+2w+y=15;
\displaystyle 3v-x+2z=1 \text{ and } 4v+2x+z=-2
\displaystyle \text{On solving, we get } u=1,\ w=5,\ y=1,\ v=1,\ x=-2,\ z=-2
\displaystyle \text{So, } A=\begin{bmatrix}1&5&1\\1&-2&-2\end{bmatrix}
\\

\displaystyle \textbf{Question 11. }\text{Evaluate :} \ \
\displaystyle \begin{bmatrix}3 & -2 & 3\\2 & 1 & -1\\4 & -3 & 2\end{bmatrix}\begin{bmatrix}-1 & -5 & -1\\-8 & -6 & 9\\-10 & 1 & 7\end{bmatrix} \ \
\displaystyle \text{Hence, solve the system of equations,} \ \
\displaystyle 3x-2y+3z=8,\ 2x+y-z=1\text{ and }4x-3y+2z=4.\text{ \hspace{0.2cm} ISC Specimen 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix},\ B=\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}
\displaystyle AB=\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix}\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}
\displaystyle =\begin{bmatrix}-3+16-30&-15+12+3&-3-18+21\\-2-8+10&-10-6-1&-2+9-7\\-4+24-20&-20+18+2&-4-27+14\end{bmatrix}
\displaystyle =\begin{bmatrix}-17&0&0\\0&-17&0\\0&0&-17\end{bmatrix}
\displaystyle \therefore AB=-17I
\displaystyle \text{On pre-multiplying by } A^{-1} \text{ on both sides, we get}
\displaystyle (A^{-1}A)B=A^{-1}(-17I)
\displaystyle \Rightarrow IB=-17A^{-1}
\displaystyle \Rightarrow B=-17A^{-1}
\displaystyle \Rightarrow A^{-1}=-\frac{1}{17}B=-\frac{1}{17}\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}
\displaystyle \text{Given, system of equation can be written as } AX=C \Rightarrow X=A^{-1}C
\displaystyle \text{where } A=\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix},\ C=\begin{bmatrix}8\\1\\4\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}
\displaystyle \Rightarrow X=-\frac{1}{17}\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}\begin{bmatrix}8\\1\\4\end{bmatrix}
\displaystyle =-\frac{1}{17}\begin{bmatrix}-8-5-4\\-64-6+36\\-80+1+28\end{bmatrix}
\displaystyle =-\frac{1}{17}\begin{bmatrix}-17\\-34\\-51\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle \therefore x=1,\ y=2,\ z=3
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\displaystyle \textbf{Question 12. }\text{Find the value of }k,\text{ if }M=\begin{pmatrix}1 & 2\\2 & 3\end{pmatrix}\text{ and} \ \
\displaystyle M^{2}-kM-I_{2}=O.\text{ \hspace{0.2cm} ISC Specimen 2018, ISC 2015} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } M=\begin{bmatrix}1&2\\2&3\end{bmatrix}
\displaystyle M^{2}=\begin{bmatrix}1&2\\2&3\end{bmatrix}\begin{bmatrix}1&2\\2&3\end{bmatrix}=\begin{bmatrix}5&8\\8&13\end{bmatrix}
\displaystyle \text{Also, } I_{2}=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \therefore M^{2}-kM-I_{2}=O
\displaystyle \Rightarrow \begin{bmatrix}5&8\\8&13\end{bmatrix}-\begin{bmatrix}k&2k\\2k&3k\end{bmatrix}-\begin{bmatrix}1&0\\0&1\end{bmatrix}=O
\displaystyle \Rightarrow \begin{bmatrix}5-k-1&8-2k-0\\8-2k-0&13-3k-1\end{bmatrix}=O
\displaystyle \Rightarrow \begin{bmatrix}4-k&8-2k\\8-2k&12-3k\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \text{On comparing the elements, we get } 4-k=0 \Rightarrow k=4
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