\displaystyle \textbf{Question 1. }\text{Assertion (A) Let the matrices }A=\begin{pmatrix}-3 & 2\\-5 & 4\end{pmatrix}\text{ and } \ \
\displaystyle B=\begin{pmatrix}4 & -2\\5 & -3\end{pmatrix}\text{ be such that }A^{100}B=BA^{100}.\text{ \hspace{4.2cm} ISC 2024} \ \
\displaystyle \text{Reason (R) }AB=BA\text{ implies }A^{n}B=BA^{n}\text{ for all positive integers }n. \ \
\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) We have, } A=\begin{bmatrix}-3&2\\-5&4\end{bmatrix} \text{ and } B=\begin{bmatrix}4&-2\\5&-3\end{bmatrix}
\displaystyle \text{Here, } AB=\begin{bmatrix}-3&2\\-5&4\end{bmatrix}\begin{bmatrix}4&-2\\5&-3\end{bmatrix}=\begin{bmatrix}-2&0\\0&-2\end{bmatrix}
\displaystyle \text{and } BA=\begin{bmatrix}4&-2\\5&-3\end{bmatrix}\begin{bmatrix}-3&2\\-5&4\end{bmatrix}=\begin{bmatrix}-2&0\\0&-2\end{bmatrix}
\displaystyle \text{We know that if } AB=BA, \text{ then by mathematical induction } A^{n}B=BA^{n}, \text{ } \\ n \text{ is positive integers.}
\displaystyle \text{For } n=100, \text{ we get } A^{100}B=BA^{100}
\displaystyle \text{Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the} \\ \text{correct explanation of Assertion (A).}
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\displaystyle \textbf{Question 2. }\text{If }A\text{ and }B\text{ are symmetric matrices of the same order, then } \\ AB-BA\text{ is \hspace{8.2cm} ISC 2024} \ \
\displaystyle \text{(a) Skew-symmetric matrix} \ \
\displaystyle \text{(b) Symmetric matrix} \ \
\displaystyle \text{(c) Diagonal matrix} \ \
\displaystyle \text{(d) Identity matrix} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } A \text{ and } B \text{ are two symmetric matrices.}
\displaystyle \text{Then, } A^{T}=A \text{ and } B^{T}=B
\displaystyle \text{Now, } (AB-BA)^{T}=(AB)^{T}-(BA)^{T}
\displaystyle =(B^{T}A^{T}-A^{T}B^{T})
\displaystyle =(BA-AB)
\displaystyle =-(AB-BA)
\displaystyle \text{Hence, } (AB-BA) \text{ is skew-symmetric matrix.}
\\

\displaystyle \textbf{Question 3. }\text{For what value of }k,\text{ the matrix }\begin{pmatrix}0 & k\\-6 & 0\end{pmatrix}\text{ is a } \ \
\displaystyle \text{skew-symmetric matrix? \hspace{8.2cm} ISC 2023} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \begin{bmatrix}0&k\\-6&0\end{bmatrix} \text{ is skew-symmetric matrix.}
\displaystyle \text{Let } A=\begin{bmatrix}0&k\\-6&0\end{bmatrix}
\displaystyle \text{Since, } A \text{ is skew-symmetric matrix.}
\displaystyle \therefore A=-A^{T}
\displaystyle \Rightarrow \begin{bmatrix}0&k\\-6&0\end{bmatrix}=\begin{bmatrix}0&6\\-k&0\end{bmatrix} \Rightarrow -k=-6 \Rightarrow k=6
\\

\displaystyle \textbf{Question 4. }\text{If }\begin{pmatrix}2 & 3\\5 & 7\end{pmatrix}\begin{pmatrix}1 & -3\\-2 & 4\end{pmatrix}=\begin{pmatrix}-4 & 6\\-9 & x\end{pmatrix},\text{ then find }x.\text{ \hspace{0.2cm} ISC 2020} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2-6&-6+12\\5-14&-15+28\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-4&6\\-9&13\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}
\displaystyle \text{On comparing the corresponding element, we get } x=13
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\displaystyle \textbf{Question 5. }\text{Show that }(A+A')\text{ is symmetric matrix, if }A=\begin{pmatrix}2 & 4\\3 & 5\end{pmatrix}.\text{ \hspace{0.2cm} ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } A=\begin{bmatrix}2&4\\3&5\end{bmatrix}
\displaystyle \text{Now, } A+A^{T}=\begin{bmatrix}2&4\\3&5\end{bmatrix}+\begin{bmatrix}2&3\\4&5\end{bmatrix}=\begin{bmatrix}4&7\\7&10\end{bmatrix}
\displaystyle \therefore (A+A^{T})^{T}=\begin{bmatrix}4&7\\7&10\end{bmatrix}=A+A^{T}
\displaystyle \text{Hence, } (A+A^{T}) \text{ is symmetric matrix.}
\\

\displaystyle \textbf{Question 6. }\text{If }A=\begin{pmatrix}5 & a\\b & 0\end{pmatrix}\text{ and }A\text{ is symmetric matrix, then show} \ \
\displaystyle \text{that }a=b.\text{ \hspace{8.2cm} ISC 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}5&a\\b&0\end{bmatrix}, A^{T}=\begin{bmatrix}5&b\\a&0\end{bmatrix}
\displaystyle A \text{ is symmetric matrix.}
\displaystyle \therefore A=A^{T}
\displaystyle \Rightarrow \begin{bmatrix}5&a\\b&0\end{bmatrix}=\begin{bmatrix}5&b\\a&0\end{bmatrix}
\displaystyle \text{On comparing the corresponding element, we get } a=b
\\

\displaystyle \textbf{Question 7. }\text{If the matrix }\begin{pmatrix}6 & -x^{2}\\2x-15 & 10\end{pmatrix}\text{ is symmetric, then find} \ \
\displaystyle \text{the value of }x.\text{ \hspace{0.2cm} ISC 2017} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=\begin{bmatrix}6&-x^{2}\\2x-15&10\end{bmatrix}
\displaystyle \text{Given, matrix } A \text{ is symmetric.}
\displaystyle \therefore A=A^{T}
\displaystyle \text{Clearly, } A^{T}=\begin{bmatrix}6&2x-15\\-x^{2}&10\end{bmatrix}
\displaystyle \text{From } A=A^{T}, \begin{bmatrix}6&-x^{2}\\2x-15&10\end{bmatrix}=\begin{bmatrix}6&2x-15\\-x^{2}&10\end{bmatrix}
\displaystyle \therefore -x^{2}=2x-15
\displaystyle \Rightarrow x^{2}+2x-15=0
\displaystyle \Rightarrow x^{2}+5x-3x-15=0
\displaystyle \Rightarrow x(x+5)-3(x+5)=0
\displaystyle \Rightarrow (x+5)(x-3)=0
\displaystyle \therefore x=-5 \text{ or } x=3
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\displaystyle \textbf{Question 8. }\text{If }A=\begin{pmatrix}3 & 1\\7 & 5\end{pmatrix},\text{ then find the values of }x\text{ and }y,\text{ such that} \ \
\displaystyle A^{2}+xI_{2}=yA.\text{ \hspace{5.2cm} ISC 2014} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}3&1\\7&5\end{bmatrix}
\displaystyle \text{Now, } A^{2}=A\cdot A=\begin{bmatrix}3&1\\7&5\end{bmatrix}\begin{bmatrix}3&1\\7&5\end{bmatrix}
\displaystyle =\begin{bmatrix}9+7&3+5\\21+35&7+25\end{bmatrix}=\begin{bmatrix}16&8\\56&32\end{bmatrix}
\displaystyle \text{Also given, } A^{2}+xI_{2}=yA
\displaystyle \therefore \begin{bmatrix}16&8\\56&32\end{bmatrix}+x\begin{bmatrix}1&0\\0&1\end{bmatrix}=y\begin{bmatrix}3&1\\7&5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}16&8\\56&32\end{bmatrix}+\begin{bmatrix}x&0\\0&x\end{bmatrix}=\begin{bmatrix}3y&y\\7y&5y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}16+x&8\\56&32+x\end{bmatrix}=\begin{bmatrix}3y&y\\7y&5y\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get } y=8 \text{ and } 32+x=5y
\displaystyle \Rightarrow 32+x=5\times 8
\displaystyle \Rightarrow 32+x=40
\displaystyle \Rightarrow x=40-32
\displaystyle \therefore x=8 \text{ and } y=8
\\

\displaystyle \textbf{Question 9. }\text{If }(A-2I)(A-3I)=O,\text{ where }A=\begin{pmatrix}4 & 2\\-1 & x\end{pmatrix}\text{ and} \ \
\displaystyle I=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix},\text{ then find the value of }x.\text{ \hspace{5.2cm} ISC 2013} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}4&2\\-1&x\end{bmatrix} \text{ and } I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \text{Also given, } (A-2I)(A-3I)=O
\displaystyle \Rightarrow \left(\begin{bmatrix}4&2\\-1&x\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}\right)\left(\begin{bmatrix}4&2\\-1&x\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}\right)=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4-2&2-0\\-1-0&x-2\end{bmatrix}\begin{bmatrix}4-3&2-0\\-1-0&x-3\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2&2\\-1&x-2\end{bmatrix}\begin{bmatrix}1&2\\-1&x-3\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2-2&4+2x-6\\-1-x+2&-2+x^{2}-5x+6\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}0&2x-2\\-x+1&x^{2}-5x+4\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get } 2x-2=0
\displaystyle \Rightarrow x-1=0 \Rightarrow x=1
\displaystyle \text{Hence, the value of } x \text{ is } 1.

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