Q.If $A=\begin{pmatrix}1&0\\2&-1\end{pmatrix}$ and $B=\begin{pmatrix}3&1\\0&2\end{pmatrix}$, verify that $(AB)'=B'A'$.

Evaluate the matrix polynomial $f(A)=A^2-2A-5I$; the scalar $5$ becomes $5I$.

Evaluating a polynomial at a matrix argument is a standard NCERT/CBSE Class 12 matrices exercise.

First $A^2$:

$A^2 = \begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix} = \begin{bmatrix}5&4\\4&5\end{bmatrix}.$

Then $2A = \begin{bmatrix}2&4\\4&2\end{bmatrix}$ and $5I = \begin{bmatrix}5&0\\0&5\end{bmatrix}$.

So

$f(A) = A^2 - 2A - 5I = \begin{bmatrix}5-2-5 & 4-4-0 \\ 4-4-0 & 5-2-5\end{bmatrix} = \begin{bmatrix}-2&0\\0&-2\end{bmatrix}.$

$X$ is a scalar matrix $3I$, so $X^5=3^5 I=243I=81X$.

Powers of a scalar matrix are an easy NCERT/CBSE Class 12 matrices computation.

The matrix $X = \begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix} = 3I$, where $I$ is the $3\times3$ identity.

Then

$X^5 = (3I)^5 = 3^5 I = 243\,I.$

Since $X = 3I \Rightarrow I = \tfrac{1}{3}X$, we can write

$243\,I = 243\cdot\frac{1}{3}X = 81X.$

Conformability of $TS$ forces $n=2$; the resulting order $3\times m=p\times4$ gives $p=3$ and $m=4$.

The rule for matrix-product order (inner dimensions match, outer dimensions give the result) is a CBSE/NCERT Class 12 matrices basic.

Here $T$ is $3\times n$ and $S$ is $2\times m$. The product $TS$ requires the inner dimensions to agree: the number of columns of $T$ ($=n$) equals the number of rows of $S$ ($=2$), so

$n = 2.$

The product $TS$ then has order (rows of $T$) $\times$ (columns of $S$) $= 3\times m$. We are told this equals $p\times4$, hence

$p = 3, \qquad m = 4.$

Diagonal matrices of the same order always commute under multiplication.

If $D_1=\text{diag}(a_1,\dots,a_n)$ and $D_2=\text{diag}(b_1,\dots,b_n)$, then $D_1D_2=\text{diag}(a_1b_1,\dots,a_nb_n)$ and $D_2D_1=\text{diag}(b_1a_1,\dots,b_na_n)$, which are equal since real-number multiplication is commutative entry by entry. So $D_1D_2=D_2D_1$.