Q.If $X = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$, then $X^5$ will be

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 …

$X$ 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

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

…

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

$n = 2.$

…

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 ar …