If X = begin{bmatrix} 3 & -4 1 & -1 end{bmat | vedclass

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(D) Given $X = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}$.We calculate $X^2 = X 	imes X = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix} \begin{

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None of these

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(D) Given $X = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}$.We calculate $X^2 = X 	imes X = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix} = \begin{bmatrix} (3)(3) + (-4)(1) & (3)(-4) + (-4)(-1) \ (1)(3) + (-1)(1) & (1)(-4) + (-1)(-1) \end{bmatrix} = \begin{bmatrix} 9-4 & -12+4 \ 3-1 & -4+1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \ 2 & -3 \end{bmatrix}$.Now,let us check the options for $n=2$:Option $(a)$ gives $\begin{bmatrix} 3(2) & -4(2) \ 2 & -2 \end{bmatrix} = \begin{bmatrix} 6 & -8 \ 2 & -2 \end{bmatrix} 
eq X^2$.Option $(b)$ gives $\begin{bmatrix} 2+2 & 5-2 \ 2 & -2 \end{bmatrix} = \begin{bmatrix} 4 & 3 \ 2 & -2 \end{bmatrix} 
eq X^2$.Option $(c)$ gives $\begin{bmatrix} 3^2 & (-4)^2 \ 1^2 & (-1)^2 \end{bmatrix} = \begin{bmatrix} 9 & 16 \ 1 & 1 \end{bmatrix} 
eq X^2$.Since none of the options match $X^2$,the correct answer is $(d)$.

If $X = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$,then the value of $X^n$ is

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