If A = begin{bmatrix} 2 & -3 -4 & 1 end{bmat | vedclass

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(D) Given: $A = \begin{bmatrix} 2 & -3 \ -4 & 1 \end{bmatrix}$ First,find the transpose of $A$: $A^T = \begin{bmatrix} 2 & -4 \ -3 & 1 \end{bmatrix}

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$\begin{bmatrix} 40 & -60 \ -45 & 25 \end{bmatrix}$

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(D) Given: $A = \begin{bmatrix} 2 & -3 \ -4 & 1 \end{bmatrix}$ First,find the transpose of $A$: $A^T = \begin{bmatrix} 2 & -4 \ -3 & 1 \end{bmatrix}$ Now,calculate $(A^T)^2$: $(A^T)^2 = \begin{bmatrix} 2 & -4 \ -3 & 1 \end{bmatrix} \begin{bmatrix} 2 & -4 \ -3 & 1 \end{bmatrix} = \begin{bmatrix} (2)(2) + (-4)(-3) & (2)(-4) + (-4)(1) \ (-3)(2) + (1)(-3) & (-3)(-4) + (1)(1) \end{bmatrix} = \begin{bmatrix} 16 & -12 \ -9 & 13 \end{bmatrix}$ Next,calculate $(12A)^T$: $12A = \begin{bmatrix} 24 & -36 \ -48 & 12 \end{bmatrix} \Rightarrow (12A)^T = \begin{bmatrix} 24 & -48 \ -36 & 12 \end{bmatrix}$ Finally,add the two matrices: $(A^T)^2 + (12A)^T = \begin{bmatrix} 16 & -12 \ -9 & 13 \end{bmatrix} + \begin{bmatrix} 24 & -48 \ -36 & 12 \end{bmatrix} = \begin{bmatrix} 40 & -60 \ -45 & 25 \end{bmatrix}$

If $A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$,then $(A^T)^2 + (12 A)^T = $

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