The matrix A^2 + 4A - 5I,where I is the ident | vedclass

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Question

(A) Given $A = \begin{bmatrix} 1 & 2 \ 4 & -3 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $A^2 = A 	imes A

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$4 \begin{bmatrix} 2 & 1 \ 2 & 0 \end{bmatrix}$

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(A) Given $A = \begin{bmatrix} 1 & 2 \ 4 & -3 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 1 & 2 \ 4 & -3 \end{bmatrix} 	imes \begin{bmatrix} 1 & 2 \ 4 & -3 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(4) & (1)(2) + (2)(-3) \ (4)(1) + (-3)(4) & (4)(2) + (-3)(-3) \end{bmatrix} = \begin{bmatrix} 9 & -4 \ -8 & 17 \end{bmatrix}$.Next,calculate $4A$:$4A = 4 \begin{bmatrix} 1 & 2 \ 4 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 8 \ 16 & -12 \end{bmatrix}$.Next,calculate $5I$:$5I = 5 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \ 0 & 5 \end{bmatrix}$.Now,compute $A^2 + 4A - 5I$:$A^2 + 4A - 5I = \begin{bmatrix} 9 & -4 \ -8 & 17 \end{bmatrix} + \begin{bmatrix} 4 & 8 \ 16 & -12 \end{bmatrix} - \begin{bmatrix} 5 & 0 \ 0 & 5 \end{bmatrix}$$= \begin{bmatrix} 9+4-5 & -4+8-0 \ -8+16-0 & 17-12-5 \end{bmatrix} = \begin{bmatrix} 8 & 4 \ 8 & 0 \end{bmatrix}$$= 4 \begin{bmatrix} 2 & 1 \ 2 & 0 \end{bmatrix}$.

The matrix $A^2 + 4A - 5I$,where $I$ is the identity matrix and $A = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix}$,equals:

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