If A = begin{bmatrix} -8 & 5 2 & 4 end{bmatr | vedclass

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(B) Given the matrix $A = \begin{bmatrix} -8 & 5 \ 2 & 4 \end{bmatrix}$.Since $A$ satisfies the equation $x^2 + 4x - p = 0$,we have $A^2 + 4A - pI =

Vedclass · Accepted Answer

$42$

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(B) Given the matrix $A = \begin{bmatrix} -8 & 5 \ 2 & 4 \end{bmatrix}$.Since $A$ satisfies the equation $x^2 + 4x - p = 0$,we have $A^2 + 4A - pI = 0$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} -8 & 5 \ 2 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \ 2 & 4 \end{bmatrix} = \begin{bmatrix} (-8)(-8) + (5)(2) & (-8)(5) + (5)(4) \ (2)(-8) + (4)(2) & (2)(5) + (4)(4) \end{bmatrix} = \begin{bmatrix} 64 + 10 & -40 + 20 \ -16 + 8 & 10 + 16 \end{bmatrix} = \begin{bmatrix} 74 & -20 \ -8 & 26 \end{bmatrix}$.Next,calculate $4A$:$4A = 4 \begin{bmatrix} -8 & 5 \ 2 & 4 \end{bmatrix} = \begin{bmatrix} -32 & 20 \ 8 & 16 \end{bmatrix}$.Now,substitute these into the equation $A^2 + 4A - pI = 0$:$\begin{bmatrix} 74 & -20 \ -8 & 26 \end{bmatrix} + \begin{bmatrix} -32 & 20 \ 8 & 16 \end{bmatrix} - \begin{bmatrix} p & 0 \ 0 & p \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Performing the matrix addition and subtraction:$\begin{bmatrix} 74 - 32 - p & -20 + 20 - 0 \ -8 + 8 - 0 & 26 + 16 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.$\begin{bmatrix} 42 - p & 0 \ 0 & 42 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Comparing the elements,we get $42 - p = 0$,which implies $p = 42$.

If $A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$ satisfies the equation $x^2 + 4x - p = 0$,then $p$ is equal to

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