Question

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

• A. 64
• B. 42
• C. 36
• D. 24

Answer 1

Answer: (B)

Given,
$A =\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} $
$A^{2} =\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \\ 2 & 4
\end{bmatrix}$
$=\begin{bmatrix} 64+10 & -40+20 \\ -16+8 & 10+16 \end{bmatrix}$
$=\begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix}$
$4 A =\begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix}$
$-p I =\begin{bmatrix} -p & 0 \\ 0 & -p \end{bmatrix}$
Since, the matrix A satisfies the equation
$x^{2}+4 x-p=0, $ then
$A^{2}+4 A-p I=0$
$\Rightarrow \begin{bmatrix}74 & -20 \\ -8 & 26\end{bmatrix}+\begin{bmatrix}-32 & 20 \\ 8 & 16\end{bmatrix}+\begin{bmatrix}-p & 0 \\ 0 & -p\end{bmatrix}$
$=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}74-32-p & -20+20+0 \\ -8+8+0 & 26+16-p\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$
$\Rightarrow \begin{bmatrix}42-p & 0 \\ 0 & 42-p\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$
On comparing, we get
$42-p=0 \Rightarrow p=42$