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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & =\left[\begin{array}{rr}64+10 & -40+20 \ -16+8 & 10+16\end{array}ight] \ & =\left[\begin{array}{ll}74 & -20 \ -8 & 26\end{array}ight] \ & 4 A=\left[\begin{array}{rr}-32 & 20 \ 8 & 16\end{array}ight] \ & -p l=\left[\begin{array}{cc}-p & 0 \ 0 & -p\end{array}ight] \ & \end{aligned}$ Since, the matrix $A$ satisfies the equation $ \begin{aligned} & x^2+4 x-p=0, 	ext { then } \ & A^2+4 A-p l=0 \ & \Rightarrow {\left[\begin{array}{rr} 74 & -20 \ -8 & 26 \end{array}ight]+\left[\begin{array}{rr} -32 & 20 \ 8 & 16 \end{array}ight]+\left[\begin{array}{cc} -p & 0 \ 0 & -p \end{array}ight] } \ & \Rightarrow {\left[\begin{array}{ll} 74-32-p & -20+20+0 \ {\left[\begin{array}{cc} 0 & 0 \ -8+8+0 & 26+16-p \end{array}ight]=\left[\begin{array}{ll} 0 & 0 \end{array}ight]} \ 0 & 0 \end{array}ight] } \ & \Rightarrow {\left[\begin{array}{cc} 42-p & 0 \ 0 & 42-p \end{array}ight]=\left[\begin{array}{ll} 0 & 0 \ 0 & 0 \end{array}ight] } \end{aligned} $ On comparing, we get $ 42-p=0 \Rightarrow p=42 $

If $A=\left[\begin{array}{rr}-8 & 5 \\ 2 & 4\end{array}\right]$ satisfies the equation $x^2+4 x-p=0$, then…

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