Question

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

• A. 64
• B. 42
• C. 36
• D. 24

Answer 1

Answer: (B)

Given,
$A=\left[\begin{array}{cc}-8& 5\\ 2& 4\end{array}\right]$
$A^2=\left[\begin{array}{cc}-8& 5\\ 2& 4\end{array}\right]\left[\begin{array}{cc}-8& 5\\ 2& 4<br/><br/>\end{array}\right]$
$=\left[\begin{array}{cc}64+10& -40+20\\ -16+8& 10+16\end{array}\right]$
$=\left[\begin{array}{cc}74& -20\\ -8& 26\end{array}\right]$
$4A=\left[\begin{array}{cc}-32& 20\\ 8& 16\end{array}\right]$
$-pI=\left[\begin{array}{cc}-p& 0\\ 0& -p\end{array}\right]$
Since, the matrix A satisfies the equation
$x^2+4x-p=0,$ then
$A^2+4A-pI=0$
$\Rightarrow \left[\begin{array}{cc}74& -20\\ -8& 26\end{array}\right]+\left[\begin{array}{cc}-32& 20\\ 8& 16\end{array}\right]+\left[\begin{array}{cc}-p& 0\\ 0& -p\end{array}\right]$
$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}74-32-p& -20+20+0\\ -8+8+0& 26+16-p\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}42-p& 0\\ 0& 42-p\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
On comparing, we get
$42-p=0\Rightarrow p=42$