Question

If $A=\left[\begin{array}{cc}5& 109\\ 1& 22\end{array}\right]$ , then det $(A^{100}-5A^{99})$ is

• A. $5$
• B. $1$
• C. $-109$
• D. $108$

Answer 1

Answer: (C)

$-109$

Given : $A=\left[\begin{array}{cc}5& 109\\ 1& 22\end{array}\right]$

Calculating determinant

$\left|A\right|=110-109=1$
$A^{100}-5A^{99}=A^{99}(A-5I)$
$|A^{100}-5A^{99}|={\left|A\right|}^{99}|A-5I|$
$=1.\left|\begin{array}{cc}0& 109\\ 1& 17\end{array}\right|$
$=1.(-109)$
$=-109$