Question

If matrix $A=\left[\begin{array}{ccc}1& 2& 3\\ -1& 0& 4\\ 6& 2& -1\end{array}\right]$, then the determinant of matrix $A^{500}-2A^{499}$ is $m$, then number of zeros at the end of $m$ is

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is C $0$

$A=\left[\begin{array}{ccc}1& 2& 3\\ -1& 0& 4\\ 6& 2& -1\end{array}\right]$
$|A|=-8+46-6=32$
Now,

$|A^{500}-2A^{499}|=|A^{499}(A-2I)|=|A^{499}|\cdot |A-2I|=|A{|}^{499}\cdot |A-2I|$ ....(1)
$|A-2I|=\left|\begin{array}{ccc}-1& 2& 3\\ -1& -2& 4\\ 6& 2& -3\end{array}\right|$
$=-(6-8)-2(3-24)+3(-2+12)=74$
$\Rightarrow |A^{500}-2A^{499}|=\left(32{)}^{499}\right(74)$
There is no factor of $5$ in this.
Hence, the number of zeros $=0$.

Hence, option C.