Question

Let $A=\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]$ satisfies $A^n=A^{n-2}+A^2-I$ for $n\ge 3.$ And trace of a square matrix $X$ is equal to the sum of elements in its principal diagonal. Further consider a matrix ${\cup }_{3\times 3}$ with its column as ${\cup }_1,{\cup }_2,{\cup }_3$ such that $A^{50}{\cup }_1=\left[\begin{array}{c}1\\ 25\\ 25\end{array}\right],A^{50}{\cup }_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],A^{50}{\cup }_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ Then,

The value of $|A^{50}|$ equals

• A. $0$
• B. $1$
• C. $-1$
• D. $25$

Answer 1

The correct option is B $1$

$A^n-A^{n-2}=A^2-I$
$\Rightarrow A^{50}=A^{48}+A^2-I$

Further,
$A^{48}=A^{46}+A^2-I$
$A^{46}=A^{44}+A^2-I$
$⋮⋮⋮$
$A^4=A^2+A^2-I$
_________________
$A^{50}=25A^2-24I$

Here,
$A^2=\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]$
$=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]$
$\Rightarrow A^{50}=\left[\begin{array}{ccc}25& 0& 0\\ 25& 25& 0\\ 25& 0& 25\end{array}\right]-24\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}1& 0& 0\\ 25& 1& 0\\ 25& 0& 1\end{array}\right]$

$\therefore |A^{50}|=1$