Question

If $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right]$, then

• A. $A^3=I$
• B. $A^{-1}=A^2$
• C. $A^n=A,$ $\forall$ $n\ne 4$
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $A^3=I$
B $A^{-1}=A^2$

Given $A=\left|\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right|$

$A^2=\left|\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right|\left|\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right|=\left|\begin{array}{ccc}0& 0& 1\\ 0& -1& 2\\ 1& -1& 1\end{array}\right|$

${\Rightarrow A}^3=\left|\begin{array}{ccc}0& 0& 1\\ 0& -1& 2\\ 1& -1& 1\end{array}\right|\left|\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right|=\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|=I$

$\Rightarrow A\left(A^2\right)=A^2\left(A\right)=I$

$A^2$ is the inverse of $\{\left|A\right|\ne 0\Rightarrow A^{-1}exists\}$

Also,

$A^4=A^{3}A=IA=A$ but $A^5=A^{4}A=A^2\ne A$