Question

If $A=\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]$, then which of the following statements is/are correct?

I) $A^n=A;\forall n\in N$
II) $A^4=I$
III) $A^{-1}=A^3$

• A. All three I, II and III
• B. Only I and II
• C. Only II and III
• D. only I

Answer 1

The correct option is C Only II and III

Given:

$A=\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]$
$A^2=\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]=\left[\begin{array}{ccc}3& -4& 4\\ 0& -1& 0\\ -2& 2& -3\end{array}\right]$
$A^3=A^{2}A=\left[\begin{array}{ccc}3& -4& 4\\ 0& -1& 0\\ -2& 2& -3\end{array}\right]\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]=\left[\begin{array}{ccc}1& -1& 0\\ -2& 3& -4\\ -2& 3& -3\end{array}\right]$
$A^4=A^{3}A=\left[\begin{array}{ccc}1& -1& 0\\ -2& 3& -4\\ -2& 3& -3\end{array}\right]\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$
Since, $A^4=I$

$\therefore$ Statement II is correct
Since, $A^2,A^3,A^4...\ne A$
So, $A^n\ne A$ for all $n\in N$
$\therefore$ Statement I is incorrect.
Now, we will find $A^{-1}$
Here, $|A|=1$. Hence, $A^{-1}$ exists.
$A^4=I$
$\Rightarrow A^{-1}{A}^4=A^{-1}I$
$\Rightarrow A^{-1}A{A}^3=A^{-1}I$
$\Rightarrow A^3=A^{-1}$

$\therefore$ Statement III is also correct.

Hence, option C.