Question

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

• A. $A^{-1}$
• B. $2A$
• C. $2A^{-1}$
• D. $\frac{3}{2}{A}^{-1}$

Answer 1

Answer: (A)

We have given, $A=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]$
$\therefore$ adj A $=\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]$ and $\left|A\right|=-1\left(-1\right)+1\cdot 1=2$

Hence, $A^{-1}=\frac{adjA}{|A|}=\frac{1}{2}\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]\dots \left(i\right)$

and $A^2=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]\cdot \left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]=\left[\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right]\dots \left(ii\right)$

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

$=\frac{1}{2}\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]=A^{-1}$ [using (i)]