Question

If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$ then $A^2-4A$

• A. $2I_3$
• B. $3I_3$
• C. $4I_3$
• D. $5I_3$

Answer 1

The correct option is D

$5I_3$

Find the value of ${\mathit{A}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathit{A}$:

Given that $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$

$A^2=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]$

$4A=4\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]=\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]$

Now, $A^2-4A=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]$

$$\begin{gathered} =\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right] \\ =5I_3 \end{gathered}$$

Hence, the correct option is D.