Question

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

• A. $\left[\begin{array}{ccc}3& 0& 0\\ 0& 1& -1\\ 0& 5& 4\end{array}\right]$
• B. $\left[\begin{array}{ccc}1& 2& -1\\ -3& 1& 1\\ 3& 5& -4\end{array}\right]$
• C. $\left[\begin{array}{ccc}3& 1& -1\\ -3& 1& -1\\ 3& 5& 4\end{array}\right]$
• D. $\left[\begin{array}{ccc}3& -1& 1\\ 3& -1& 1\\ 3& 5& 4\end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{ccc}3& 1& -1\\ -3& 1& -1\\ 3& 5& 4\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]$
$A^2=\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]=\left[\begin{array}{ccc}4& 3& 0\\ -3& 2& -2\\ 6& 4& 5\end{array}\right]$
Now, $A^2-A=\left[\begin{array}{ccc}4& 3& 0\\ -3& 2& -2\\ 6& 4& 5\end{array}\right]-\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]$
$=\left[\begin{array}{ccc}3& 1& -1\\ -3& 1& -1\\ 3& 5& 4\end{array}\right]$