Question

If $A=\left[\begin{array}{ccc}1& 2& 3\\ -1& 1& 2\\ 2& 2& 4\end{array}\right]$ then $(A^2-5A)A^{-1}=$

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

Answer 1

The correct option is B $\left[\begin{array}{ccc}-4& 2& 3\\ -1& -4& 2\\ 2& 2& -1\end{array}\right]$

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

To find $(A^2-5A)A^{-1}$

$(A^2-5A)A^{-1}=A^2\cdot A^{-1}-5A{A}^{-1}$

$=A\cdot A{A}^{-1}-5I$

$=A-5I$

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

$(A^2-5A)A^{-1}=\left[\begin{array}{ccc}-4& 2& 3\\ -1& -4& 2\\ 2& 2& -1\end{array}\right]$

$\therefore (A^2-5A)A^{-1}=\left[\begin{array}{ccc}-4& 2& 3\\ -1& -4& 2\\ 2& 2& -1\end{array}\right]$.