Question

If$A=\left[\begin{array}{cc}2& -1\\ 1& 2\end{array}\right]$then $A^2+2A-3I=$

• A. $\left[\begin{array}{cc}4& -6\\ 6& 4\end{array}\right]$
• B. $0$
• C. $\left[\begin{array}{cc}-6& 2\\ 2& 6\end{array}\right]$
• D. $5I$

Answer 1

The correct option is A

$\left[\begin{array}{cc}4& -6\\ 6& 4\end{array}\right]$

Find the value of ${\mathit{A}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{2}\mathit{A}\mathbf{-}\mathbf{3}\mathit{I}$:

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

$$\begin{gathered} A^2+2A-3I=\left[\begin{array}{cc}2& -1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}2& -1\\ 1& 2\end{array}\right]+2\left[\begin{array}{cc}2& -1\\ 1& 2\end{array}\right]-3\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A^2+2A-3I=\left[\begin{array}{cc}4-1& -2-2\\ 2+2& -1+4\end{array}\right]+\left[\begin{array}{cc}4& -2\\ 2& 4\end{array}\right]-\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right] \\ \Rightarrow A^2+2A-3I=\left[\begin{array}{cc}3& -4\\ 4& 3\end{array}\right]+\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right] \\ \Rightarrow A^2+2A-3I=\left[\begin{array}{cc}4& -6\\ 6& 4\end{array}\right] \end{gathered}$$

Hence, the correct option is (A).