Question

If $A+I=\left[\begin{array}{cc}3& -2\\ 4& 1\end{array}\right]$, then $(A+I)(A-I)=$

• A. $\left[\begin{array}{cc}-5& -4\\ 8& -9\end{array}\right]$
• B. $\left[\begin{array}{cc}-5& -4\\ -8& 9\end{array}\right]$
• C. $\left[\begin{array}{cc}5& 4\\ 8& 9\end{array}\right]$
• D. $\left[\begin{array}{cc}-5& -4\\ -8& -9\end{array}\right]$

Answer 1

The correct option is A

$\left[\begin{array}{cc}-5& -4\\ 8& -9\end{array}\right]$

Explanation for the correct option:

Step 1. Find the value of $\mathbf{(}\mathit{A}\mathbf{+}\mathit{I}\mathbf{)}\mathbf{(}\mathit{A}\mathbf{-}\mathit{I}\mathbf{)}\mathbf{:}$

Given: $A+I=\left[\begin{array}{cc}3& -2\\ 4& 1\end{array}\right]$

We know that,$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Now,

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

Step 2. Multiply the value of $\left(A+I\right)$ and $\left(A-I\right)$

$\begin{array}{rcl}\therefore (A+I)(A-I)& =& \left[\begin{array}{cc}3& -2\\ 4& 1\end{array}\right]\left[\begin{array}{cc}1& -2\\ 4& -1\end{array}\right]\\ & =& \left[\begin{array}{cc}-5& -4\\ 8& -9\end{array}\right]\\ & =& \left[\begin{array}{cc}-5& -4\\ 8& -9\end{array}\right]\end{array}$

Hence, the option $\mathbf{\left(}\mathit{A}\mathbf{\right)}$ is correct.