Question

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

• A. $\left[\begin{array}{cc}12& 4\\ 12& 4\end{array}\right]$
• B. $\left[\begin{array}{cc}12& -4\\ 4& 12\end{array}\right]$
• C. $\left[\begin{array}{cc}4& 12\\ 12& 4\end{array}\right]$
• D. $\left[\begin{array}{cc}4& 12\\ -12& -4\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}12& 4\\ 12& 4\end{array}\right]$

$A^2+2A+I=(A+I{)}^2$

Now, $A+I=\left[\begin{array}{cc}3& 1\\ 3& 1\end{array}\right]$

$(A+I{)}^2=A^2+2A+I=\left[\begin{array}{cc}3& 1\\ 3& 1\end{array}\right]\left[\begin{array}{cc}3& 1\\ 3& 1\end{array}\right]=\left[\begin{array}{cc}12& 4\\ 12& 4\end{array}\right]$