Question

If $A=\left(\begin{array}{cc}6& 5\\ 7& 6\end{array}\right)$, show that $A^2-12A+I=0$.

Answer 1

Step1: Calculating the matrix $A^2$.

Multiply the matrix $A$ by itself to obtain the matrix $A^2$.

$\begin{array}{rcl}{A}^2& =& A\times A\\ & \Rightarrow & A^2=\left(\begin{array}{cc}6& 5\\ 7& 6\end{array}\right)\times \left(\begin{array}{cc}6& 5\\ 7& 6\end{array}\right)\\ & \Rightarrow & A^2=\left(\begin{array}{cc}6\left(6\right)+5\left(7\right)& 6\left(5\right)+5\left(6\right)\\ 7\left(6\right)+6\left(7\right)& 7\left(5\right)+6\left(6\right)\end{array}\right)\\ \therefore A^2& =& \left(\begin{array}{cc}71& 60\\ 84& 71\end{array}\right)\end{array}$

Step2: Evaluating $A^2-12A+I$.

The identity matrix of order 2 is defined as $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.

Substitute $\begin{array}{rcl}{A}^2& =& \left(\begin{array}{cc}71& 60\\ 84& 71\end{array}\right)\end{array}$, $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ and $A=\left(\begin{array}{cc}6& 5\\ 7& 6\end{array}\right)$ in $A^2-12A+I$ and simplify.

$\begin{array}{rcl}{A}^2-12A+I& =& \left(\begin{array}{cc}71& 60\\ 84& 71\end{array}\right)-12\left(\begin{array}{cc}6& 5\\ 7& 6\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ & \Rightarrow & A^2-12A+I=\left(\begin{array}{cc}71& 60\\ 84& 71\end{array}\right)-\left(\begin{array}{cc}72& 60\\ 84& 72\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ & \Rightarrow & A^2-12A+I=\left(\begin{array}{cc}71-72+1& 60-60+0\\ 84-84+0& 71-72+1\end{array}\right)\\ \therefore A^2-12A+I& =& \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\end{array}$

Conclusion: The relation $A^2-12A+I=0$ is proved.