Question

If $A=\left(\begin{array}{cc}3& -5\\ -4& 2\end{array}\right)$, show that $A^2-5A-14I=0$.

Answer 1

Step1: Calculation of the matrix $A^2$.

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

$\begin{array}{rcl}{A}^2& =& A\times A\\ & \Rightarrow & A^2=\left(\begin{array}{cc}3& -5\\ -4& 2\end{array}\right)\times \left(\begin{array}{cc}3& -5\\ -4& 2\end{array}\right)\\ & \Rightarrow & A^2=\left(\begin{array}{cc}3\left(3\right)-5\left(-4\right)& 3\left(-5\right)-5\left(2\right)\\ -4\left(3\right)+2\left(-4\right)& -4\left(-5\right)+2\left(2\right)\end{array}\right)\\ \therefore A^2& =& \left(\begin{array}{cc}29& -25\\ -20& 24\end{array}\right)\end{array}$

Step2: Evaluation of $A^2-5A-14I$.

The identity matrix of order 2 is defined as $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$. Therefore, $14I=\left[\begin{array}{cc}14·1& 14·0\\ 14·0& 14·1\end{array}\right]=\left[\begin{array}{cc}14& 0\\ 0& 14\end{array}\right]$.

Substitute $\begin{array}{rcl}{A}^2& =& \left(\begin{array}{cc}29& -25\\ -20& 24\end{array}\right)\end{array}$, $14I=\left[\begin{array}{cc}14& 0\\ 0& 14\end{array}\right]$ and $A=\left(\begin{array}{cc}3& -5\\ -4& 2\end{array}\right)$ in $A^2-5A-14I$ and simplify.

$\begin{array}{rcl}{A}^2-5A-14I& =& \left(\begin{array}{cc}29& -25\\ -20& 24\end{array}\right)-5\left(\begin{array}{cc}3& -5\\ -4& 2\end{array}\right)-\left(\begin{array}{cc}14& 0\\ 0& 14\end{array}\right)\\ & \Rightarrow & A^2-5A-14I=\left(\begin{array}{cc}29& -25\\ -20& 24\end{array}\right)+\left(\begin{array}{cc}-15& 25\\ 8& -10\end{array}\right)-\left(\begin{array}{cc}14& 0\\ 0& 14\end{array}\right)\\ & \Rightarrow & A^2-5A-14I=\left[\begin{array}{cc}29-15-14& -25+25+0\\ -20+8+0& 24-10-14\end{array}\right]\\ \therefore A^2-5A-14I& =& \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Conclusion: The equation $A^2-5A-14I=0$ is verified.