Question

If $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$, find $f\left(A\right)$ where $f\left(x\right)=2x^2-3x+5$.

Answer 1

Step1: Calculating the matrix $A^2$.

For $x=A$, $f\left(x\right)=2x^2-3x+5$ becomes $f\left(A\right)=2A^2-3A+5I$.

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

$\begin{array}{rcl}{A}^2& =& A\times A\\ & \Rightarrow & A^2=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\times \left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\\ & \Rightarrow & A^2=\left(\begin{array}{cc}1\left(1\right)+2\left(3\right)& 1\left(2\right)+2\left(4\right)\\ 3\left(1\right)+4\left(3\right)& 3\left(2\right)+4\left(4\right)\end{array}\right)\\ \therefore A^2& =& \left(\begin{array}{cc}7& 10\\ 15& 22\end{array}\right)\end{array}$

Step2: Evaluating $2A^2-3A+5I$.

The identity matrix of order 2 is defined as $I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$. Therefore, $5I=\left(\begin{array}{cc}5·1& 5·0\\ 5·0& 5·1\end{array}\right)=\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)$.

Substitute $\begin{array}{rcl}{A}^2& =& \left(\begin{array}{cc}7& 10\\ 15& 22\end{array}\right)\end{array}$, $5I=\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)$ and $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$ in $2A^2-3A+5I$ and simplify to obtain $f\left(A\right)$.

$\begin{array}{rcl}2A^2-3A+5I& =& 2\left(\begin{array}{cc}7& 10\\ 15& 22\end{array}\right)-3\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)+\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)\\ & \Rightarrow & 2A^2-3A+5I=\left(\begin{array}{cc}14& 20\\ 30& 44\end{array}\right)+\left(\begin{array}{cc}-3& -6\\ -9& -12\end{array}\right)+\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)\\ & \Rightarrow & 2A^2-3A+5I=\left(\begin{array}{cc}14-3+5& 20-6+0\\ 30-9+0& 44-12+5\end{array}\right)\\ \therefore 2A^2-3A+5I& =& \left(\begin{array}{cc}16& 14\\ 21& 37\end{array}\right)\end{array}$

Final Answer: The value is $f\left(A\right)=\left(\begin{array}{cc}16& 14\\ 21& 37\end{array}\right)$.