Question

If $A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$, f (x) = x² − 2x − 3, show that f (A) = 0

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ f\left(x\right)=x^2-2x-3 \\ \Rightarrow f\left(A\right)=A^2-2A-3I_2 \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}1+4& 2+2\\ 2+2& 4+1\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}5& 4\\ 4& 5\end{array}\right] \\ f\left(A\right)=A^2-2A-3I_2 \\ \Rightarrow f\left(A\right)=\left[\begin{array}{cc}5& 4\\ 4& 5\end{array}\right]-2\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]-3\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow f\left(A\right)=\left[\begin{array}{cc}5& 4\\ 4& 5\end{array}\right]-\left[\begin{array}{cc}2& 4\\ 4& 2\end{array}\right]-\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right] \\ \Rightarrow f\left(A\right)=\left[\begin{array}{cc}5-2-3& 4-4-0\\ 4-4-0& 5-2-3\end{array}\right] \\ \Rightarrow f\left(A\right)=\left[\begin{array}{cc}5-5& 0\\ 0& 5-5\end{array}\right] \\ \Rightarrow f\left(A\right)=0 \end{gathered}$$