Question

Show that the matrix $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$ satisfies the equation $A^2-4A+I=0$, where $I$ is $2\times 2$ identity matrix and $0$ is $2\times 2$ zero matrix. Using this equation,find $A^{-1}$.

Answer 1

We have , $A^2=A\cdot A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}4+3& 6+6\\ 2+2& 3+4\end{array}\right]=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]$

$4A=4\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}8& 12\\ 4& 8\end{array}\right]$

Hence, $A^2-4A+I=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]-\left[\begin{array}{cc}8& 12\\ 4& 8\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$

Now , $A^2-4A+I=0$

or $A.A\left(A^{-1}\right)-4A{A}^{-1}=-I{A}^{-1}$ (Post multiplying by $A^{-1}$ because $\left|A\right|\ne 0$)

or$A\left(A{A}^{-1}\right)-4I=-A^{-1}$

or $A-4I=-A^{-1}$ $[A\cdot A^{-1}=IandIA=AI=A]$

or $A^{-1}=4I-A=\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]-\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$

Hence, $A^{-1}=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$