Question

If $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]$ show that $A^3-6A^2+5A+11I=0$. Hence find $A^{-1}$

Answer 1

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]$

$A^2=\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]$

$A^3=\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]$

Now,

$A^3-6A^2+5A+11I$

$=\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]+(-6)\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]+5\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]+11\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]+\left[\begin{array}{ccc}-24& -12& -6\\ 18& -48& 84\\ -42& 18& -84\end{array}\right]+\left[\begin{array}{ccc}5& 5& 5\\ 5& 10& -15\\ 10& -5& 15\end{array}\right]+\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& 11\end{array}\right]$

$=\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]+\left[\begin{array}{ccc}-8& -7& -1\\ 23& -27& 69\\ -32& 13& -58\end{array}\right]$

$=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=0$

Now,

$A^3-6A^2+5A+11I=0$

Operating $A^{-1}$ on both sides

$A^2-6A+5I+11A^{-1}=0$ $(A{A}^{-1}=I)$

$11A^{-1}=-A^2+6A-5I$

$11A^{-1}=-\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]+6\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]-5\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$11A^{-1}=\left[\begin{array}{ccc}-3& 4& 5\\ 9& -1& -4\\ 5& -3& -1\end{array}\right]$

$A^{-1}=\left[\begin{array}{ccc}\frac{-3}{11}& \frac{4}{11}& \frac{5}{11}\\ \frac{9}{11}& \frac{-1}{11}& \frac{-4}{11}\\ \frac{5}{11}& \frac{-3}{11}& \frac{-1}{11}\end{array}\right]$