# If verify that A3 − 6A2 + 9A − 4I = O and hence find A−1

Question:

If $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$ verify that $A^{3}-6 A^{2}+9 A-4 I=0$ and hence find $A^{-1}$

Solution:

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

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

$=\left[\begin{array}{lll}4+1+1 & -2-2-1 & 2+1+2 \\ -2-2-1 & 1+4+1 & -1-2-2 \\ 2+1+2 & -1-2-2 & 1+1+4\end{array}\right]$

$=\left[\begin{array}{lll}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right]$

\begin{aligned} A^{3}=A^{2} A &=\left[\begin{array}{lll}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right]\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right] \\ &=\left[\begin{array}{ccc}12+5+5 & -6-10-5 & 6+5+10 \\ -10-6-5 & 5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12\end{array}\right] \\ &=\left[\begin{array}{ccc}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{array}\right] \end{aligned}

Now,

$A^{3}-6 A^{2}+9 A-4 I$

$=\left[\begin{array}{ccc}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{array}\right]-6\left[\begin{array}{ccc}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right]+9\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]-4\left[\begin{array}{llll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=$

$=\left[\begin{array}{ccc}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{array}\right]-\left[\begin{array}{ccc}36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36\end{array}\right]+\left[\begin{array}{ccc}18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18\end{array}\right]-\left[\begin{array}{cccc}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$

$=\left[\begin{array}{ccc}40 & -30 & 30 \\ -30 & 40 & -30 \\ 30 & -30 & 40\end{array}\right]-\left[\begin{array}{ccc}40 & -30 & 30 \\ -30 & 40 & -30 \\ 30 & -30 & 40\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

$\therefore A^{3}-6 A^{2}+9 A-4 I=O$

Now,

$A^{3}-6 A^{2}+9 A-4 I=O$

$\Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+9 A A^{-1}-4 I A^{-1}=O \quad$ [Post-multiplying by $A^{-1}$ as $\left.|A| \neq 0\right]$

$\Rightarrow A A\left(A A^{-1}\right)-6 A\left(A A^{-1}\right)+9\left(A A^{-1}\right)=4\left(I A^{-1}\right)$

$\Rightarrow A A I-6 A I+9 I=4 A^{-1}$

$\Rightarrow A^{2}-6 A+9 I=4 A^{-1}$

$\Rightarrow A^{-1}=\frac{1}{4}\left(A^{2}-6 A+9 I\right)$

$A^{2}-6 A+9 I$

$=\left[\begin{array}{lrl}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right]-6\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]+9\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

$=\left[\begin{array}{lll}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right]-\left[\begin{array}{ccc}12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12\end{array}\right]+\left[\begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]$

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

From equation (1), we have:

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