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# $\text{For the matrix A = } \begin{bmatrix} 2&-1&1 \\ -1& 2&- 1 \\ 1 &-1& 2 \end{bmatrix}$ $\text{Show that } A^{3} - 6A^{2} + 9A - 4I = O. \text{ Hence, find } A^{-1}$

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Toolbox:
• (i)A matrix is said to be singular if $|A|$ =0.
• (ii)A matrix is said to be invertible only if $|A|\neq 0$.
• (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
• (iv)The adjoint of a square matrix A=[$a_{ij}]_{n\times n}$ is defined as the transpose of the matrix$[A_{ij}]n\times n$
• where $A_{ij}$ is the cofactor of the element $[a_{ij}].Adjoint of the matrix A is denoted by adj A.$
Given $A=\begin{bmatrix}2 & -1& 1\\-1 & 2& -1\\1 & -1 & 2\end{bmatrix}$
Let us find $A^2=A.A$
Matrix multiplication can be done by multiplying the rows and corresponding columns.
$A^2=A.A=\begin{bmatrix}2 & -1& 1\\-1 & 2& -1\\1 & -1 & 2\end{bmatrix}\begin{bmatrix}2 & -1& 1\\-1 & 2& -1\\1 & -1 & 2\end{bmatrix}$
$\;\;\;=\begin{bmatrix}4+1+1 & -2-2-1& 2+1+2\\-2-2-1 & 1+4+1& -1-2-2\\2+1+2 & -1-2-2 & 1+1+4\end{bmatrix}$
$\;\;\;=\begin{bmatrix}6 & -5& 5\\-5 & 6& -5\\5 & -5 & 6\end{bmatrix}$
$A^3=A^2.A=\begin{bmatrix}6 & -5& 5\\-5 & 6& -5\\5 & -5 & 6\end{bmatrix}\begin{bmatrix}2 & -1& 1\\-1 & 2& -1\\1 & -1 & 2\end{bmatrix}$
Multiplying as before
$A^3=\begin{bmatrix}12+5+5 & -6-10-5& 6+5+10\\-10-6-5 & 5+12+5& -5-6-10\\10+5+6 & -5-10-6 & 5+5+12\end{bmatrix}$
$\;\;\;=\begin{bmatrix}22 & -21& 21\\-21 & 22& -21\\21 & -21 & 22\end{bmatrix}$
Let us find $A^3-6A^2+9A-4I$
(i.e)$\begin{bmatrix}22 & -21& 21\\-21 & 22& -21\\21 & -21 & 22\end{bmatrix}-6\begin{bmatrix}6 & -5& 5\\-5 & 6& -5\\5 & -5 & 6\end{bmatrix}+9\begin{bmatrix}2 & -1& 1\\-1 & 2& -1\\1 & -1 & 2\end{bmatrix}-4\begin{bmatrix}1 & 0& 0\\0 & 1& 0\\0 & 0 & 1\end{bmatrix}$
=$\begin{bmatrix}22 & -21& 21\\-21 & 22& -21\\21 & -21 & 22\end{bmatrix}-\begin{bmatrix}36 & -30& 30\\-30 & 36& -30\\30 & -30 & 36\end{bmatrix}+\begin{bmatrix}18 & -9& 9\\-9 & 18& -9\\9 & -9 & 18\end{bmatrix}+\begin{bmatrix}4 & 0& 0\\0 & 4& 0\\0 & 0 & 4\end{bmatrix}$
On simplifying we get,
$\begin{bmatrix}22-36+18-4 & -21+30-9+0 & 21-30-9+0\\-21+30-9-0 & 22-36+18-4 & -21+30-9+0\\21-30+9-0 & -21+30-9+0 & 22-36+18+4\end{bmatrix}$
$\;\;\;=\begin{bmatrix} 0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$
Hence $A^3-6A^2+9A-4I$=0
Now let us find $A^{-1}$
Given $A^3-6A^2+9A-4I$=0
Premultiply $A^{-1}$
$(AAA)A^{-1})-6(AA)A^{-1}+9(A)A^{-1}-4IA^{-1}=0$
This can be written as
$AA(AA^{-1})-6A(AA^{-1})+9(AA^{-1})-4A^{-1}=0$
But we know $AA^{-1}=I$
$AA(I)-6A(I)+9I-4IA^{-1}=0.$
$\Rightarrow A^2-6A+9I=4A^{-1}$
$A^{-1}=\frac{1}{4}(A^2-6A+9I)$------(1)
Let us evaluate $A^2-6A+9I.$
$\;\;\;=\begin{bmatrix} 6 & -5 & 5\\-5 & 6 & -5\\5 & -5 & 6\end{bmatrix}-6\begin{bmatrix}2 & -1& 1\\-1 & 2& -1\\1 & -1 & 2\end{bmatrix}+9\begin{bmatrix}1 & 0& 0\\0 & 1& 0\\0 & 0 & 1\end{bmatrix}$
$\;\;\;=\begin{bmatrix}6-12+9 & -5+6+0& 5-6+0\\-5+6+0 & 6-12+9& -5+6+0\\5-6+0 & -5+6+0 & 6-12+9\end{bmatrix}$
$\;\;\;=\begin{bmatrix}3 & 1& -1\\1 & 3& 1\\-1 & 1 & 3\end{bmatrix}$
Now substituting the above matrix for $A^{-1}$(eq(1)) we get,
$A^{-1}=\frac{1}{4}\begin{bmatrix}3 & 1& -1\\1 & 3& 1\\-1 & 1 & 3\end{bmatrix}$
answered Mar 5, 2013