Question

If A=\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)verify that A³-6A²+9A-4 I=0 and hence find A^-1 

Answer 1

A=\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)

A²=\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)

=\(\begin{bmatrix}4+1+1&-2-2-1&2+1+2\\-2-2-1&1+4+1&-1-2-2\\2+1+2&-1-2-2&1+1+4\end{bmatrix}\)

=\(\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\)

A³=A². A

=\(\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\)\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)

=\(\begin{bmatrix}12+5+5&-6-10-5&6+5+10\\-10-6-5&5+12+5&-5-6-10\\10+5+6&-5-10-6&5+5+12\end{bmatrix}\)

=\(\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}\)

Now  A³-6A²+9A-4 \(I\)
\(\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}\)-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\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

=\(\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}\)-\(\begin{bmatrix}36&-30&30\\-30&36&-30\\30&-30&36\end{bmatrix}\)+\(\begin{bmatrix}18&-9&9\\-9&18&-9\\9&-9&18\end{bmatrix}\)-\(\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}\)

=[40 -30 30 -30 40 -30 30 -30 40]-[40 -30 30 -30 40 -30 30 -30 40]=\(\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)
so  A³-6A²+9A-4I=0
(AAA)A^-1-6(AA)A^-1+9AA^-1-4IA^-1=0    [post multiplying by A^-1 as IAI≠0]
\(\Rightarrow\) AA(AA^-1)-6A(AA^-1)+9(AA^-1)=4(IA^-1)
\(\Rightarrow\) AAi-6AI+9I=4A^-1
\(\Rightarrow\) A²-6A+9I=4A^-1
\(\Rightarrow\) A^-1=\(\frac{1}{4}\)(A²-6A+9I)  ..(1)
A²-6A+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\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

=\(\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\)-\(\begin{bmatrix}12&-6&6\\-6&12&-6\\6&-6&12\end{bmatrix}\)+\(\begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix}\)

=\(\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}\)

From equation (1), we have: 

A^-1=\(\frac{1}{4}\)\(\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}\)