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
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
Solution and Explanation
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=\(\frac{1}{4}\)\(\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}\)
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