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Mathematics Question on Determinants

If A=[2−11−12−1\1−12]\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

A=[2−11−12−1\1−12]\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}

=[4+1+1−2−2−12+1+2−2−2−11+4+1−1−2−2\2+1+2−1−2−21+1+4]\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}

=[6−55−56−5\5−56]\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}

=[12+5+5−6−10−56+5+10−10−6−55+12+5−5−6−10\10+5+6−5−10−65+5+12]\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}

=[22−2121−2122−21\21−2122]\begin{bmatrix}22&-21&21\\\\-21&22&-21\\\21&-21&22\end{bmatrix}

Now A3-6A2+9A-4 II
[22−2121−2122−21\21−2122]\begin{bmatrix}22&-21&21\\\\-21&22&-21\\\21&-21&22\end{bmatrix}-6[6−55−56−5\5−56]\begin{bmatrix}6&-5&5\\\\-5&6&-5\\\5&-5&6\end{bmatrix}+9[2−11−12−1\1−12]\begin{bmatrix}2&-1&1\\\\-1&2&-1\\\1&-1&2\end{bmatrix}-4[100\010\001]\begin{bmatrix}1&0&0\\\0&1&0\\\0&0&1\end{bmatrix}

=[22−2121−2122−21\21−2122]\begin{bmatrix}22&-21&21\\\\-21&22&-21\\\21&-21&22\end{bmatrix}-[36−3030−3036−30\30−3036]\begin{bmatrix}36&-30&30\\\\-30&36&-30\\\30&-30&36\end{bmatrix}+[18−99−918−9\9−918]\begin{bmatrix}18&-9&9\\\\-9&18&-9\\\9&-9&18\end{bmatrix}-[400\040\004]\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]=[000\000\000]\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=14\frac{1}{4}(A2-6A+9I) ..(1)
A2-6A+9I

=[6−55−56−5\5−56]\begin{bmatrix}6&-5&5\\\\-5&6&-5\\\5&-5&6\end{bmatrix}-6[2−11−12−1\1−12]\begin{bmatrix}2&-1&1\\\\-1&2&-1\\\1&-1&2\end{bmatrix}+9[100\010\001]\begin{bmatrix}1&0&0\\\0&1&0\\\0&0&1\end{bmatrix}

=[6−55−56−5\5−56]\begin{bmatrix}6&-5&5\\\\-5&6&-5\\\5&-5&6\end{bmatrix}-[12−66−612−6\6−612]\begin{bmatrix}12&-6&6\\\\-6&12&-6\\\6&-6&12\end{bmatrix}+[900\090\009]\begin{bmatrix}9&0&0\\\0&9&0\\\0&0&9\end{bmatrix}

=[31−1\131−113]\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}