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Mathematics Question on Determinants

For the matrix A=[111 12−3 2−13]A=\begin{bmatrix}1&1&1\\\ 1&2&-3\\\ 2&-1&3\end{bmatrix}show that A3−6A2+5A+11I=OA^3−6A^2+5A+11I=O.Hence,find A−1.

Answer

A=[111 12−3 2−13]A=\begin{bmatrix}1&1&1\\\ 1&2&-3\\\ 2&-1&3\end{bmatrix}
A2=[111 12−3 2−13][111 12−3 2−13]A^2=\begin{bmatrix}1&1&1\\\ 1&2&-3\\\ 2&-1&3\end{bmatrix}\begin{bmatrix}1&1&1\\\ 1&2&-3\\\ 2&-1&3\end{bmatrix}
=[1+1+21+2−11−3+3 1+2−61+4+31−6−9 2−1+62−2−32+3+9]=[421 −38−14 7−314]=\begin{bmatrix}1+1+2& 1+2-1& 1-3+3\\\ 1+2-6& 1+4+3& 1-6-9\\\ 2-1+6& 2-2-3& 2+3+9\end{bmatrix}=\begin{bmatrix}4&2&1\\\ -3&8&-14\\\ 7&-3&14\end{bmatrix}
A3=[421 −38−14 7−314][111 12−3 2−13]A^3=\begin{bmatrix}4&2&1\\\ -3&8&-14\\\ 7&-3&14\end{bmatrix}\begin{bmatrix}1&1&1\\\ 1&2&-3\\\ 2&-1&3\end{bmatrix}
=[4+2+24+4−14−6+3 −3+8−28−3+16+14−3−24−42 7−3+287−6−147+9+12]=\begin{bmatrix}4+2+2& 4+4-1& 4-6+3\\\ -3+8-28& -3+16+14& -3-24-42\\\ 7-3+28& 7-6-14& 7+9+12\end{bmatrix}
=[871 −2327−69 32−1358]=\begin{bmatrix}8&7&1\\\ -23& 27& -69\\\ 32& -13& 58\end{bmatrix}
therefore A3−6A2+5A+11IA^3−6A^2+5A+11I
=[871 −2327−69 32−1358]−6[421 −38−14\7−314]+5[111 12−3 2−13]+11[100 010 001]=\begin{bmatrix}8&7&1\\\ -23& 27& -69\\\ 32& -13& 58\end{bmatrix}-6\begin{bmatrix}4&2&1\\\ -3&8&-14 \\\7&-3&14\end{bmatrix}+5\begin{bmatrix}1&1&1\\\ 1&2&-3\\\ 2&-1&3\end{bmatrix}+11\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix}
=[871 −2327−69 32−1358]−[24126 −1848−84\42−1884]+[555 510−15 10−515]+[1100 0110 0011]=\begin{bmatrix}8&7&1\\\ -23& 27& -69\\\ 32& -13& 58\end{bmatrix}-\begin{bmatrix}24&12&6\\\ -18&48&-84 \\\42&-18&84\end{bmatrix}+\begin{bmatrix}5&5&5\\\ 5&10&-15\\\ 10&-5&15\end{bmatrix}+\begin{bmatrix}11&0&0\\\ 0&11&0\\\ 0&0&11\end{bmatrix}
=[24126 −1848−84 42−1884]−[24126−1848−84 42−1884]=\begin{bmatrix}24& 12& 6\\\ -18& 48& -84\\\ 42& -18& 84\end{bmatrix}-\begin{bmatrix}24& 12& 6 \\\\-18& 48& -84\\\ 42& -18& 84\end{bmatrix}
=0=0
Thus A3−6A2+5A+11I=0A^3−6A^2+5A+11I=0
(AAA)A−1−6(AA)A−1+5AA−1+11IA−1=0(AAA)A^{-1}-6(AA)A^{-1}+5AA^{-1}+11IA^{-1}=0 [post multipying by A−1as∣A∣≠0A^{-1} as |A|≠0]
  ⟹  AA(AA−1)−6A(AA−1)+5(AA−1)=−11(IA−1)\implies AA(AA^{-1})-6A(AA^{-1})+5(AA^{-1})=-11(IA^{-1})
=>A2−6A+5I=−11A−1=>A^2-6A+5I=-11A^{-1}
  ⟹  A−1=−111(A2−6A+5I)...(1)\implies A^{-1}= \frac{-1}{11}(A^2-6A+5I) ...(1)
Now A2−6A+5IA^2-6A+5I
=[421 −38−14 7−314]−6[111 12−3 2−13]+5[100 010 001]=\begin{bmatrix}4&2&1\\\ -3&8&-14\\\ 7&-3&14\end{bmatrix}-6\begin{bmatrix}1&1&1\\\ 1&2&-3\\\ 2&-1&3\end{bmatrix}+5\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix}
=[421 −38−14 7−314]−[666 612−18 12−618]+[500 050 005]=\begin{bmatrix}4&2&1\\\ -3&8&-14\\\ 7&-3&14\end{bmatrix}-\begin{bmatrix}6&6&6\\\ 6&12&-18\\\ 12&-6&18\end{bmatrix}+\begin{bmatrix}5&0&0\\\ 0&5&0\\\ 0&0&5\end{bmatrix}
=[921 −313−14 7−319]−[666 612−18 12−618]=\begin{bmatrix}9&2&1\\\ -3& 13& -14\\\ 7& -3& 19\end{bmatrix}-\begin{bmatrix}6&6&6\\\ 6& 12& -18\\\ 12& -6& 18\end{bmatrix}
=[3−4−5 −914 −531]=\begin{bmatrix}3&-4&-5\\\ -9&1&4\\\ -5&3&1\end{bmatrix}
Form equation (1) we have
A−1=−111[3−4−5 −914 −531]A^{-1}=\frac{-1}{11}\begin{bmatrix}3&-4&-5\\\ -9&1&4\\\ -5&3&1\end{bmatrix}
=111[−345 9−1−4 5−3−1]=\frac{1}{11}\begin{bmatrix}-3&4&5\\\ 9&-1&-4\\\ 5&-3&-1\end{bmatrix}