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  • For the matrix A = 1 1 1 1 2 -3 2 -1 3 Show that A^3-6A^2+5A+11 I =O. Hence, find A^-1.

Q : 15     For the matrix  \small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}   Show that \small A^3-6A^2+5A+11I=O     Hence, find \small A^-^1.
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Answers (1)

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Given matrix: \small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix};

To show: \small A^3-6A^2+5A+11I=O

Finding each term:

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}

= \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}

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}

= \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+42 \end{bmatrix}

= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}

So now we have, \small A^3-6A^2+5A+11I

= \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}

= \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}

= \begin{bmatrix} 8-24+5+11 &7-12+5 &1-6+5 \\ -23+18+5&27-48+10+11 &-69+84-15 \\ 32-42+10&-13+18-5 & 58-84+15+11 \end{bmatrix}

= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = 0

Now finding the inverse of A;

Post-multiplying by A^{-1} as, |A| \neq 0

\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +5AA^{-1}+11IA^{-1} = 0

\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +5(AA^{-1})=- 11IA^{-1}

\Rightarrow A^{2}-6A +5I=- 11A^{-1}

A^{-1} = \frac{-1}{11}(A^{2}-6A+5I)                                        ...................(1)

Now,

From equation (1) we get;

A^{-1} = \frac{-1}{11}( \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})

 

A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4-6+5 &&2-6 &&1-6 \\ -3-6 &&8-12+5 &&-14+18 \\ 7-12 &&-3+6 && 14-18+5 \end{bmatrix}

 

A^{-1} = \frac{-1}{11}( \begin{bmatrix} 3 &&-4 &&-5 \\ -9 &&1 &&4 \\ -5 &&3 && 1 \end{bmatrix}

 

 

Posted by

Divya Prakash Singh

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