Show that for the matrix    A= \begin{bmatrix} 1 &1 & 1\\ 1& 2& -3\\ 2& -1 & 3 \end{bmatrix},A^{3}-6A^{2}+5A+11\: I= 0. 
Hence,find A^{-1}.

 

 

 

 
 
 
 
 

Answers (1)

given   A= \begin{bmatrix} 1 & 1 & 1\\ 1& 2 & -3\\ 2& -1 & 3 \end{bmatrix}
           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}
Now putting A^{3},A^{2}\: in\: 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}
     \Rightarrow \begin{bmatrix} 8-24 & 7-12 & 1-6\\ -23+18& 27-48 &-69+84 \\ 32-42& -13+18 & 58-84 \end{bmatrix}+\begin{bmatrix} 5+11 & 5+0 &5+0 \\ 5+0& 10+1 &-15+0 \\ 10-0&-5+0 & 15-11 \end{bmatrix}
   \Rightarrow \begin{bmatrix} -16& -5 & -5\\ -5& -21 & 15\\ -10& 5 & -26 \end{bmatrix}+\begin{bmatrix} 16& 5 & 5\\ 5& 21 & -15\\ 10& -5 & 26 \end{bmatrix}
   \Rightarrow \begin{bmatrix} 0 & 0 & 0\\ 0 &0 & 0\\ 0&0 &0 \end{bmatrix}= 0 \: \: Hence \: proved
we have, A^{3}-6A^{2}+5A+11I= 0
Multiplying by A^{-1}  on both sides,
A^{-1} \left ( A^{3} -6A^{2}+5A+11\, I\right )= 0\cdot A^{-1}
\Rightarrow A^{3}\cdot A^{-1}-6A^{2}A^{-1}+5A\cdot A^{-1}+11IA^{-1}= 0
\Rightarrow A^{2}\left ( AA^{-1} \right )-6A\left ( AA^{-1} \right )+5\left ( AA^{-1} \right )+11IA^{-1}= 0

\Rightarrow A^{2}\left ( AA^{-1} \right )-6A\left ( A\, A^{-1} \right )+5\left ( A\, A^{-1} \right )+11IA^{-1}= 0

\Rightarrow A^{2}I-6AI+5I+11A^{-1}= 0
 \left [ 11IA^{-1}= 11A^{-1} \right ]\left [ \because AA^{-1}=I \right ]

\Rightarrow A^{2}I-6AI+5I+11A^{-1}= 0\: \: \left [ \because AA^{-1}= I\, \& \: 11IA^{-1} = 11A^{-1}\right ]
\Rightarrow A^{2}-6A+5I+11\, A^{-1}= 0
\Rightarrow 11\, A^{-1}= -A^{2}+6A-5I
\Rightarrow A^{-1}= \frac{1}{11}\left ( -A^{2} +6A-5I\right )
A^{-1}= \frac{-\left ( A^{2} -6A+5I\right )}{11}
A^{-1}=\frac{1}{11}\left ( \begin{bmatrix} 4 & 2 &1 \\ -3&8 &-14 \\ 7& -3& 14 \end{bmatrix} \right )-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}\left ( \begin{bmatrix} 4 & 2 &1 \\ -3&8 &-14 \\ 7&-3 &14 \end{bmatrix} \right )-\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}
A^{-1}= \frac{-1}{11}\left ( \begin{bmatrix} 4-6 & 2-6 &1-6 \\ -3-6&8-12 &-14+18 \\ 7+12& -3+6 & -14+18 \end{bmatrix}+\begin{bmatrix} 5 & 0 &0 \\ 0&5 &0 \\ 0 & 0 & 5 \end{bmatrix} \right )
A^{-1}= \frac{-1}{11}\begin{bmatrix} -2 &-4 &-5 \\ -9& -4 &4 \\ -5&3 &-4 \end{bmatrix}+\begin{bmatrix} 5 & 0 &0 \\ 0 & 5 & 0\\ 0 & 0 & 5 \end{bmatrix}
A^{-1}= \frac{-1}{11}\begin{bmatrix} 3 & -4 &-5 \\ -9 & 1 & 4\\ -5 & 3 & 1 \end{bmatrix}
A^{-1}= \frac{1}{11}\begin{bmatrix} -3 & 4 &5 \\ 9 & -1 &- 4\\ 5 & -3 & -1 \end{bmatrix}
         

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