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  • If A = [1 0 2 0 2 1], prove that A^3 - 6A^2 + 7A + 2I = 0

Q16.    If

 A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}

prove that A^3 - 6A^2 + 7A + 2I = 0.

Answers (1)

best_answer

A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}

First, find the square of matrix A and then multiply it with A to get the cube of matrix A

A\times A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}\times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}

A^{2} = \begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9 \end{bmatrix}

A^{2} = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}

A^{3}=A^{2}\times A

A^{2}\times A = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}   \times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}

A^{3} = \begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39 \end{bmatrix}

A^{3} = \begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}

I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}

    \therefore     A^3 - 6A^2 + 7A + 2I = 0

         L.H.S :    

       \begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}- 6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}+7 \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}+2 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}

       =\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}  - \begin{bmatrix}30&0&48\\12&24&30\\48&0&78 \end{bmatrix} + \begin{bmatrix}7&0&14\\0&14&7\\14&0&21 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 &0 \\ 0 & 0 & 2 \end{bmatrix}

      =\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2 \end{bmatrix}

     =\begin{bmatrix}30-30&0&48-48\\12-12&24-24&30-30\\48-48&0&78-78 \end{bmatrix}

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

Hence, L.H.S = R.H.S   

i.e.A^3 - 6A^2 + 7A + 2I = 0.

 

 

 

 

 

 

Posted by

seema garhwal

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