If

A= \begin{bmatrix} 2 & 0 &1 \\ 2&1 & 3\\ 1&-1 &0 \end{bmatrix} then find (A^{2}-5A) 

 

 

 

 
 
 
 
 

Answers (1)

  given :\: A= \begin{bmatrix} 2 &0 &1 \\ 2& 1&3 \\ 1&-1 & 0 \end{bmatrix}
A^{2}= \begin{bmatrix} 2 &0 &1 \\ 2& 1&3 \\ 1&-1 & 0 \end{bmatrix} \begin{bmatrix} 2 &0 &1 \\ 2& 1&3 \\ 1&-1 & 0 \end{bmatrix}
A^{2}= \begin{bmatrix} 4+0+1 &0 +0-1&2+0+0 \\ 4+2+3& 0+1-3&2+3+0 \\ 2-2 -0&0-1-0 & 1-3+0 \end{bmatrix}
A^{2}= \begin{bmatrix} 5 & -1 &2 \\ 9& -2 & 5\\ 0& -1 &-2 \end{bmatrix}
and \: 5A= 5\begin{bmatrix} 2 &0 &1 \\ 2 & 1 &3 \\ 1 &-1 &0 \end{bmatrix}\Rightarrow \begin{bmatrix} 10 & 0 & 5\\ 10& 5 &15 \\ 5& -5 &0 \end{bmatrix}
\therefore A^{2}-5A= \begin{bmatrix} 5 & -1 &2 \\ 9&-2 &5 \\ 0 & -1 &-2 \end{bmatrix}\, - \begin{bmatrix} 10 & 0 &5 \\ 10&5 &15 \\ 5& -5 &0 \end{bmatrix}
\Rightarrow \begin{bmatrix} -5 & -1&-3 \\ -1 & -7 &-10 \\ -5&4 & -2 \end{bmatrix}
 

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