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Q : 14      For the matrix \small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix} , find the numbers \small a and \small b such that A^2+aA+bI=0.

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Given \small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix} then we have the relation A^2+aA+bI=O 

So, calculating each term;

A^2 = \begin{bmatrix} 3& 2\\ 1& 1 \end{bmatrix}\begin{bmatrix} 3&2 \\ 1& 1 \end{bmatrix} = \begin{bmatrix} 9+2 &6+2 \\ 3+1&2+1 \end{bmatrix} = \begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}

therefore  A^2+aA+bI=O;

=\begin{bmatrix}11 &8 \\ 4& 3 \end{bmatrix} + a\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + b \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}

\begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}

So, we have equations; 

11+3a+b = 0,\ 8+2a = 0   and 4+a = 0,and\ \ 3+a+b = 0

We get a = -4\ and\ b= 1.

 

Posted by

Divya Prakash Singh

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