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# If A=$\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix}$ find the values of a and b, such that $A^2+aA+bI=0.$

$\begin{array}{1 1}(A)\;a=4,b=-1\\(B)\;a=-1,b=4\\(C)\;a=-1,b=-4\\(D)\;a=-4,b=1\end{array}$

Can you answer this question?

Here $A=\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix}$
$A^2=A.A=\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix}\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}3\times 3+2\times 1 & 3\times 2+2\times 1\\1\times 3+1\times 1 & 1\times 2+1\times 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}9+2 & 6+2\\ 3+1 & 2+1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}11 & 8\\ 4 & 3\end{bmatrix}$
$A^2+aA+bI=0$
$\begin{bmatrix}11 & 8\\ 4 & 3\end{bmatrix}+a\begin{bmatrix}3 & 2\\ 1 & 1\end{bmatrix}+b\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}$
$\begin{bmatrix}11+3a+b & 8+2a\\4+a &3+a+b\end{bmatrix}=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}$
11+3a+b=0-----(1)
8+2a=0------(2)
4+a=0-------(3)
3+a+b=0-----(4)
From equation (3) we have
4+a=0
a=-4.
From equation (4) we have
3+a+b=0
3-4+b=0.
-1+b=0
b=1.
answered Apr 4, 2013