# $\text{For the matrix A = } \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}, \text{ find the numbers a and b such that } A^{2} + aA + bI = O$

$\begin{array}{1 1} a=4,b=1 \\ a=-4,b=1 \\ a=-4,b=-1 \\ a=4,b=-1 \end{array}$

Toolbox:
• (i)A matrix is said to be singular if $|A|$ =0.
• (ii)A matrix is said to be invertible only if $|A|\neq 0$.
• (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
Let $A=\begin{bmatrix}3& 2\\1 & 1\end{bmatrix}$

$A^2=A.A$

$\;\;\;\;=\begin{bmatrix}3& 2\\1 & 1\end{bmatrix}\begin{bmatrix}3& 2\\1 & 1\end{bmatrix}$

Matrix multiplication can be done by multiplying the rows and corresponding columns.

Hence $A^2=\begin{bmatrix}3\times 3+2\times 1 & 3\times 2+1\times 2\\3\times 1+1\times 1 & 2\times 1+1\times 1\end{bmatrix}$

$\;\;\;\quad=\begin{bmatrix}9+2 & 6+2\\3+1 & 2+1 \end{bmatrix}$

$\;\;\;\quad=\begin{bmatrix}11 & 8\\4 & 3 \end{bmatrix}$

Given $A^2+aA+bI=0.$

Post multiplying $A^{-1}$ we get,

$(AA^{-1})+a(AA^{-1})+b(IA^{-1})$=0.

But we know $AA^{-1}=I$ and $IA^{-1}=A^{-1}.$

Hence,

$AI+aI+bA^{-1}=0.$

$\Rightarrow AI+aI=-bA^{-1}$ (But AI=A)

$A+aI=-bA^{-1}$

$A^{-1}=\frac{-1}{b}(A+aI)$----(1)

We also know $A^{-1}=\frac{1}{|A|}(adj A)$

The value of the determinant is

$|A|=3\times 1-(-2\times 1)$

$\;\;\;=3-2=1.$

Adjoint of A can be obtained by interchanging $a_{11}$ and $a_{22}$ and changing the symbols of $a_{12}$ and $a_{21}$

$adj A=\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}$

$A^{-1}=\frac{1}{|A|}(adj A)=\frac{1}{1}\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}$-----(2)

Now equating eq(1) and eq(2) we get,

$\frac{-1}{b}[A+aI]=\frac{1}{1}\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}$

$\Rightarrow \frac{1}{-b}\bigg(\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix}+\begin{bmatrix}a & 0\\0 & a\end{bmatrix}\bigg)=\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}$

On simplifying we get,

$\frac{-1}{b}\begin{bmatrix}3+a & 2\\1 & 1+a\end{bmatrix}=\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}$

$\begin{bmatrix}-(3+a)/b & -2/b\\-1/b &-( 1+a)/b\end{bmatrix}=\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}$

Now comparing the corresponding elements of the two matrices,

$\frac{-(3+a)}{b}=1\Rightarrow -3-a=b\Rightarrow a-b=3$-----(1)

$\frac{-2}{b}=-2\Rightarrow -2=-2b\Rightarrow b=1.$

Substituting for b in eq(1) we get,

a-1=3$\Rightarrow a=-4$

Hence the value of a=-4 and b=1

edited Feb 28, 2013