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Home  >>  CBSE XII  >>  Math  >>  Determinants
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\[ \text{For the matrix A = } \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}, \text{ find the numbers a and b such that } A^{2} + aA + bI = O \]

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

Can you answer this question?
 
 

1 Answer

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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

 

answered Feb 26, 2013 by sreemathi.v
edited Feb 28, 2013 by vijayalakshmi_ramakrishnans
 

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