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Home  >>  CBSE XII  >>  Math  >>  Determinants
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\[ \text{If A = } \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, \text{show that } A^{2} -5A +7I = 0. \text{ Hence find } A^{-1}\]

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Toolbox:
  • A matrix is said to be singular if $|A|$ =0.
  • A matrix is said to be invertible only if $|A|\neq 0$.
  •  $A^{-1}=\frac{1}{|A|}adj \;A$
  • The adjoint of a square matrix A=[$a_{ij}]_{n\times n}$ is defined as the transpose of the matrix$ [A_{ij}]n\times n$
  • where $A_{ij}$ is the cofactor of the element $[a_{ij}].$
Let $A=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}$
 
Let us find $A^2=A.A$
 
Matrix multiplication can be done by multiplying the rows and corresponding columns.
 
$A^2=A.A=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}$
 
5A can be obtained by multiplying each element by 5
 
$5A=\begin{bmatrix}5\times3 & 5\times1\\5\times(-1) & 5\times 2\end{bmatrix}=\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}$
 
We know$ I=\begin{bmatrix}1& 0\\0 &1\end{bmatrix}$
 
$7I=7\begin{bmatrix}1& 0\\0 &1\end{bmatrix}=\begin{bmatrix}7& 0\\0 &7\end{bmatrix}$
 
Now $A^2-5A+7I=\begin{bmatrix}8& 5\\-5 &3\end{bmatrix}-\begin{bmatrix}15& 5\\-5 &10\end{bmatrix}+\begin{bmatrix}7& 0\\0 &7\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}8-15+7& 5-5+0\\-5-5+0 &3-10+7\end{bmatrix}=\begin{bmatrix}0& 0\\0 &0\end{bmatrix}$
 
Hence $A^2-5A+7I=0.$
 
now $A^2=5A-7I$ or $A^2-5A=-7I.$
 
A.A-5A=-7I.
 
Multiply $A^{-1}$ on both sides,
 
$A.A(A^{-1})-5A.A^{-1}=-7IA^{-1}$
 
$A(AA^{-1})-5(AA^{-1})=-7IA^{-1}$
 
Because $AA^{-1}=I$ and $IA^{-1}=A^{-1}$
 
We get,
 
$AI-5I=-7A^{-1}$
 
$A^{-1}=\frac{-1}{7}(AI-5I)$
 
$A^{-1}=\frac{1}{7}(5I-AI)$
 
Now substituting for I and A we get,
 
$A^{-1}=\frac{1}{7}\bigg(5\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}-\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\bigg)$
 
$A^{-1}=\frac{1}{7}\bigg(\begin{bmatrix}5 & 0\\0 & 5\end{bmatrix}-\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\bigg)$
 
$\;\;\;=\frac{1}{7}\begin{bmatrix}2 & -1\\1 & 3\end{bmatrix}$
 
Hence $A^{-1}=\frac{1}{7}\begin{bmatrix}2 & -1\\1 & 3\end{bmatrix}$

 

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

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