logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Determinants
0 votes

Find the inverse of the matrix (if it exists): \[ \begin{bmatrix} 2&-2 \\ 4&3 \\ \end{bmatrix} \]

$\begin{array}{1 1} -14\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix} \\ 14\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix} \\ \frac{1}{14}\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix} \\ \frac{-1}{14}\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix} \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)A square matrix is said to be singular if $|A|$ =0.
  • (ii)A square matrix is invertible only if it is a non-singular matrix.
  • (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
  • (iv)the adjoint of a square matrix A=[$a_{ij}]_{n\times n}$
  • where $A_{ij}$ is the cofactor of the element $a_{ij}.$
  • Adjoint is denoted by adj A.
Let A=$\begin{bmatrix}2 & -2\\4 & 3\end{bmatrix}$
The value of the determinant ,
|A|=$(2\times 3)-(4\times -2)$
$\;\;\;=6+8=14.$
Since |A|$\neq 0$ inverse exists,
Now let us find the adjoint of the matrix.
$M_{11}=3$
$M_{12}=4$
$M_{21}=-2$
$M_{22}=2$
$A_{11}=(-1)^{1+1}.3=3.$
$A_{12}=(-1)^{1+2}.4=-4.$
$A_{21}=(-1)^{2+1}.(-2)=2.$
$A_{22}=(-1)^{2+2}.2=2.$
Hence (adj A)=$\begin{bmatrix}a_{11} & a_{21}\\a_{12} &a_{22}\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix}$
The inverse of the matrix is $A^{-1}=\frac{1}{|A|}adj \;A$
We have already found |A|=14.
Therefore inverse =$\frac{1}{14}\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix}$
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...