logo

Ask Questions, Get Answers

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

Find the matrix A satisfying the matrix equation $\begin{bmatrix}2 & 1\\3 & 2\end{bmatrix}A\begin{bmatrix}3 & 2\\5 & 3\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 12\end{bmatrix}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
  • In order to find the inverse of a square matrix first we must find |A| if |A|$\neq 0$ (i.e) matrix A is non singular then only $A^{-1}$ exists.
Step1:
Let $B=\begin{bmatrix}2 & 1\\3 & 2\end{bmatrix}$ and $C=\begin{bmatrix}-3 & 2\\5 & -3\end{bmatrix}$
Let us find
|B|=[4-3]=1$\neq 0.$
|C|=[9-10]=-1$\neq 0.$
In order to find the inverse of a square matrix first we must find |A| if |A|$\neq 0$ (i.e) matrix A is non singular then only $A^{-1}$ exists.
$\Rightarrow B^{-1}$ and $C^{-1}$ exist.
The given matrix equation becomes
BAC=$I_2$
$\Rightarrow B^{-1}(BAC)C^{-1}$
$\Rightarrow B^{-1}I_2C^{-1}$
$\Rightarrow B^{-1}(BAC)C^{-1}=B^1I_2C^{-1}$
($B^{-1}B)ACC^{-1}=B^{-1}(I_2C^{-1})$
$I_2AI_2=B^{-1}C^{-1}$
$A=B^{-1}C^{-1}$
Step2:
Now the cofactors of the elements of B are
$B_{11}=2,B_{12}=-1,B_{12}=-3,B_{22}=2.$
adj B=transpose of $\begin{bmatrix}2 & -3\\-1 & 2\end{bmatrix}$
$\qquad\qquad\qquad=\begin{bmatrix}2 & -1\\-3 & 2\end{bmatrix}$
We get by changing the rows & column.
$B^{-1}=\frac{1}{|B|}.adj B=\begin{bmatrix}2 & -1\\-3 & 2\end{bmatrix}$
Again the co-factors of the elements of C are
$C_{11}=-3,C_{12}=-5,C_{21}=-2,C_{22}=-3.$
adj C=transpose of $\begin{bmatrix}-3 & -5\\-2 & -3\end{bmatrix}$
$\qquad\qquad\qquad=\begin{bmatrix}-3 & -2\\-5 & -3\end{bmatrix}$
$C^{-1}=\frac{1}{|C|}.adj C=\begin{bmatrix}3 & 2\\5 & 3\end{bmatrix}$
Step3:
We got
$A=B^{-1}C^{-1}$
$\;\;=\begin{bmatrix}2 & -1\\-3 & 2\end{bmatrix}\begin{bmatrix}3 & 2\\5 & 3\end{bmatrix}$
$\;\;=\begin{bmatrix}6-5 & 4-3\\-9+10 & -6+6\end{bmatrix}$
$\;\;=\begin{bmatrix}1 & 1\\1 & 0\end{bmatrix}$
answered Mar 21, 2013 by sharmaaparna1
 

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
...