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

Given $A=\begin{bmatrix}2 & 2 & 4\\4 & 2 & 4\\2 & 1 & 5\end{bmatrix},B=\begin{bmatrix}1 & 1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix}$,find BA and use this to solve the system of equations y+2z=7,x-y=3,2x+3y+4z=17

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If |A|$\neq 0$,then it is a non-singular matrix.
  • $A^{-1}=\frac{1}{|A|}adj \;A$
  • AX=B$\Rightarrow X=A^{-1}B.$
Given $A=\begin{bmatrix}2 & 2 & 4\\4 & 2& 4\\2 & 1 & 5\end{bmatrix}B=\begin{bmatrix}1 & 1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix}$
Let us find BA
Matrix multiplication can be done by multiplying rows of Matrix A by the columns of matrix B.
Therefore $BA=\begin{bmatrix}2+4+0 & 2-2+0 & -4+4+0\\4-12+2 &4+6-4 & 8-12+20\\0-4+4 & 0+2-2 & 0-4+10\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}6 & 0 & 0\\0 & 6& 0\\0 & 0 & 6\end{bmatrix}=6\begin{bmatrix}1 & 0 &0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$
BA=6I.
$\Rightarrow B.\frac{1}{6}A=I.$
The given system of equations are
y+2z=7
x-y=3
2x+3y+4z=17.
The given system of equation can be written in the form CX=D.
$\begin{bmatrix}1 & -1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\17\\7\end{bmatrix}$
Where $C=\begin{bmatrix}1 & -1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix},X=\begin{bmatrix}x\\y \\z\end{bmatrix}$ and $B=\begin{bmatrix}3\\17\\7\end{bmatrix}$
Now $|C|=\begin{bmatrix}1 & -1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix}$
Expanding along $R_1$ we get,
|C|=1($3\times 2-4\times 1)-(-1)(2\times 2-0)+0.$
$\;\;\;=6-4+4=6\neq 0$.
Therefore $C^{-1}$ exists.
Now C and B are same matrices and $B.\frac{1}{6}A$=Identity matrix.
$\Rightarrow \frac{1}{6}A$ is inverse of B and hence inverse of C.
Pre multiplying by $C^{-1}$ on both sides of CX=D.
We get,
$X=C^{-1}D.$
$\Rightarrow X=\frac{1}{6}A.D$
$X=\frac{1}{6}\begin{bmatrix}2 & 2 & -4\\-4 & 2 & -4\\2 & -1 & 5\end{bmatrix}\begin{bmatrix}3\\17 \\7\end{bmatrix}$
$\;\;=\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}$
$\Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}=\begin{bmatrix}2\\-1\\4\end{bmatrix}$
Therefore x=2,y=-1,z=4.
answered Mar 20, 2013 by sreemathi.v
 

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
...