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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Solve the system of equations: $\large \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=$$4.\quad $ $\large \frac{4}{x}-\frac{6}{y}+\frac{5}{z}$$=1.\quad $ $\large \frac{6}{x}+\frac{9}{y}-\frac{20}{z}$$=2.$

This question is appeared in model paper 2012
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Toolbox:
  • A matrix is said to be singular if |A|=0.
  • A matrix is said to be invertible if |A|$\neq 0.$
  • $A^{-1}=\frac{1}{|A|}(adj A)$
  • A system of linear equations can be expressed of the form $AX=B,$ then $X=A^{-1}B$.
Given:
$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4.$\[\] $\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1.$ \[\] $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2.$
 
Let $\frac{1}{x}=a,\frac{1}{y}=b,\frac{1}{z}=c.$
 
2a+3b+10c=4
 
4a-6b+5c=1.
 
6a+9b-20c=2
 
This can be written in the form AX=B.
 
$\begin{bmatrix}2 & 3 & 10\\4 & -6 & 5\\6 & 9 & -20\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}4\\1\\2\end{bmatrix}$
 
Where $\begin{bmatrix}2 & 3 & 10\\4 & -6 & 5\\6 & 9 & -20\end{bmatrix} X=\begin{bmatrix}a\\b\\c\end{bmatrix} B=\begin{bmatrix}4\\1\\2\end{bmatrix}$
 
Let us first see if 'A' is singular or non-singular .To evaluate the value of the determinants,let us expand along $R_1$
 
$|A|=2(-6\times -20-5\times 9)-3(4\times -20-6\times 5)+10(4\times 9-6\times -6)$
 
$\;\;\;\;=2(120-45)-3(-80-30)+10(36+36)$
 
$\;\;\;\;=150+330+720=1200.$
 
$\;\;\;\;\neq 0.$
 
Hence it is a non-singular matrix.
 
$A^{-1}\;exists.$
 
Let us now find the adj of A.
 
The cofactors are
 
$M_{11}=\begin{vmatrix}-6& 5\\9 & -20\end{vmatrix}=120-45=75.$
 
$M_{12}=\begin{vmatrix}4& 5\\6 & -20\end{vmatrix}=-80-30=-110.$
 
$M_{13}=\begin{vmatrix}4& -6\\6 & 9\end{vmatrix}=36+36=72.$
 
$M_{21}=\begin{vmatrix}3& 10\\9 & -20\end{vmatrix}=-60-90=-150.$
 
$M_{22}=\begin{vmatrix}2& 10\\6 & -20\end{vmatrix}=-40-60=-100.$
 
$M_{23}=\begin{vmatrix}2& 3\\6 & 9\end{vmatrix}=18-18=0.$
 
$M_{31}=\begin{vmatrix}3& 10\\-6 & 5\end{vmatrix}=15+60=75.$
 
$M_{32}=\begin{vmatrix}2& 10\\4 & 5\end{vmatrix}=10-40=-30.$
 
$M_{33}=\begin{vmatrix}2& 3\\4 & -6\end{vmatrix}=-12-12=-24.$
 
$A_{11}=(-1)^{1+1}.75=75.$
 
$A_{12}=(-1)^{1+2}.(-110)=110.$
 
$A_{13}=(-1)^{1+3}.72=72.$
 
$A_{21}=(-1)^{2+1}.(-150)=150.$
 
$A_{22}=(-1)^{2+2}.(-100)=-100.$
 
$A_{23}=(-1)^{2+3}.0=0.$
 
$A_{31}=(-1)^{3+1}.75=75.$
 
$A_{32}=(-1)^{3+2}.(-30)=30.$
 
$A_{33}=(-1)^{3+3}.-24=-24.$
 
adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}75 & 150 & 75\\110 & -100 & 30\\72 & 0 & -24\end{bmatrix}$
 
We know |A|=1200.
 
$A^{-1}=\frac{1}{|A|}adj A.$
 
$\;\;\;=\frac{1}{1200}\begin{bmatrix}75 & 150 & 75\\110 & -100 & 30\\72 & 0 & -24\end{bmatrix}$
 
We know X=$A^{-1}B$
 
Therefore $\begin{bmatrix}a\\b\\c\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}75 & 150 & 75\\110 & -100 & 30\\72 & 0 & -24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}$
 
Matrix multiplication can be done by multiplying the rows of Matrix A by column of B.
 
Therefore $\begin{bmatrix}a\\b\\c\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}300+150+150\\440-100+60\\280+0+48\end{bmatrix}$
 
$\;\;\;=\frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}=\begin{bmatrix}600/1200\\600/1200\\240/1200\end{bmatrix}$
 
$\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}1/2\\1/3\\1/5\end{bmatrix}$
 
Therefore a=1/2$\rightarrow x=2.$
 
b=1/3$\rightarrow y=3.$
 
c=1/5$\rightarrow z=5.$

 

answered Mar 4, 2013 by sreemathi.v
edited Mar 4, 2013 by sreemathi.v
 

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