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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Solve system of linear equations, using matrix method:\[\] \[x-y+2z=7\] \[3x+4y-5z=-5\] \[2x-y+3z=12\]

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Toolbox:
  • A matrix is said to be singular if |A|= 0.
  • A matrix is said to be invertible if |A|$\neq 0$.
  • If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
  • Using this we can solve the system of equation which has unique solution.
This given system can be written in the form AX=B.
$\begin{bmatrix}2 & 3&3\\1 & -2&1\\3 &-1 & -2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\-4\\3\end{bmatrix}$
Where $A=\begin{bmatrix}2 & 3& 3\\1 &-2&1\\3 &-1 &-2\end{bmatrix}\;X=\begin{bmatrix}x\\y\\z\end{bmatrix}\;B=\begin{bmatrix}5\\-4\\3\end{bmatrix}$
Let us now find if A is singular or non-singular.The value of determinant A,|A| can be found by expanding along $R_1$
$|A|=2(-2\times -2 -1\times -1)-3(1\times -2-1\times 3)+3(1\times -1-3\times -2)$
$\;\;\;=2(4+1)-3(-2-3)+3(-1+6)$
$\;\;\;=10+15+15=40\neq 0$
Hence it is a non-singular matrix.
Therefore inverse exists
We know $A^{-1}=\frac{1}{|A|}(adj \;A)$
Let us find the (adj A) by finding the minors and cofactors.
$M_{11}=\begin{vmatrix}4 & -5\\-1 & 3\end{vmatrix}=12-5=7$
$M_{12}=\begin{vmatrix}3 &-5\\2 & 3\end{vmatrix}=9+10=19$
$M_{13}=\begin{vmatrix}3 &4\\2 & -1\end{vmatrix}=-3-8=-11$
$M_{21}=\begin{vmatrix}-1 &2\\-1 & 3\end{vmatrix}=-3+2=-1$
$M_{22}=\begin{vmatrix}1 &2\\2 & 3\end{vmatrix}=3-4=-1$
$M_{23}=\begin{vmatrix}1 &-1\\2 & -1\end{vmatrix}=-1+2=1$
$M_{31}=\begin{vmatrix}-1 &2\\4 & -5\end{vmatrix}=5-8=-3$
$M_{32}=\begin{vmatrix}1 &2\\3 & -5\end{vmatrix}=-5-6=-11$
$M_{33}=\begin{vmatrix}1 &-1\\3 & 4\end{vmatrix}=4+3=7$
$A_{11}=(-1)^{1+1}.7=7$
$A_{12}=(-1)^{1+2}.19=-19$
$A_{13}=(-1)^{1+3}.-11=-11$
$A_{21}=(-1)^{2+1}.-1=1$
$A_{22}=(-1)^{2+2}.-1=-1$
$A_{23}=(-1)^{2+3}.1=-1$
$A_{31}=(-1)^{3+1}.-3=-3$
$A_{32}=(-1)^{3+2}.-11=11$
$A_{33}=(-1)^{3+3}.7=7$
Therefore adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}7 & 1 & -3\\-19 & -1 & 11\\-11 & -1 & 7\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}(adj \;A)$ We know |A|=4
$A^{-1}=\frac{1}{4}\begin{bmatrix}7 & 1 & -3\\-19 & -1 & 11\\-11 & -1 & 7\end{bmatrix}$
We know $X=A^{-1}B$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}7 & 1 & -3\\-19 & -1 & 11\\-11 & 1 & 7\end{bmatrix}\begin{bmatrix}7 \\-5\\12\end{bmatrix}$
We can do matrix multiplication by multiplying the rows of matrix A by the column of matrix B
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}49-5-36\\-133+5+132\\-77+5+84\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$
Hence x=2,y=1,z=3.
answered Mar 5, 2013 by vijayalakshmi_ramakrishnans
 

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