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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Solve the following system of equations by matrix method:x+2y+z=7,x+3z=11,2x-3y=1.

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Toolbox:
  • If the value of the determinant of a square matrix is not equal to zero,then it is a non-singular matrix.
  • If it is a non-singular matrix,then inverse exists.
  • $A^{-1}=\frac{1}{|A|}$adjoint of A
  • $A^{-1}B=X$
Step 1:
The given system of equation is of the form AX=B.
(i.e)$\begin{bmatrix}1 & 2& 1\\1 &0 & 3\\2 & -3 & 0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}7\\11\\1\end{bmatrix}$
Where $A=\begin{bmatrix}1 & 2&1\\1 & 0 & 3\\2 & -3 &0\end{bmatrix},X=\begin{bmatrix}x\\y\\z\end{bmatrix}$ and $B=\begin{bmatrix}7\\11\\1\end{bmatrix}$
Let us now find the determinant value of A
|A|=$\begin{bmatrix}1 & 2 & 1\\1 & 0 &3\\2 & -3 & 0\end{bmatrix}$
$\;\;=1(0+9)-2(0-6)+1(-3-0)$
$\;\;=9+12-3=18\neq 0.$
It is non-singular.Hence inverse exists.
Step 2:
Next let us find the adjoint of A
$A_{11}=(-1)_{1+1}\begin{vmatrix}0 & 3\\-3 & 0\end{vmatrix}$=9.
$A_{12}=(-1)_{1+2}\begin{vmatrix}1 & 3\\2 & 0\end{vmatrix}$=6.
$A_{13}=(-1)_{1+3}\begin{vmatrix}1 & 0\\2 & -3\end{vmatrix}$=-3.
$A_{21}=(-1)_{2+1}\begin{vmatrix}2 & 1\\-3 & 0\end{vmatrix}$=-3.
$A_{22}=(-1)_{2+2}\begin{vmatrix}1 & 1\\2 & 0\end{vmatrix}$=-2.
$A_{23}=(-1)_{2+3}\begin{vmatrix}1 & 2\\2 & -3\end{vmatrix}$=7.
$A_{31}=(-1)_{3+1}\begin{vmatrix}2 & 1\\0 & 3\end{vmatrix}$=6.
$A_{32}=(-1)_{3+2}\begin{vmatrix}1 & 1\\1 & 3\end{vmatrix}$=-2.
$A_{33}=(-1)_{3+3}\begin{vmatrix}0 & 3\\-3 & 0\end{vmatrix}$=-2.
Hence the adjoint of A is $\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
$\qquad\qquad\qquad\qquad=\begin{bmatrix}9 & -3 & 6\\6 & -2 & -2\\-3 & 7& -2\end{bmatrix}$
$A^{-1}=\frac{1}{18}\begin{bmatrix}9 & -3 & 6\\6 & -2 & -2\\-3 & 7 & -2\end{bmatrix}$
Step 3:
$A^{-1}B=X$,substituting for $A^{-1}$,B and X we get
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{18}\begin{bmatrix}9 & -3& 6\\6 &-2 & -2\\-3 & 7 & -2\end{bmatrix}\begin{bmatrix}7\\11\\1\end{bmatrix}$
$\qquad=\frac{1}{18}\begin{bmatrix}63-33+6\\42-22-2\\-21+77-2\end{bmatrix}=\begin{bmatrix}\frac{36}{18}\\\frac{18}{18}\\\frac{54}{18}\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$
Hence x=2,y=1,z=3.
answered Apr 5, 2013 by sreemathi.v
edited Apr 8, 2013 by sreemathi.v
 
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