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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Using matrix method ,solve the system of equations 3x+2y-2z=3,x+2y+3z=6,2x-y+z=2.

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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:3x+2y-2z=3
$\qquad$x+2y+3z=6.
$\qquad$2x-y+z=2.
This system of equations is of the form
AX=B.
$\begin{bmatrix}3 & 2 & -2\\1 & 2 & 3\\2 & -1 & 1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\6\\2\end{bmatrix}$
Where $A=\begin{bmatrix}3 & 2 & -2\\1 & 2 & 3\\2 & -1 & 1\end{bmatrix},X=\begin{bmatrix}x\\y \\z\end{bmatrix}$ and $B=\begin{bmatrix}3\\6\\2\end{bmatrix}$
$A^{-1}B=X.$
To find $A^{-1}$,let us find if A is a singular or a non-singular matrix.
To check if A is singular or non-singular,let us find the value of the determinant.
By expanding along $R_1$,we get
|A|=3(2$\times 1-3\times (-1))-2(1\times 1-3\times 2)+(-2)(1\times -1-2\times 2)$
$\;\;=3(2+3)-2(1-6)-2(-1-4)$
$\;\;=15+10+0=35\neq 0$.
Hence inverse exists.
Let us next find the cofactors of the elements of matrix A
$A_{11}=(-1)^{1+1}\begin{vmatrix}2 & 3\\-1 & 1\end{vmatrix}$=2+3=5.
$A_{12}=(-1)^{1+2}\begin{vmatrix}1 & 3\\2 & 1\end{vmatrix}$=-(1-6)=-(-5)=5.
$A_{13}=(-1)^{1+3}\begin{vmatrix}1 & 2\\2 & -1\end{vmatrix}$=-1-4=-5.
$A_{21}=(-1)^{2+1}\begin{vmatrix}2 & -2\\-1 & 1\end{vmatrix}$=2-2=0.
$A_{22}=(-1)^{2+2}\begin{vmatrix}3 & -2\\2 & 1\end{vmatrix}$=3+4=7.
$A_{23}=(-1)^{2+3}\begin{vmatrix}3 & 2\\2 & -1\end{vmatrix}$=-(-3-4)=-(-7)=7.
$A_{31}=(-1)^{3+1}\begin{vmatrix}2 & -2\\2 & 3\end{vmatrix}$=6+4=10.
$A_{32}=(-1)^{3+2}\begin{vmatrix}3 & -2\\1 & 3\end{vmatrix}$=-(9+2)=-11.
$A_{33}=(-1)^{3+3}\begin{vmatrix}3 & 2\\1 & 2\end{vmatrix}$=6-2=4.
$Adj\;A=\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
$\qquad=\begin{bmatrix}5 & 0 & 10\\5 & 7 & -11\\-5 & 7 & 4\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}adj\;A$ we know |A|=35.
Therefore $A^{-1}=\frac{1}{35}\begin{bmatrix}5 & 0 & 10\\5 & 7 & -11\\-5 & 7 & 4\end{bmatrix}$
X=$A^{-1}B.$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{35}\begin{bmatrix}5 & 0 &10\\5 & 7 & -11\\-5 & 7 & 4\end{bmatrix}\begin{bmatrix}3\\6\\2\end{bmatrix}$
Matrix multiplication can be done by multiplying the columns of matrix A with column of matrix B.
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{35}\begin{bmatrix}15+0+20\\15+42-22\\-15+42+8\end{bmatrix}$
$\qquad=\frac{1}{35}\begin{bmatrix}35\\35\\35\end{bmatrix}=\begin{bmatrix}35/35\\35/35\\35/35\end{bmatrix}$
$\begin{bmatrix}x\\y \\z\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}$
Hence x=1,y=1,z=1.
answered Mar 20, 2013 by sreemathi.v
 

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