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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using matrices, solve the following system of equation \[ x+y-z=3\] \[ 2x+3y+z=10 \] \[ 3x-y-7z=1\]

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Toolbox:
  • If A is a non singular matrix,then its inverse exists.Hence X=$A^{-1}B.$
  • $A^{-1}=\frac{1}{|A|}adj \;A$
Step 1:
The given equation are of the form AX=B.
(i.e)$\begin{bmatrix}1 & 1 & -1\\2 & 3 & 1\\3 & -1 & -7\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\10\\1\end{bmatrix}$
Where A=$\begin{bmatrix}1 & 1 & -1\\2 & 3 & 1\\3 & -1 & -7\end{bmatrix};X=\begin{bmatrix}x\\y\\z\end{bmatrix}\; and\; B=\begin{bmatrix}3\\10 \\1\end{bmatrix}$
We know $X=A^{-1}B.$
To determine $A^{-1}$ ,let us find the adj A.To determine adj A,let us find the minors and cofactors of the respective elements.
$M_{11}=\begin{vmatrix}3 & 1\\-1 & -7\end{vmatrix}=-21+1=-20.$
$M_{12}=\begin{vmatrix}2 & 1\\3 & -7\end{vmatrix}=-14-3=-17.$
$M_{13}=\begin{vmatrix}2 & 3\\3 & -1\end{vmatrix}=-2-9=-11$.
$M_{21}=\begin{vmatrix}1 & -1\\-1 & -7\end{vmatrix}=-7-1=-8.$
$M_{22}=\begin{vmatrix}1 & -1\\3 & -7\end{vmatrix}=-7+3=-4.$
$M_{23}=\begin{vmatrix}1 & 1\\3 & -1\end{vmatrix}=-1-3=-4.$
$M_{31}=\begin{vmatrix}1 & -1\\3 & 1\end{vmatrix}=1+3=4.$
$M_{32}=\begin{vmatrix}1 & -1\\2 & 1\end{vmatrix}=1+2=3.$
$M_{33}=\begin{vmatrix}1 & 1\\2 & 3\end{vmatrix}=3-2=1.$
$A_{11}=(-1){1+1}(-20)=-20.$
$A_{12}=(-1){1+2}(-17)=17.$
$A_{13}=(-1){1+3}(-11)=-11$
$A_{21}=(-1){2+1}(-8)=8.$
$A_{22}=(-1){2+2}(-4)=-4.$
$A_{23}=(-1){2+3}(-4)=4.$
$A_{31}=(-1){3+1}(4)=4.$
$A_{32}=(-1){3+2}(3)=-3$
$A_{33}=(-1){3+3}(1)=1.$
$adj A=\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}-20 & 8 & 4\\17 & -4 & -3\\-11 & 4 & 1\end{bmatrix}$
Step 2:
The value of determinant can be obtained by expanding along $R_1$
$|A|=1(3\times -7-1\times -1)-1(2\times -7-3\times 1)+(-1)(2\times -1-3\times 3)$
$\;\;\;\;=-21+1+14+3+2+9.$
$\;\;\;\;=8.$
We know $A^{-1}=\frac{1}{|A|}adj A$
$\qquad\qquad=1/8\begin{bmatrix}-20 & 8 & 4\\17 & -4 & -3\\-11 & 4 & 1\end{bmatrix}$
Step 3:
We know $X=A^{-1}B.$
$\begin{bmatrix}x\\y \\z\end{bmatrix}=1/8\begin{bmatrix}-20 & 8 & 4\\17 & -4 & -3\\-11 & 4 & 1\end{bmatrix}\begin{bmatrix}3\\10 \\1\end{bmatrix}$
Matrix multiplication be done by multiplying rows of matrix A by column of B.
$\begin{bmatrix}x\\y\\z\end{bmatrix}=1/8\begin{bmatrix}-60+80+4\\51-40-3\\-33+40+1\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=1/8\begin{bmatrix}24\\8\\8\end{bmatrix}=\begin{bmatrix}24/8\\8/8\\8/8\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\1\\1\end{bmatrix}$
Hence x=3,y=1 and z=1.
answered Mar 11, 2013 by sreemathi.v
edited Apr 9, 2013 by sreemathi.v
 
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