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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using matrices, solve the following system of equation \[ 2x+8y+5z=5\]\[x+y+z=-2\]\[x+2y-z=2\]

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Toolbox:
  • If |A|$\neq $ 0,then it is a non-singular matrix.
  • Hence it is invertible.
  • $A^{-1}=\frac{1}{|A|}.adj\; A$
  • X=$A^{-1}B.$
Given:
2x+8y+5z=5.
x+y+z=-2.
x+2y-z=2.
The given system of equation is of the form
AX=B.
(i.e)$\begin{bmatrix}2 & 8 & 5\\1 & 1 & 1\\1 & 2 & -1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\-2\\2\end{bmatrix}$
Therefore x=$A^{-1}B.$
To find $A^{-1}$,let us first see if [A] is singular or non-singular.
|A| can be determined by expanding along $R_1$
|A|=$2(1\times -1-2\times 1)-8(1\times -1-1\times -1)+5(1\times 2-1\times 1)$
$\;\;=2(-1-2)-8(-1-1)+5(2-1)$
$\;\;=-6+16+5=15\neq 0$
Since $|A| \neq 0$ [A] is non-singular.
Now let us find the adj A.
To find adj A,let us find the minors and cofactors of the elements of [A].
$M_{11}=\begin{vmatrix}1 & 1\\2 & -1\end{vmatrix}$=-1-2=-3.
$M_{12}=\begin{vmatrix}1 & 1\\1 & -1\end{vmatrix}$=-1-1=-2.
$M_{13}=\begin{vmatrix}1 & 1\\1 & 2\end{vmatrix}$=2-1=1.
$M_{21}=\begin{vmatrix}8 & 5\\2 & -1\end{vmatrix}$=-8-10=-18.
$M_{22}=\begin{vmatrix}2 & 5\\1 & -1\end{vmatrix}$=-2-5=-7.
$M_{23}=\begin{vmatrix}2 & 8\\1 & 2\end{vmatrix}$=4-8=-4.
$M_{31}=\begin{vmatrix}8 & 5\\1 & 1\end{vmatrix}$=8-5=3.
$M_{32}=\begin{vmatrix}2 & 5\\1 & 1\end{vmatrix}$=2-5=-3.
$M_{33}=\begin{vmatrix}2 & 8\\1 & 1\end{vmatrix}$=2-8=-6.
$A_{11}=(-1)^{1+1}$.-3=-3.
$A_{12}=(-1)^{1+2}$.-2=2.
$A_{13}=(-1)^{1+3}$.1=1.
$A_{21}=(-1)^{2+1}$.-18=18.
$A_{22}=(-1)^{2+2}$.-7=-7.
$A_{23}=(-1)^{2+3}$.-4=4.
$A_{31}=(-1)^{3+1}$.3=3.
$A_{32}=(-1)^{3+2}$.-3=3.
$A_{33}=(-1)^{3+3}$.-6=-6.
Now adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}-3 & 18 & 3\\2 & -7 & 3\\1 & 4 & -6\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}Adj \;A$,where |A|=15.
$A^{-1}=\frac{1}{15}\begin{bmatrix}-3 & 18 & 3\\2 & -7 & 3\\1 & 4 & -6\end{bmatrix}$
X=$A^{-1}B$,substituting for x,$A^{-1}$ and B,
$\begin{bmatrix}x\\y \\z\end{bmatrix}=1/15\begin{bmatrix}-3 & 18 & 3\\2 & -7 & 3\\1 & 4 & -6\end{bmatrix}\begin{bmatrix}5\\-2\\2\end{bmatrix}$
Matrix multiplication can be done by multiplying rows of matrix A by column of matrix B.
$\begin{bmatrix}x\\y\\z\end{bmatrix}=1/15\begin{bmatrix}-15-36+6\\10+14+6\\5-8-12\end{bmatrix}=1/15\begin{bmatrix}-45\\30\\-15\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-45/15\\30/15\\-15/15\end{bmatrix}=\begin{bmatrix}-3\\2\\-1\end{bmatrix}$
Hence x=-3,y=2,z=-1.
answered Mar 12, 2013 by sreemathi.v
edited Mar 28, 2013 by sreemathi.v
 

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