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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Show that the given system of equations are consistent and non infinite number of a solutions $ 2x+3y-z=7 \\ x+2y-z=4 \\ 3x-y+4z=5$

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Toolbox:
  • If the value of the determinant of square matrix is zero, then it is a non-singular matrix
  • If it is a singular matrix, then $A^{-1}$ does not exists
  • $|A|=0$ and $(adj A)$ B=0,then the system consistent and has infinite number of solutions.
Step 1:
Given $ 2x+3y-z=7, \: x+2y-z=4, \: 3x-y+4z=5 $
The given system of the equation is of the form $AX=B$
where $A=\begin{bmatrix} 2 & 3 & -1 \\ 1 & 2 & -1 \\ 3 & -1 & 4 \end{bmatrix}\qquad X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}\;and \; B=\begin{bmatrix} 7 \\ 4 \\ 5 \end{bmatrix}$
Let us first find the determinant value of A (ie) $A^{-1}$
$A=\begin{vmatrix} 2 & 3 & -1 \\ 1 & 2 & -1 \\ 3 & -1 & 4 \end{vmatrix}$
$|A|=2(8 - 1) - 3(4+3)-1(-1-6)$
$=14-21+7=0$
Since $|A|=0$ it is a singular matrix
$A^{-1}$ does not exist
Next let us find the adjoint of A
Step 2:
$A_{11}=(-1)^{1+1} \begin{vmatrix} 2 & -1 \\ -1 & 4 \end{vmatrix}=7$
$A_{12}=(-1)^{1+2} \begin{vmatrix} 1 & -1 \\ 3 & 4 \end{vmatrix}=-7$
$A_{13}=(-1)^{1+3} \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix}=-7$
$A_{21}=(-1)^{2+1} \begin{vmatrix} 3 & -1 \\ -1 & 4 \end{vmatrix}=-11$
$A_{22}=(-1)^{2+2} \begin{vmatrix} 2 & -1 \\ 3 & 4 \end{vmatrix}=11$
$A_{23}=(-1)^{2+3} \begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix}=11$
$A_{31}=(-1)^{3+1} \begin{vmatrix} 3 & -1 \\ 2 & -1 \end{vmatrix}=-1$
$A_{32}=(-1)^{3+2} \begin{vmatrix} 2 & -1 \\ 1 & -1 \end{vmatrix}=1$
$A_{33}=(-1)^{3+3} \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix}=1$
Adj $A=\begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$
$=\begin{bmatrix} 7 & -11 & -1 \\ -7 & 11 & 1 \\ -7 & 11 & 1 \end{bmatrix}$
Let us find $(adj A).B$
$\begin{bmatrix} 7 & -11 & -1 \\ -7 & 11 & 1 \\ -7 & 11 & 1 \end{bmatrix}=\begin{bmatrix} 7 \\ 4 \\ 5 \end{bmatrix}=\begin{bmatrix} 49-44-5 \\ -49+44+5 \\ -49+44+5 \end{bmatrix}$
=$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
Step 3:
Therefore the system of equation is consistent and has infinitely many no of solution
Now put $z=k,$ and write the first two equations are
$2x+3y=7+k; x+2y=4+k$
The reduced system can be written as
$\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end {bmatrix}=\begin{bmatrix} 7+k \\ 4+k \end{bmatrix}$
$=>\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}^{-1}\begin{bmatrix} 7+k \\ 4+k \end{bmatrix}$
Inverse of $\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
Therefore $\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 7+k \\ 4+k \end{bmatrix}=\begin{bmatrix} 14+2k-12-3k \\ -7-k+8+2k \end{bmatrix}$
$=\begin{bmatrix} 2-k \\ k+1 \end{bmatrix}$
Hence $ x=2-k,y=k+1,and \; z=k$
answered Apr 10, 2013 by meena.p
 

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