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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Using matrices, solve the following system of equations:$x+\frac{2}{y}+3xz=-1,2x-\frac{4}{y}-3xz=3,3x+\frac{6}{y}-2xz=4$

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Toolbox:
  • If the value of the determinant of a $3 \times 3$ square matrix is not equal to zero, then it is a non-singular matrix
  • If it is a nonsingular matrix, then inverse exists
  • $A^{-1}=\frac{1}{|A|} (adjoint A)$
  • $A^{-1}B=X$
Step 1:
Here the given system of equation is non linear equation, Let us first change into linear equation by substitution
Let $ \large\frac{1}{y}=v,\;and\;xz=p$
Now the system of equations become,
$x+2v+3p=-1$
$2x-4v-3p=3$
$3x+6v-2p=4$
The given system is of the form $AX=B$
where $A=\begin{bmatrix} 1 & 2 & 3 \\ 2 & -4 & -3 \\ 3 & 6 & -2 \end{bmatrix},\qquad X=\begin{bmatrix} x \\ v \\ p \end{bmatrix}\;and \; B=\begin{bmatrix} -1 \\ 3 \\ 4 \end{bmatrix}$
now first Let us find the determinant value of A
$|A|=\begin{vmatrix} 1 & 2 & 3 \\ 2 & -4 & -3 \\ 3 & 6 & -2 \end{vmatrix}$
$|A|=1(8+8)-2(-4+9)+3(12+12)$
$=26-10+72$
$=88$
$|A| \neq 0$
Hence it is a non singular matrix and $A^{-1}$ exists
Let us now find the adjoint of A
Step 2:
$A_{11}=(-1)^{1+1} \begin{vmatrix} -4 & -3 \\ 6 & -2 \end{vmatrix}=26$
$A_{12}=(-1)^{1+2} \begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix}=-5$
$A_{13}=(-1)^{1+3} \begin{vmatrix} 2 & -4 \\ 3 & 6 \end{vmatrix}=24$
$A_{21}=(-1)^{2+1} \begin{vmatrix} 2 & 3 \\ 6 & -2 \end{vmatrix}=22$
$A_{22}=(-1)^{2+2} \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix}=-11$
$A_{23}=(-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix}=0$
$A_{31}=(-1)^{3+1} \begin{vmatrix} 2 & 3 \\ -4 & -3 \end{vmatrix}=6$
$A_{32}=(-1)^{3+2} \begin{vmatrix} 1 & 3 \\ 2 & -3 \end{vmatrix}=9$
$A_{33}=(-1)^{3+3} \begin{vmatrix} 1 & 2 \\ 2 & -4 \end{vmatrix}=-8$
Hence Adj of A is $\begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$
$=\begin{bmatrix} 26 & 22 & 6 \\ -5 & -11 & 9 \\ 24 & 0 & -8 \end{bmatrix}$
$A^{-1}=\frac{1}{|A|} adj (A),$
we know $|A|=88$
Hence $A^{-1}=\frac{1}{88}\begin{bmatrix} 26 & 22 & 6 \\ -5 & -11 & 9 \\ 24 & 0 & -8 \end{bmatrix}$
Step 3:
$AX=B \qquad => X=A^{-1}B$
Now Substituting for X, $A^{-1}$ and B we get,
$\begin{bmatrix} x \\ v \\ p \end{bmatrix}=\large\frac{1}{88}\begin{bmatrix} 26 & 22 & 6 \\ -5 & -11 & 9 \\ 24 & 0 & -8 \end{bmatrix} \begin{bmatrix} -1 \\ 3 \\ 4 \end{bmatrix}$
$=\frac{1}{88}\begin{bmatrix} -26+66+24 \\ 5-33+36 \\ -24+0-32 \end{bmatrix}=\frac{1}{88}\begin{bmatrix} 64 \\ 8 \\ -56 \end{bmatrix}$
$\begin{bmatrix} x \\ v \\ p \end{bmatrix}=\frac{8}{88} \begin{bmatrix} 8 \\ 1 \\ 7 \end{bmatrix}=\frac{1}{11} \begin{bmatrix} 8 \\ 1 \\ 7 \end{bmatrix}$
Therefore $ x=\large \frac{8}{11},v=\frac{1}{11},\; p=\frac{-7}{11}$
(ie) $x=\large\frac{8}{11},v=\frac{1}{y}=\frac{1}{11}=>y=11,p=xz=\frac{-7}{11}=>\frac{8}{11}.z=\frac{7}{11}=>z=\frac{-7}{8}$
Therefore $x=\large\frac{8}{11},y=11\; and\; z=\frac{-7}{8}$
answered Apr 9, 2013 by meena.p
 

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