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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using matrices, solve the following system of equations: $ 4x+3y+3z=60, x+2y+3z=45 \: and \: 6x+2y+3z=70 $

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Toolbox:
  • If determinant value of the given matrix is not equal to zero, then it is a non-singular matrix
  • If it is a nonsingular matrix, then inverse exists
  • $A^{-1}=\frac{1}{|A|} adj (A)$
  • $A^{-1}B=X$
Step 1:
Given $4x+3y+3z=60,x+2y+3z=45,6x+2y+3z=70$
This system of the equation is of the form
$AX=B$
$(ie)\begin{bmatrix} 4 & 3 & 3 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix}$
Here $A=\begin{bmatrix} 4 & 3 & 3 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{bmatrix}\qquad X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}\;and \; B=\begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix}$
Let us find |A| by expanding $\begin{bmatrix} 4 & 3 & 3 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{bmatrix}$ along $R_1$
$|A|=4(2 \times 3 - 3 \times 2)-3(1 \times 3-6 \times 3)+3(1 \times 2- 6 \times 2)$
$=4(6-6)-3(3-18)+3(2-12)$
$=0+45-30$
$=15 \neq 0$
Hence this is a non singular matrix and $A^{-1}$ exists
Step 2:
let us now find the adjoint of A
$A_{11}=(-1)^{1+1} \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix}=6-6=0$
$A_{12}=(-1)^{1+2} \begin{vmatrix} 1 & 3 \\ 6 & 3 \end{vmatrix}=-(3-18)=+15$
$A_{13}=(-1)^{1+3} \begin{vmatrix} 1 & 2 \\ 6 & 2 \end{vmatrix}=2-12=-10$
$A_{21}=(-1)^{2+1} \begin{vmatrix} 3 & 3 \\ 2 & 3 \end{vmatrix}=-(9-6)=-3$
$A_{22}=(-1)^{2+2} \begin{vmatrix} 4 & 3 \\ 6 & 3 \end{vmatrix}=12-18=-6$
$A_{23}=(-1)^{2+3} \begin{vmatrix} 4 & 3 \\ 6 & 2 \end{vmatrix}=-(8-18)=10$
$A_{31}=(-1)^{3+1} \begin{vmatrix} 3 & 3 \\ 2 & 3 \end{vmatrix}=9-6=6$
$A_{32}=(-1)^{3+2} \begin{vmatrix} 4 & 3 \\ 1 & 3 \end{vmatrix}=-(12-3)=-9$
$A_{33}=(-1)^{3+3} \begin{vmatrix} 4 & 3 \\ 1 & 2 \end{vmatrix}=8-3=5$
Adj $A=\begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$
$=\begin{bmatrix} 0 & -3 & 6 \\ 15 & -6 & -9 \\ -10 & 10 & 5 \end{bmatrix}$
$A^{-1}=\frac{1}{|A|} adj (A),$ we know $|A|=15$
Step 3:
Hence $A^{-1}=\frac{1}{15}\begin{bmatrix} 0 & -3 & 6 \\ 15 & -6 & -9 \\ -10 & 10 & 5 \end{bmatrix}$
$AX=B \qquad => X=A^{-1}B$
Now Substituting for X, A^{-1} and B we get
$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{15}\begin{bmatrix} 0 & -3 & 6 \\ 15 & -6 & -9 \\ -10 & 10 & 5 \end{bmatrix} \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix}$
$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{15}\begin{bmatrix} 0-135+210 \\ 900-270-630 \\ -600+450+350 \end{bmatrix}=\begin{bmatrix} \frac{75}{15} \\ 0 \\ \frac{200}{15} \end{bmatrix}$
Therefore $ x=5,y=0,z=\frac{40}{3}$
answered Apr 5, 2013 by meena.p
 

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