Solve the system of the following equations: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $\frac{1}{x}=p, \frac{1}{y}=q, \frac{1}{z}=r$. The given system of equations becomes: $2p+3q+10r=4$ $4p-6q+5r=1$ $6p+9q-20r=2$ This system can

Vedclass · Accepted Answer

$x=2, y=3, z=5$

Vedclass · Answer

(C) Let $\frac{1}{x}=p, \frac{1}{y}=q, \frac{1}{z}=r$. The given system of equations becomes: $2p+3q+10r=4$ $4p-6q+5r=1$ $6p+9q-20r=2$ This system can be written as $AX=B$,where $A=\begin{bmatrix} 2 & 3 & 10 \ 4 & -6 & 5 \ 6 & 9 & -20 \end{bmatrix}, X=\begin{bmatrix} p \ q \ r \end{bmatrix}, B=\begin{bmatrix} 4 \ 1 \ 2 \end{bmatrix}$. Calculating the determinant $|A|$: $|A| = 2(120-45) - 3(-80-30) + 10(36+36) = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200$. Since $|A| 
eq 0$,the inverse $A^{-1}$ exists. Calculating the adjoint of $A$: $A_{11} = 120-45 = 75, A_{12} = -(-80-30) = 110, A_{13} = 36+36 = 72$ $A_{21} = -(-60-90) = 150, A_{22} = -40-60 = -100, A_{23} = -(18-18) = 0$ $A_{31} = 15+60 = 75, A_{32} = -(10-40) = 30, A_{33} = -12-12 = -24$ $A^{-1} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \ 110 & -100 & 30 \ 72 & 0 & -24 \end{bmatrix}$. $X = A^{-1}B = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \ 110 & -100 & 30 \ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \ 1 \ 2 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 300+150+150 \ 440-100+60 \ 288+0-48 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 600 \ 400 \ 240 \end{bmatrix} = \begin{bmatrix} 1/2 \ 1/3 \ 1/5 \end{bmatrix}$. Thus,$p=1/2, q=1/3, r=1/5$. Therefore,$x=2, y=3, z=5$.

Solve the system of the following equations: $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$,$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$,and $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$.

Similar Questions

If $\begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 0 & -6 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix}$,then $(x, y, z) = $

The following system of equations $x+y+z=9$,$2x+5y+7z=52$,$x+7y+11z=77$ has

The number of values of $\alpha$ for which the system of equations: $x+y+z=\alpha$,$\alpha x+2 \alpha y+3 z=-1$,and $x+3 \alpha y+5 z=4$ is inconsistent,is:

If $x=\alpha, y=\beta, z=\gamma$ is the solution of the system of equations $2x+3y+z=-1$,$3x+y+z=4$,and $x-3y-2z=1$,then the value of $\beta$ is:

Consider the system of linear equations $x+y+z=4\mu$,$x+2y+2\lambda z=10\mu$,and $x+3y+4\lambda^2 z=\mu^2+15$,where $\lambda, \mu \in \mathbb{R}$. Which one of the following statements is $NOT$ correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module