Solve the following equations
$5 x - 6 y + 4 z = 15$
$7 x + 4 y - 3 z = 19$
$2 x + y + 6 z = 46$

Open in App

Solution

The above equations can be written in matrix form as
$⎡ ⎢ ⎣ \begin{matrix} 5 & ; - 6 & ; 4 7 & ; 6 & ; - 3 2 & ; 1 & ; 6 \end{matrix} ⎤ ⎥ ⎦$ $⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ = $⎡ ⎢ ⎣ \begin{matrix} 151946 \end{matrix} ⎤ ⎥ ⎦$ or AX = B
|A| = D = $∣ ∣ ∣ ∣ \begin{matrix} 5 & ; - 6 & ; 4 7 & ; 4 & ; - 3 2 & ; 1 & ; 6 \end{matrix} ∣ ∣ ∣ ∣$
Apply $C_{1} - 2 C_{2}, C_{3} - 6 C_{2}$ to make two zeros in $R_{3}$
$D = ∣ ∣ ∣ ∣ \begin{matrix} 17 & ; - 6 & ; 40 - 1 & ; 4 & ; - 27 0 & ; 1 & ; 0 \end{matrix} ∣ ∣ ∣ ∣ = - ∣ ∣ ∣ \begin{matrix} 17 & ; 40 - 1 & ; - 27 \end{matrix} ∣ ∣ ∣ = - [- 459 + 40] = 419 \neq 0$ .
$∴$ The matrix A is non-singular or rank of matrix A is
3. We will have a unique solution and the equations are consistent.
By Crammer's rule
$\frac{x}{D_{1}} = \frac{y}{D_{2}} = \frac{z}{D_{3}} = \frac{1}{D}$
Where $D_{1}$ is obtained from D by replacing the first column of D by b's i.e. $15, 19, 46$
$D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 15 & ; - 6 & ; 4 19 & ; 4 & ; - 3 46 & ; 1 & ; 6 \end{matrix} ∣ ∣ ∣ ∣ = 15 (27) - 19 (- 40) + 46 (2) = 1257$
$∴ \frac{x}{1257} = \frac{y}{1676} = \frac{z}{2514} = \frac{1}{419}$
$∴ x = 3, y = 4, z = 6$ .

Suggest Corrections

Similar questions

5 x - 6 y + 4 z = 157 x + 4 y - 3 z = 192 x + y + 6 z = 46

\begin{matrix} 5 x - 6 y + 4 z = 15 7 x + 4 y - 3 z = 19 2 x + y + 6 z = 46 \end{matrix}

x, y, z

\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0

x + 3 y + 3 z = 16

x + 4 y + 4 z = 21

x + 3 y + 4 z = 19

2 x - 4 y + 3 z = 1, x - 2 y + 4 z = 3, 3 x - y + 5 z = 2

Join BYJU'S Learning Program