Consider the following system of equations:
$3 x + 2 y = 1$
$4 x + 7 z = 1$
$x + y + z = 3$
$x - 2 y + 7 z = 0$
The number of solutions for this system is _______ .
1

Open in App

Solution

The correct option is A 1
$[A : B] = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 3 & 2 & 0 & ⋮ & 1 4 & 0 & 7 & ⋮ & 1 1 & 1 & 1 & ⋮ & 3 1 & - 2 & 7 & ⋮ & 0 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$
$R_{1} \leftrightarrow R_{3}$
$[A : B] = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 1 & 1 & 1 & ⋮ & 3 4 & 0 & 7 & ⋮ & 1 3 & 2 & 0 & ⋮ & 1 1 & - 2 & 7 & ⋮ & 0 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$

$R_{2} \to R_{2} - 4 R_{1}; R_{3} \to R_{3} - 3 R_{1}; R_{4} \to R_{4} - R_{1}$

$[A : B] = ⎛ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 1 & 1 & 1 & 3 0 & - 4 & 3 & - 11 0 & - 1 & - 3 & - 8 0 & - 3 & 6 & - 3 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎠$
$R_{2} \leftrightarrow R_{3}$
[A:B]= $⎛ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 1 & 1 & 1 & 3 0 & - 1 & - 3 & - 8 0 & - 4 & 3 & - 11 0 & - 3 & 6 & - 3 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎠$

$R_{3} \to R_{3} - 4 R_{2}; R_{4} \to R_{4} - 3 R_{2}$

$[A : B] \sim ⎛ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 1 & 1 & 1 & 3 0 & - 1 & - 3 & - 8 0 & 0 & 15 & 21 0 & 0 & 15 & 21 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎠$

$R_{4} \to R_{4} - R_{3}$

$[A : B] \sim ⎛ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 1 & 1 & 1 & 3 0 & - 1 & - 3 & - 8 0 & 0 & 15 & 21 0 & 0 & 0 & 0 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎠$

$∴ ρ (A) = ρ (A : B) = 3$ no. of variables
Hence, unique solution exist.

Suggest Corrections

Similar questions

x + y + z = 6 x + 2 y + 3 z = 14 x + 4 y + 7 z = 30

3 x + 4 y + 7 z = 4, 2 x - y + 3 z = - 3; x + 2 y - 3 z = 8

\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

Join BYJU'S Learning Program