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What values of x, y and z satisfy the following system of linear equations?

$$\begin{bmatrix} 1 &2 &3 \\ 1& 3 &4 \\ 2& 2 &3 \end{bmatrix} \begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 6\\8 \\ 12 \end{bmatrix}$$

1. x = 6, y = 3, z = 2
2. x = 12, y = 3, z = - 4
3. x = 6, y = 6, z = - 4
4. x = 12, y = - 3, z = 0
Just apply row transformation on (A | B) and directly check option.

Correct answer is (C). It can be easily verified by keeping the value of variables in the equations.
answered by Boss (5.3k points) 10 41 74
selected by
$\begin{bmatrix} 1 &2 &3 \\ 1& 3 &4 \\ 2& 2 &3 \end{bmatrix} \begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 6\\8 \\ 12 \end{bmatrix}$

$x+2y+3z=6........(1)$

$x+3y+4z=8........(2)$

$2x+2y+3z=12........(3)$

Apply (3)-(1)

x=6

Put value of x in any of 2 equations.Let's take (1) and (2)

$2y+3z=0$

$3y +4z=2$

$y=6 , z=-4$

Hence,Option(C) $x=6,y=6,z=-4$.
answered by Veteran (35.3k points) 11 49 231
edited ago
How x = 6 satisfied if we put x=6 in equation no 2 then we get 3y + 4z = 0 but according to u 3y + 4z = 2 ????

@LeenSharma
Btw u have taken equation no 2 wrong it is x + 3y + 4z = 8 not x + 3y + 4z =6