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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
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Correct answer is (C). It can be easily verified by keeping the value of variables in the equations.
answered by Boss (5k points)
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$\begin{bmatrix} 1 &2 &3 \\ 1& 3 &4 \\ 2& 2 &3 \end{bmatrix} \begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 6\\8 \\ 12 \end{bmatrix}$

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

$x+3y+4z=6........(2)$

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

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 (31.7k points)