If $3 x - 4 y + 2 z = - 1$ , $2 x + 3 y + 5 z = 7$ , $x + z = 2$ , then $x = ?$

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Solution

The correct option is A $3$
Given equations are $3 x - 4 y + 2 z = - 1, 2 x + 3 y + 5 z = 7$
$x + z = 2$
So, augmented matrix is $⎡ ⎢ ⎣ \begin{matrix} 3 & - 4 & 2 & - 1 2 & 3 & 5 & 7 1 & 0 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦$
By using Gauss Jordan's method $u p =↓ R_{3} \leftrightarrow R_{1}$
$⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 2 0 & 3 & 3 & 3 0 & - 4 & - 1 & - 7 \end{matrix} ⎤ ⎥ ⎦$ $R_{2} \to R_{2} - 2 R_{1} \leftarrow ------- - R_{3} \to R_{3} - 3 R_{1}$ $⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 2 2 & 3 & 5 & 7 3 & - 4 & 2 & - 1 \end{matrix} ⎤ ⎥ ⎦$
$R_{3} \to \frac{R_{3}}{- 1} ↓ R_{2} \to \frac{R_{2}}{3}$
$⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 2 0 & 1 & 1 & 1 0 & 4 & 1 & 7 \end{matrix} ⎤ ⎥ ⎦$ $R_{3} \to R_{3} \to 4 R_{2} - ------- \to$ $⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 2 0 & 1 & 1 & 1 0 & 0 & - 3 & 3 \end{matrix} ⎤ ⎥ ⎦$ $\begin{matrix} R_{3} \to \frac{R_{3}}{- 3} R_{1} \to R_{1} - R_{3} - ---------- \to R_{2} \to R_{2} - R_{3} \end{matrix}$ $⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 3 0 & 1 & 0 & 2 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎦$
$∴ x = 3, y = 2, z = - 1$

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