# 3x + 5y + 5z = −2 −3x + −1y + 0z = 7 −6x + 6y + 10z = k For what value of k is this system consistent?

## 3x + 5y + 5z = −2 −3x + −1y + 0z = 7 −6x + 6y + 10z = k For what value of k is this system consistent?

Sep 21, 2017

$k = 24$

#### Explanation:

Let's create a matrix from the system and then perform Gaussian elimination.

$\left(\begin{matrix}3 & 5 & 5 & - 2 \\ - 3 & - 1 & 0 & 7 \\ - 6 & 6 & 10 & k\end{matrix}\right)$

Let's do ${R}_{2} \implies {R}_{2} + {R}_{3}$ and ${R}_{3} \implies {R}_{3} + 2 {R}_{1}$.

$= \left(\begin{matrix}3 & 5 & 5 & - 2 \\ 0 & 4 & 5 & 5 \\ 0 & 16 & 20 & k - 4\end{matrix}\right)$

Now let's do ${R}_{1} \implies \frac{1}{3} {R}_{1}$.

$= \left(\begin{matrix}1 & \frac{5}{3} & \frac{5}{3} & - \frac{2}{3} \\ 0 & 4 & 5 & 5 \\ 0 & 16 & 20 & k - 4\end{matrix}\right)$

${R}_{1} \implies {R}_{1} - \frac{5}{12} {R}_{2}$

$= \left(\begin{matrix}1 & 0 & - \frac{5}{12} & - \frac{33}{12} \\ 0 & 4 & 5 & 5 \\ 0 & 16 & 20 & k - 4\end{matrix}\right)$

${R}_{2} \implies \frac{1}{4} {R}_{2}$

$= \left(\begin{matrix}1 & 0 & - \frac{5}{12} & - \frac{33}{12} \\ 0 & 1 & \frac{5}{4} & \frac{5}{4} \\ 0 & 16 & 20 & k - 4\end{matrix}\right)$

${R}_{3} \implies {R}_{3} - 16 {R}_{2}$

$= \left(\begin{matrix}1 & 0 & - \frac{5}{12} & - \frac{33}{12} \\ 0 & 1 & \frac{5}{4} & \frac{5}{4} \\ 0 & 0 & 0 & k - 24\end{matrix}\right)$

We can stop now: recall that the bottom row translates into $0 x + 0 y + 0 x = k - 24$, that is, $0 = k - 24$. So $k = 24$.