# How do you solve 3u + v + w = 9, u + v - w = 5 and u + 2v + w = 9 using matrices?

Mar 5, 2017

$\left(\begin{matrix}u \\ v \\ w\end{matrix}\right) = \left(\begin{matrix}\frac{7}{4} \\ \frac{7}{2} \\ \frac{1}{4}\end{matrix}\right)$

#### Explanation:

In matrix form:

$\left(\begin{matrix}3 & 1 & 1 \\ 1 & 1 & - 1 \\ 1 & 2 & 1\end{matrix}\right) \left(\begin{matrix}u \\ v \\ w\end{matrix}\right) = \left(\begin{matrix}9 \\ 5 \\ 9\end{matrix}\right)$

Using Gaussian elimination, we can row reduce, so we set up the augmented matrix:

$\left(\begin{matrix}3 & 1 & 1 \\ 1 & 1 & - 1 \\ 1 & 2 & 1\end{matrix}\right)$ $\left(\begin{matrix}9 \\ 5 \\ 9\end{matrix}\right)$

$R 2 \to R 2 - R 3$

$\left(\begin{matrix}3 & 1 & 1 \\ 0 & - 1 & - 2 \\ 1 & 2 & 1\end{matrix}\right)$ $\left(\begin{matrix}9 \\ - 4 \\ 9\end{matrix}\right)$

$R 3 \to R 3 - \frac{1}{3} R 1$

$\left(\begin{matrix}3 & 1 & 1 \\ 0 & - 1 & - 2 \\ 0 & \frac{5}{3} & \frac{2}{3}\end{matrix}\right)$ $\left(\begin{matrix}9 \\ - 4 \\ 6\end{matrix}\right)$

$R 3 \to R 3 + \frac{5}{3} R 2$

$\left(\begin{matrix}3 & 1 & 1 \\ \textcolor{red}{0} & - 1 & - 2 \\ \textcolor{red}{0} & \textcolor{red}{0} & - \frac{8}{3}\end{matrix}\right)$ $\left(\begin{matrix}9 \\ - 4 \\ - \frac{2}{3}\end{matrix}\right)$

This is now upper triangular, we are in row echelon form. Now we back-substitute:

From the bottom row:

$- \frac{8}{3} w = - \frac{2}{3}$

$\implies w = \frac{1}{4}$

From the middle row:

$- v - 2 w = - 4$

$\implies v = 4 - 2 w$

$\implies v = \frac{7}{2}$

From the top row:

$3 u + v + w = 9$

$\implies u = \frac{9 - v - w}{3}$

$= \frac{9 - \frac{7}{2} - \frac{1}{4}}{3}$

$\implies u = \frac{7}{4}$

So:

$\left(\begin{matrix}u \\ v \\ w\end{matrix}\right) = \left(\begin{matrix}\frac{7}{4} \\ \frac{7}{2} \\ \frac{1}{4}\end{matrix}\right)$