# How do you solve the following linear system 5x - 4y + 4z = 18, -x + 3y - 2z = 0, 4x - 2y + 7z = 3 ?

Apr 4, 2016

Either solve using matrices or use elimination

#### Explanation:

The following system of linear equations, when written in matrix form, is

$\left(\begin{matrix}5 & - 4 & 4 \\ - 1 & 3 & - 2 \\ 4 & - 2 & 7\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}18 \\ 0 \\ 3\end{matrix}\right)$

Assuming you know how to find the determinant of a $3 \times 3$ matrix, we can apply Cramer's Rule.

$x = \frac{\det \left(\begin{matrix}18 & - 4 & 4 \\ 0 & 3 & - 2 \\ 3 & - 2 & 7\end{matrix}\right)}{\det \left(\begin{matrix}5 & - 4 & 4 \\ - 1 & 3 & - 2 \\ 4 & - 2 & 7\end{matrix}\right)}$

$= \frac{294}{49}$

$= 6$

$y = \frac{\det \left(\begin{matrix}5 & 18 & 4 \\ - 1 & 0 & - 2 \\ 4 & 3 & 7\end{matrix}\right)}{\det \left(\begin{matrix}5 & - 4 & 4 \\ - 1 & 3 & - 2 \\ 4 & - 2 & 7\end{matrix}\right)}$

$= \frac{0}{49}$

$= 0$

$z = \frac{\det \left(\begin{matrix}5 & - 4 & 18 \\ - 1 & 3 & 0 \\ 4 & - 2 & 3\end{matrix}\right)}{\det \left(\begin{matrix}5 & - 4 & 4 \\ - 1 & 3 & - 2 \\ 4 & - 2 & 7\end{matrix}\right)}$

$= \frac{- 147}{49}$

$= - 3$

Therefore,

• $x = 6$
• $y = 0$
• $z = - 3$