How do you solve ((5, 1, 4), (2, -3, -4), (7, 2, -6))X=((5), (2), (5))?

Sep 19, 2016

$X = \left(\begin{matrix}\frac{99}{107} \\ - \frac{28}{107} \\ \frac{17}{107}\end{matrix}\right)$

Explanation:

Starting with the augmented matrix:

$\left(\begin{matrix}5 & 1 & 4 & | & 5 \\ 2 & - 3 & - 4 & | & 2 \\ 7 & 2 & - 6 & | & 5\end{matrix}\right)$

Perform a sequence of row operations until the left hand side $3 \times 3$ matrix is the identity matrix. Then the right hand column will be $X$.

Subtract row 1 and 2 from row 3 to get:

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

Divide row 1 by $5$ to get:

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

Subtract twice row 1 from row 2 to get:

$\left(\begin{matrix}1 & \frac{1}{5} & \frac{4}{5} & | & 1 \\ 0 & - \frac{17}{5} & - \frac{28}{5} & | & 0 \\ 0 & 4 & - 6 & | & - 2\end{matrix}\right)$

Multiply row $2$ by $- \frac{5}{17}$ to get:

$\left(\begin{matrix}1 & \frac{1}{5} & \frac{4}{5} & | & 1 \\ 0 & 1 & \frac{28}{17} & | & 0 \\ 0 & 4 & - 6 & | & - 2\end{matrix}\right)$

Subtract $4$ times row 2 from row 3 to get:

$\left(\begin{matrix}1 & \frac{1}{5} & \frac{4}{5} & | & 1 \\ 0 & 1 & \frac{28}{17} & | & 0 \\ 0 & 0 & - \frac{214}{17} & | & - 2\end{matrix}\right)$

Multiply row $3$ by $- \frac{17}{214}$ to get:

$\left(\begin{matrix}1 & \frac{1}{5} & \frac{4}{5} & | & 1 \\ 0 & 1 & \frac{28}{17} & | & 0 \\ 0 & 0 & 1 & | & \frac{17}{107}\end{matrix}\right)$

Subtract $\frac{1}{5}$ row 2 from row 1 to get:

$\left(\begin{matrix}1 & 0 & \frac{8}{17} & | & 1 \\ 0 & 1 & \frac{28}{17} & | & 0 \\ 0 & 0 & 1 & | & \frac{17}{107}\end{matrix}\right)$

Subtract $\frac{8}{17}$ row 3 from row 1 to get:

$\left(\begin{matrix}1 & 0 & 0 & | & \frac{99}{107} \\ 0 & 1 & \frac{28}{17} & | & 0 \\ 0 & 0 & 1 & | & \frac{17}{107}\end{matrix}\right)$

Subtract $\frac{28}{17}$ row 3 from row 2 to get:

$\left(\begin{matrix}1 & 0 & 0 & | & \frac{99}{107} \\ 0 & 1 & 0 & | & - \frac{28}{107} \\ 0 & 0 & 1 & | & \frac{17}{107}\end{matrix}\right)$

So:

$X = \left(\begin{matrix}\frac{99}{107} \\ - \frac{28}{107} \\ \frac{17}{107}\end{matrix}\right)$