Question

Solve the system of linear equations using the Gauss-Jordan elimination method. 2x1 − x2 + 3x3 = −10 x1 − 2x2 + x3 = −3 x1 − 5x2 + 2x3 = −7 (x1, x2, x3) =

Answer 1

Answer:

The solution is: $(x_{1}, x_{2}, x_{3}) = (1,0,-4)$

Step-by-step explanation:

The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

We have the following system:

$2x_{1} - x_{2} + 3x_{3} = -10$

$x_{1} - 2x_{2} + x_{3} = -3$

$x_{1} - 5x_{2} + 2x_{3} = -7$

This system has the following augmented matrix:

$\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right]$

To make the reductions easier, i am going to swap the first two lines. So

$L1 L2$

Now the matrix is:

$\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right]$

Now we reduce the first row, doing the following operations

$L2 = L2 - 2L1$

$L3 = L3 - L1$

So, the matrix is:

$\left[\begin{array}{ccc}1&-2&1|-3\\0&3&1|-4\\0&-3&1| -4\end{array}\right]$

Now we divide L2 by 3

$L2 = \frac{L2}{3}$

So we have

$\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&-3&1| -4\end{array}\right]$

Now we have:

$L3 = 3L2 + L3$

So, now we have our row reduced matrix:

$\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right]$

We start from the bottom line, where we have:

$2x_{3} = -8$

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

$x_{3} = -4$

At second line:

$x_{2} + \frac{x_{3}}{3} = \frac{-4}{3}$

$x_{2} - \frac{4}{3} = -\frac{4}{3}$

$x_{2} = 0$

At the first line

$x_{1} -2x_{2} + x_{3} = -3$

$x_{1} - 4 = -3$

$x_{1} = 1$

The solution is: $(x_{1}, x_{2}, x_{3}) = (1,0,-4)$