Question

Solve the systems using the Gaussian elimination method.
x+2y+z=4
2x-y-3z=2
x-8y-9z=-8
5x+5y =14

Answer 1

Answer on Question #61276 – Math – Linear Algebra

Question

Solve the systems using the Gaussian elimination method.

$$x + 2y + z = 4$$ $$2x - y - 3z = 2$$ $$x - 8y - 9z = -8$$ $$5x + 5y = 14$$

Solution

I’ll convert the system to its upper triangular form.

1) Multiply the first row by -2 and add it to the second row:

$$-2(x + 2y + z) + (2x - y - 3z) = -2x - 4y - 2z + 2x - y - 3z = -5y - 5z;$$ $$-2 \cdot 4 + 2 = -6.$$

The result is written in the second row:

$$\begin{array}{r c l} x & +2y & +z = 4 \\ & -5y & -5z = -6 \\ x & -8y & -9z = -8 \\ 5x & +5y & = 14 \end{array}$$

2) Multiply the first row by -1 and add it to the third row:

$$-(x + 2y + z) + (x - 8y - 9z) = -x - 2y - z + x - 8y - 9z = -10y - 10z;$$ $$-4 + (-8) = -12.$$

The result is written in the third row:

$$\begin{array}{r c l} x & +2y & +z = 4 \\ & -5y & -5z = -6 \\ & -10y & -10z = -12 \\ 5x & +5y & = 14 \end{array}$$

3) Multiply the first row by -5 and add it to the fourth row:

$$-5(x + 2y + z) + (5x + 5y) = -5x - 10y - 5z + 5x + 5y = -5y - 5z;$$ $$-5 \cdot 4 + 14 = -6.$$

The result is written in the fourth row:

$x + 2y + z = 4$

-5y -5z = -6

-10y -10z = -12

-5y -5z = -6

4) Divide the third row by 2:

$x + 2y + z = 4$

-5y -5z = -6

-5y -5z = -6

-5y -5z = -6

5) Hence we have the system:

$x + 2y + z = 4$

-5y -5z = -6

6) Divide the third row by -5:

$x + 2y + z = 4$

$y + z = 6/5$

7) Subtract the second row from the first row:

$x + y = 4 - 6/5$

$y + z = 6/5$

obtain

$x + y = 14/5$

$y + z = 6/5$

This system has infinite number of solutions, because the number of equations $(k = 2)$ is less than the number of variables $(n = 3)$.

If $y$ is an independent variable, then it follows from the first equation of the system that $x = 14/5 - y$, it follows from the second equation of the system that $z = \frac{6}{5} - y$. Thus, $(x, y, z) = \left( \frac{14}{5} - s, s, \frac{6}{5} - s \right)$, where $s \in \mathbb{R}$.

If $z$ is an independent variable, then it follows from the second equation of the system that

$$y = \frac{6}{5} - z.$$

Substituting it into the first equation of the system

$$x + y = \frac{14}{5};$$ $$x + \frac {6}{5} - z = \frac {1 4}{5};$$ $$x = \frac {1 4}{5} - \frac {6}{5} + z;$$ $$x = \frac {8}{5} + z.$$

Thus, $(x,y,z) = \left(t + \frac{8}{5},\frac{6}{5} -t,t\right)$, where $t\in \mathbb{R}$.

If $x$ is an independent variable, then it follows from the first equation of the system that

$$y = \frac {1 4}{5} - x.$$

Substituting it into the second equation of the system

$$y + z = \frac {6}{5}$$ $$\frac {1 4}{5} - x + z = \frac {6}{5};$$ $$z = \frac {6}{5} - \frac {1 4}{5} + x = x - \frac {8}{5};$$

Thus, $(x,y,z) = \left(u,\frac{14}{5} -u,u - \frac{8}{5}\right)$, where $u\in \mathbb{R}$.

Answer: $(x,y,z) = \left(u,\frac{14}{5} -u,u - \frac{8}{5}\right)$, where $u\in \mathbb{R}$

or

$$(x, y, z) = \left(\frac {1 4}{5} - s, s, \frac {6}{5} - s\right), \text{ where } s \in \mathbb {R},$$

or

$$(x, y, z) = \left(t + \frac {8}{5}, \frac {6}{5} - t, t\right), \text{ where } t \in \mathbb {R}.$$

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