Question

Consider the following system of equations:2x1 − 4x2 + 3x3 − x4 + 2x5 = 03x1 − 6x2 + 5x3 − 2x4 + 4x5 = 05x1 − 10x2 + 7x3 − 3x4 + 18x5 = 0.Find the general solution of the system.

Answer 1

The system has a corresponding augmented matrix

$\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\3&-6&5&-2&4&0\\5&-10&7&-3&18&0\end{array}\right]$

Let's get this into reduced row echelon form.

Add 3(row 1) to -2(row 2), and add 5(row 1) to -2(row 3):

$\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&-1&1&-2&0\\0&0&1&1&-26&0\end{array}\right]$

Multiply through row 2 by -1:

$\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&1&-1&2&0\\0&0&1&1&-26&0\end{array}\right]$

Add -1(row 2) to row 3:

$\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&1&-1&2&0\\0&0&0&2&-28&0\end{array}\right]$

Multiply through row 3 by 1/2:

$\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&1&-1&2&0\\0&0&0&1&-14&0\end{array}\right]$

Add row 3 to row 2:

$\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&1&0&-12&0\\0&0&0&1&-14&0\end{array}\right]$

Add -3(row 2) and row 3 to row 1:

$\left[\begin{array}{ccccc|c}2&-4&0&0&24&0\\0&0&1&0&-12&0\\0&0&0&1&-14&0\end{array}\right]$

Multiply through row 1 by 1/2:

$\left[\begin{array}{ccccc|c}1&-2&0&0&12&0\\0&0&1&0&-12&0\\0&0&0&1&-14&0\end{array}\right]$

This tells us

$\begin{cases}x_1-2x_2+12x_5=0\\x_3-12x_5=0\\x_4-14x_5=0\end{cases}$

The matrix has rank 3, so we can pick 2 free variables, say $x_1$ and $x_2$. Then the general solution to the system is

$\begin{cases}x_1=s\\x_2=t\\x_3=2t-s\\x_4=\frac{7(2t-s)}6\\x_5=\frac{2t-s}{12}\end{cases}$

where $s,t\in\Bbb R$.