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$\newcommand{\Sp}{\phantom{0}}$There is a system of linear equations: \begin{alignat*}{4} &x - &&y - 2&&z = &&1, \\ 2&x + 3&&y - &&z =-&&2. \end{alignat*}

I create the matrix of the system: $$ \left[\begin{array}{rrr|r} 1 & -1 & -2 & 1 \\ 2 & 3 & -1 & -2 \end{array}\right] $$ then with GEM, $$ \left[\begin{array}{rrr|r} 1 & -1 & -2 & 1 \\ 0 & 5 & 3 & -4 \end{array}\right] $$

I don't know how to proceed after that? I have found the correction of this exercise but I still don't understand the way to solve it. Can someone help me please?

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Sorry, it was a 3. –  rusol Jun 1 at 17:10

2 Answers 2

I would apply two more row operations and present a solution with a free variable:

$$ \begin{align*} & \left[\begin{array}{rrr|r} 1 & -1 & -2 & 1 \\ 2 & 3 & -1 & -2 \end{array}\right] R_2+(-2)R_1 \rightarrow R_2 \\ \equiv & \left[\begin{array}{rrr|r} 1 & -1 & -2 & 1 \\ 0 & 5 & 3 & -4 \end{array}\right] \frac{1}{5}R_2 \rightarrow R_2 \\ \equiv & \left[\begin{array}{rrr|r} 1 & -1 & -2 & 1 \\ 0 & 1 & \frac{3}{5} & -\frac{4}{5} \end{array}\right] R_1+R_2 \rightarrow R_1 \\ \equiv & \left[\begin{array}{rrr|r} 1 & 0 & -\frac{7}{5} & \frac{1}{5} \\ 0 & 1 & \frac{3}{5} & -\frac{4}{5} \end{array}\right]. \end{align*} $$

Read the solution straight from the matrix as

$$ \begin{align*} x &=\frac{1}{5}+\frac{7}{5}r \\ y&=-\frac{4}{5}-\frac{3}{5}r \\ z&=r, \end{align*} $$

with $r\in \mathbb{R}$.

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This is the problem of finding the complete solution to $Ax=b$. To learn how to do this, you may want to watch this video or Gilbert Strang's full video lecture or notes on the topic.

The idea is that we need to find

(a) the null space $N$ of $ \left[\begin{array}{rrr|r} 1 & -1 & -2 \\ 2 & 3 & -1 \end{array}\right]$ and

(b) a single solution $c$ to $ \left[\begin{array}{rrr} 1 & -1 & -2 \\ 2 & 3 & -1 \end{array}\right]c = \left[\begin{array}{r} 1 \\-2 \end{array}\right] $.

Then all the solutions are exactly $x = N + c$.


Row reduction is the first step regardless of what we do next.

(a) We have $\left[\begin{array}{rrr} 1 & -1 & -2 \\ 0 & 5 & 3 \end{array}\right] \left[\begin{array}{r} 7/5 \\ -3/5 \\ 1 \end{array}\right] = \left[\begin{array}{r} 0 \\ 0 \end{array}\right]$. Thus the null-space is $N = \left[\begin{array}{r} 7/5 \\ -3/5 \\ 1 \end{array}\right]a$.

(b) Notice that $ \left[\begin{array}{rrr} 1 & -1 & -2 \\ 0 & 5 & 3 \end{array}\right] \left[\begin{array}{r} 1/5 \\ -4/5 \\ 0 \end{array}\right] = \left[\begin{array}{r} 1 \\ -4 \end{array}\right] $.

Hence the solutions are exactly all $x$ such that

$$ x = \left[\begin{array}{r} 7/5 \\ -3/5 \\ 1 \end{array}\right]a + \left[\begin{array}{r} 1/5 \\ -4/5 \\ 0 \end{array}\right].$$

Another way of writing this is

\begin{align*} x &= \frac{7}{5}a + \frac{1}{5} \\ y &= -\frac{3}{5}a - \frac{4}{5} \\ z &= a. \end{align*}

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