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Solution set of x + 2y - z + w = 3, y + z - w = 1

Step by step, with the reason for every step · Solution sets · Linear Algebra

Step by step

The question

Question: find all the solutions of x + 2y − z + w = 3 and y + z − w = 1.

x + 2y − z + w = 3①

We want every x, y, z and w that make both true.

There are 4 letters, but only 2 equations.

y + z − w = 1②

1 Write the augmented matrix

Each row is one equation: the numbers in front of x, y, z and w, then the number on the right, after the bar.

= [1, 2, −1, 1, 3; 0, 1, 1, −1, 1]

We call the rows R 1 and R 2.

2 Row reduce

We use the three row moves: swap two rows, times a row by a number that is not 0, add a multiple of one row to another.

They never change the answers.

Our goal: a leading 1 in each row, with 0s above and below it.

swap two rows
× a row by a number (not 0)
add a multiple of one row to another
→ [1, 0, −3, 3, 1; 0, 1, 1, −1, 1]

3 Find the free letters

The first number that is not 0 in each row is a 1.

We call it a leading 1, and ring it.

The letters x and y have a leading 1 in their columns.

The letters z and w have none: they are free.

A free letter can be any number.

leading 1: x, y
free: z, w

4 Write the leading letters with the free ones

Row 1 as an equation: x − 3z + 3w = 1.

x − 3z + 3w = 1

We want x alone.

To remove − 3z and + 3w, we add 3z and take away 3w on both sides.

x − 3z + 3w + 3z − 3w = 1 + 3z − 3w

They cancel on the left.

So x = 1 + 3z − 3w.

x = 1 + 3z − 3w

Careful!

The − 3z moves to the other side, so its sign changes.

So x is not 1 − 3z.

Careful: not x = 1 − 3z ✗

Row 2 as an equation: y + z − w = 1.

y + z − w = 1

We want y alone.

To remove + z and − w, we take away z and add w on both sides.

y + z − w − z + w = 1 − z + w

They cancel on the left.

So y = 1 − z + w.

y = 1 − z + w

5 Write it as vectors

We stack the letters in a column.

The free letters stay as they are.

[x; y; z; w] = [1 + 3z − 3w; 1 − z + w; z; w]

We split it up.

The numbers alone make one vector: (1, 1, 0, 0).

What goes with z makes another: (3, −1, 1, 0), and the same for w.

[x; y; z; w] = [1; 1; 0; 0] + z[3; −1; 1; 0] + w[−3; 1; 0; 1], z and w any numbers

So every solution is one point, plus any amount of each direction.

Two free letters: the solutions make a flat plane.

one point + any amount of each direction

6 Check

Let's check.

First the point: all free letters 0.

We put x = 1, y = 1, z = 0 and w = 0 into every equation.

① 1 + 2(1) − 0 + 0 = 1 + 2 + 0 + 0 = 3
② 1 + 0 − 0 = 1 + 0 + 0 = 1

The point works in every equation.

Now the direction (3, −1, 1, 0).

Its left sides must give 0, so adding it changes nothing.

① 3 + 2(−1) − 1 + 0 = 3 − 2 − 1 + 0 = 0
② (−1) + 1 − 0 = −1 + 1 + 0 = 0

Now the direction (−3, 1, 0, 1).

Its left sides must give 0, so adding it changes nothing.

① (−3) + 2(1) − 0 + 1 = −3 + 2 + 0 + 1
= 0
② 1 + 0 − 1 = 0

Every left side gives 0.

So adding any amount of it keeps every equation true.

The answer is right!

Let's look back.

First: write the augmented matrix.

Then: row reduce.

Then: find the free letters.

Then: write the leading letters with the free ones.

Then: write it as vectors.

Last: check.

Answer
one point + any amount of each direction

Your turn

Try this one yourself, then check:

Find all the solutions of x + 2y − z + w = 1, y + z − w = 1
Show the answer
[x; y; z; w] = [−1; 1; 0; 0] + z[3; −1; 1; 0] + w[−3; 1; 0; 1], z and w any numbers

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