Solution set of x + y + z + w = 4, x - y + z

Step by step

The question

Question: find all the solutions of x + y + z + w = 4 and x − y + z − w = 0.

x + y + z + w = 4①

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

There are 4 letters, but only 2 equations.

x − y + z − w = 0②

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, 1, 1, 1, 4; 1, −1, 1, −1, 0]

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, 1, 1, 1, 4; 0, −2, 0, −2, −4]

→ [1, 1, 1, 1, 4; 0, 1, 0, 1, 2]

→ [1, 0, 1, 0, 2; 0, 1, 0, 1, 2]

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 + z = 2.

x + z = 2

We want x alone.

To remove + z, we take away z on both sides.

x ~~+ z − z~~ = 2 − z

They cancel on the left.

So x = 2 − z.

x = 2 − z

Careful!

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

So x is not 2 + z.

Careful: not x = 2 + z ✗

Row 2 as an equation: y + w = 2.

y + w = 2

We want y alone.

To remove + w, we take away w on both sides.

y ~~+ w − w~~ = 2 − w

They cancel on the left.

So y = 2 − w.

y = 2 − w

5 Write it as vectors

We stack the letters in a column.

The free letters stay as they are.

[x; y; z; w] = [2 − z; 2 − w; z; w]

We split it up.

The numbers alone make one vector: (2, 2, 0, 0).

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

[x; y; z; w] = [2; 2; 0; 0] + z[−1; 0; 1; 0] + w[0; −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 = 2, y = 2, z = 0 and w = 0 into every equation.

① 2 + 2 + 0 + 0 = 4

② 2 − 2 + 0 − 0 = 2 − 2 + 0 + 0 = 0

The point works in every equation.

Now the direction (−1, 0, 1, 0).

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

① (−1) + 0 + 1 + 0 = −1 + 0 + 1 + 0 = 0

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

Now the direction (0, −1, 0, 1).

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

① 0 + (−1) + 0 + 1 = 0 − 1 + 0 + 1 = 0

② 0 − (−1) + 0 − 1 = 0 + 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