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Solution set of x + y + z = 6, x + 2y + 3z = 10, 2x + 3y + 4z = 16

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 + y + z = 6, x + 2y + 3z = 10 and 2x + 3y + 4z = 16.

x + y + z = 6①
x + 2y + 3z = 10②

We want every x, y and z that make all of them true.

2x + 3y + 4z = 16③

1 Write the augmented matrix

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

= [1, 1, 1, 6; 1, 2, 3, 10; 2, 3, 4, 16]

We call the rows R 1, R 2 and R 3.

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, 6; 0, 1, 2, 4; 0, 1, 2, 4]
→ [1, 0, −1, 2; 0, 1, 2, 4; 0, 0, 0, 0]

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 letter z has none: it is free.

A free letter can be any number.

A row of 0s says 0 = 0: always true, so it tells us nothing.

leading 1: x, y
free: z

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 add 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 + 2z = 4.

y + 2z = 4

We want y alone.

To remove + 2z, we take away 2z on both sides.

y + 2z − 2z = 4 − 2z

They cancel on the left.

So y = 4 − 2z.

y = 4 − 2z

5 Write it as vectors

We stack the letters in a column.

The letter z is free, so it stays z.

[x; y; z] = [2 + z; 4 − 2z; z]

We split it up.

The numbers alone make one vector: (2, 4, 0).

What goes with z makes another: (1, −2, 1).

[x; y; z] = [2; 4; 0] + z[1; −2; 1], z any number

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

One free letter: the solutions make a line.

one point + any amount of the direction

6 Check

Let's check.

First the point: all free letters 0.

We put x = 2, y = 4 and z = 0 into every equation.

① 2 + 4 + 0 = 6
② 2 + 2(4) + 3(0) = 2 + 8 + 0 = 10
③ 2(2) + 3(4) + 4(0) = 4 + 12 + 0 = 16

The point works in every equation.

Now the direction (1, −2, 1).

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

① 1 + (−2) + 1 = 1 − 2 + 1 = 0
② 1 + 2(−2) + 3(1) = 1 − 4 + 3 = 0
③ 2(1) + 3(−2) + 4(1) = 2 − 6 + 4 = 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 the direction

Your turn

Try this one yourself, then check:

Find all the solutions of x + y + z = 0, x + 3y + 3z = 2, 2x + 4y + 4z = 2
Show the answer
[x; y; z] = [−1; 1; 0] + z[0; −1; 1], z any number

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