Steply

Solve x + y + 2z = 11, x + 3z = 11, x + 2z = 7 by Cramer's rule

Step by step, with the reason for every step · Cramer's rule with three equations · Algebra 2

Step by step

The question

Question: solve x + y + 2z = 11, x + 3z = 11, x + 2z = 7 by Cramer's rule.

x + y + 2z = 11①
x + 3z = 11②

Three equations, three letters: x, y and z.

x + 2z = 7③

Cramer's rule, like with two equations: x is D x over D, y is D y over D, z is D z over D.

Four determinants, each by the diagonals.

x = DxD, y = DyD, z = DzD

1 Find D

D: the numbers in front of x, y and z, in their places.

D = |1, 1, 2; 1, 0, 3; 1, 0, 2|1, 1; 1, 0; 1, 0

We write the first two columns again, on the right.

Then every diagonal has three numbers.

Down: 1 × 0 × 2, 1 × 3 × 1, 2 × 1 × 0.

down = 1 × 0 × 2 + 1 × 3 × 1 + 2 × 1 × 0

0, 3, 0.

Together: 3.

= 0 + 3 + 0 = 3

Up: 1 × 0 × 2, 0 × 3 × 1, 2 × 1 × 1.

up = 1 × 0 × 2 + 0 × 3 × 1 + 2 × 1 × 1

0, 0, 2.

Together: 2.

= 0 + 0 + 2 = 2

Down minus up: 3 minus 2 is 1.

So D = 1.

D = 3 − 2 = 1

2 Find Dx

D x: the x column is replaced by the right sides, 11, 11, 7.

Dx = |11, 1, 2; 11, 0, 3; 7, 0, 2|11, 1; 11, 0; 7, 0

Down: 11 × 0 × 2, 1 × 3 × 7, 2 × 11 × 0.

down = 11 × 0 × 2 + 1 × 3 × 7 + 2 × 11 × 0

0, 21, 0.

Together: 21.

= 0 + 21 + 0 = 21

Up: 7 × 0 × 2, 0 × 3 × 11, 2 × 11 × 1.

up = 7 × 0 × 2 + 0 × 3 × 11 + 2 × 11 × 1

0, 0, 22.

Together: 22.

= 0 + 0 + 22 = 22

Down minus up: 21 minus 22 is −1.

So D x = −1.

Dx = 21 − 22 = −1

3 Find Dy

D y: the y column is replaced by the right sides, 11, 11, 7.

Dy = |1, 11, 2; 1, 11, 3; 1, 7, 2|1, 11; 1, 11; 1, 7

Down: 1 × 11 × 2, 11 × 3 × 1, 2 × 1 × 7.

down = 1 × 11 × 2 + 11 × 3 × 1 + 2 × 1 × 7

22, 33, 14.

Together: 69.

= 22 + 33 + 14 = 69

Up: 1 × 11 × 2, 7 × 3 × 1, 2 × 1 × 11.

up = 1 × 11 × 2 + 7 × 3 × 1 + 2 × 1 × 11

22, 21, 22.

Together: 65.

= 22 + 21 + 22 = 65

Down minus up: 69 minus 65 is 4.

So D y = 4.

Dy = 69 − 65 = 4

4 Find Dz

D z: the z column is replaced by the right sides, 11, 11, 7.

Dz = |1, 1, 11; 1, 0, 11; 1, 0, 7|1, 1; 1, 0; 1, 0

Careful!

For D z we replace the z column, the third one.

The other two stay.

Careful: Dz: the third column is replaced

Down: 1 × 0 × 7, 1 × 11 × 1, 11 × 1 × 0.

down = 1 × 0 × 7 + 1 × 11 × 1 + 11 × 1 × 0

0, 11, 0.

Together: 11.

= 0 + 11 + 0 = 11

Up: 1 × 0 × 11, 0 × 11 × 1, 7 × 1 × 1.

up = 1 × 0 × 11 + 0 × 11 × 1 + 7 × 1 × 1

0, 0, 7.

Together: 7.

= 0 + 0 + 7 = 7

Down minus up: 11 minus 7 is 4.

So D z = 4.

Dz = 11 − 7 = 4

5 Divide

x is D x over D: −1/1 is −1.

x = DxD = −11 = −1

y is D y over D: 4/1 is 4.

y = DyD = 41 = 4

z is D z over D: 4/1 is 4.

z = DzD = 41 = 4

So the answer is (−1, 4, 4).

(x, y, z) = (−1, 4, 4)

6 Check

Let's check in every equation.

We put x = −1, y = 4, z = 4 into ①.

① (−1) + 4 + 2(4) = −1 + 4 + 8 = 11

We get 11, the same as the right side.

Now ②.

② (−1) + 3(4) = −1 + 12 = 11

We get 11, the same as the right side.

Now ③.

③ (−1) + 2(4) = −1 + 8 = 7

We get 7, the same as the right side.

All of them work.

So the answer is right!

Let's look back.

First: find D.

Then: find Dx.

Then: find Dy.

Then: find Dz.

Then: divide.

Last: check.

Answer
(x, y, z) = (−1, 4, 4)

Your turn

Try this one yourself, then check:

Solve x + y + z = −7, x + 3z = −7, x + 2z = −5
Show the answer
(x, y, z) = (−1, −4, −2)

More from Cramer's rule with three equations

The whole lesson: Cramer's rule with three equations →