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Cramer's rule with three equations

Algebra 2 · 20 problems, easy to harder

▶ Watch the first one solved

How we do it

  1. Find D
  2. Find Dx
  3. Find Dy
  4. Find Dz
  5. Divide
  6. Check

The steps for Solve x + y + z = 6, 2x - y + z = 3, x + 2y - z = 2 by Cramer's rule. Every solution names its steps, so you always know what comes next.

Worked examples

Solve x + y + z = 6, 2x - y + z = 3, x + 2y - z = 2 by Cramer's rule Easy

The question

Question: solve x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2 by Cramer's rule.

x + y + z = 6①
2x − y + z = 3②

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

x + 2y − z = 2③

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, 1; 2, −1, 1; 1, 2, −1|1, 1; 2, −1; 1, 2

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

Then every diagonal has three numbers.

Down: 1 × (−1) × (−1), 1 × 1 × 1, 1 × 2 × 2.

down = 1 × (−1) × (−1) + 1 × 1 × 1 + 1 × 2 × 2

1, 1, 4.

Together: 6.

= 1 + 1 + 4 = 6

Up: 1 × (−1) × 1, 2 × 1 × 1, −1 × 2 × 1.

up = 1 × (−1) × 1 + 2 × 1 × 1 + (−1) × 2 × 1

−1, 2, −2.

Together: −1.

= −1 + 2 + (−2) = −1

Down minus up: 6 minus (−1) is 7.

So D = 7.

D = 6 − (−1) = 7

2 Find Dx

D x: the x column is replaced by the right sides, 6, 3, 2.

Dx = |6, 1, 1; 3, −1, 1; 2, 2, −1|6, 1; 3, −1; 2, 2

Down: 6 × (−1) × (−1), 1 × 1 × 2, 1 × 3 × 2.

down = 6 × (−1) × (−1) + 1 × 1 × 2 + 1 × 3 × 2

6, 2, 6.

Together: 14.

= 6 + 2 + 6 = 14

Up: 2 × (−1) × 1, 2 × 1 × 6, −1 × 3 × 1.

up = 2 × (−1) × 1 + 2 × 1 × 6 + (−1) × 3 × 1

−2, 12, −3.

Together: 7.

= −2 + 12 + (−3) = 7

Down minus up: 14 minus 7 is 7.

So D x = 7.

Dx = 14 − 7 = 7

3 Find Dy

D y: the y column is replaced by the right sides, 6, 3, 2.

Dy = |1, 6, 1; 2, 3, 1; 1, 2, −1|1, 6; 2, 3; 1, 2

Down: 1 × 3 × (−1), 6 × 1 × 1, 1 × 2 × 2.

down = 1 × 3 × (−1) + 6 × 1 × 1 + 1 × 2 × 2

−3, 6, 4.

Together: 7.

= −3 + 6 + 4 = 7

Up: 1 × 3 × 1, 2 × 1 × 1, −1 × 2 × 6.

up = 1 × 3 × 1 + 2 × 1 × 1 + (−1) × 2 × 6

3, 2, −12.

Together: −7.

= 3 + 2 + (−12) = −7

Down minus up: 7 minus (−7) is 14.

So D y = 14.

Dy = 7 − (−7) = 14

4 Find Dz

D z: the z column is replaced by the right sides, 6, 3, 2.

Dz = |1, 1, 6; 2, −1, 3; 1, 2, 2|1, 1; 2, −1; 1, 2

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 × (−1) × 2, 1 × 3 × 1, 6 × 2 × 2.

down = 1 × (−1) × 2 + 1 × 3 × 1 + 6 × 2 × 2

−2, 3, 24.

Together: 25.

= −2 + 3 + 24 = 25

Up: 1 × (−1) × 6, 2 × 3 × 1, 2 × 2 × 1.

up = 1 × (−1) × 6 + 2 × 3 × 1 + 2 × 2 × 1

−6, 6, 4.

Together: 4.

= −6 + 6 + 4 = 4

Down minus up: 25 minus 4 is 21.

So D z = 21.

Dz = 25 − 4 = 21

5 Divide

x is D x over D: 7/7 is 1.

x = DxD = 77 = 1

y is D y over D: 14/7 is 2.

y = DyD = 147 = 2

z is D z over D: 21/7 is 3.

z = DzD = 217 = 3

So the answer is (1, 2, 3).

(x, y, z) = (1, 2, 3)

6 Check

Let's check in every equation.

We put x = 1, y = 2, z = 3 into ①.

① 1 + 2 + 3 = 6

We get 6, the same as the right side.

Now ②.

② 2(1) − 2 + 3 = 2 − 2 + 3 = 3

We get 3, the same as the right side.

Now ③.

③ 1 + 2(2) − 3 = 1 + 4 − 3 = 2

We get 2, 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, 2, 3)

▶ Watch it solved

Solve x + 2z = 5, 2x + y + z = 11, x - y + 3z = 2 by Cramer's rule Medium

The question

Question: solve x + 2z = 5, 2x + y + z = 11, x − y + 3z = 2 by Cramer's rule.

x + 2z = 5①
2x + y + z = 11②

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

x − y + 3z = 2③

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, 0, 2; 2, 1, 1; 1, −1, 3|1, 0; 2, 1; 1, −1

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

Then every diagonal has three numbers.

Down: 1 × 1 × 3, 0 × 1 × 1, 2 × 2 × (−1).

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

3, 0, −4.

Together: −1.

= 3 + 0 + (−4) = −1

Up: 1 × 1 × 2, −1 × 1 × 1, 3 × 2 × 0.

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

2, −1, 0.

Together: 1.

= 2 + (−1) + 0 = 1

Down minus up: −1 minus 1 is −2.

So D = −2.

D = −1 − 1 = −2

2 Find Dx

D x: the x column is replaced by the right sides, 5, 11, 2.

Dx = |5, 0, 2; 11, 1, 1; 2, −1, 3|5, 0; 11, 1; 2, −1

Down: 5 × 1 × 3, 0 × 1 × 2, 2 × 11 × (−1).

down = 5 × 1 × 3 + 0 × 1 × 2 + 2 × 11 × (−1)

15, 0, −22.

Together: −7.

= 15 + 0 + (−22) = −7

Up: 2 × 1 × 2, −1 × 1 × 5, 3 × 11 × 0.

up = 2 × 1 × 2 + (−1) × 1 × 5 + 3 × 11 × 0

4, −5, 0.

Together: −1.

= 4 + (−5) + 0 = −1

Down minus up: −7 minus (−1) is −6.

So D x = −6.

Dx = −7 − (−1) = −6

3 Find Dy

D y: the y column is replaced by the right sides, 5, 11, 2.

Dy = |1, 5, 2; 2, 11, 1; 1, 2, 3|1, 5; 2, 11; 1, 2

Down: 1 × 11 × 3, 5 × 1 × 1, 2 × 2 × 2.

down = 1 × 11 × 3 + 5 × 1 × 1 + 2 × 2 × 2

33, 5, 8.

Together: 46.

= 33 + 5 + 8 = 46

Up: 1 × 11 × 2, 2 × 1 × 1, 3 × 2 × 5.

up = 1 × 11 × 2 + 2 × 1 × 1 + 3 × 2 × 5

22, 2, 30.

Together: 54.

= 22 + 2 + 30 = 54

Down minus up: 46 minus 54 is −8.

So D y = −8.

Dy = 46 − 54 = −8

4 Find Dz

D z: the z column is replaced by the right sides, 5, 11, 2.

Dz = |1, 0, 5; 2, 1, 11; 1, −1, 2|1, 0; 2, 1; 1, −1

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 × 1 × 2, 0 × 11 × 1, 5 × 2 × (−1).

down = 1 × 1 × 2 + 0 × 11 × 1 + 5 × 2 × (−1)

2, 0, −10.

Together: −8.

= 2 + 0 + (−10) = −8

Up: 1 × 1 × 5, −1 × 11 × 1, 2 × 2 × 0.

up = 1 × 1 × 5 + (−1) × 11 × 1 + 2 × 2 × 0

5, −11, 0.

Together: −6.

= 5 + (−11) + 0 = −6

Down minus up: −8 minus (−6) is −2.

So D z = −2.

Dz = −8 − (−6) = −2

5 Divide

x is D x over D: −6 over (−2) is 3.

x = DxD = −6−2 = 3

y is D y over D: −8 over (−2) is 4.

y = DyD = −8−2 = 4

z is D z over D: −2 over (−2) is 1.

z = DzD = −2−2 = 1

So the answer is (3, 4, 1).

(x, y, z) = (3, 4, 1)

6 Check

Let's check in every equation.

We put x = 3, y = 4, z = 1 into ①.

① 3 + 2(1) = 3 + 2 = 5

We get 5, the same as the right side.

Now ②.

② 2(3) + 4 + 1 = 6 + 4 + 1 = 11

We get 11, the same as the right side.

Now ③.

③ 3 − 4 + 3(1) = 3 − 4 + 3 = 2

We get 2, 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) = (3, 4, 1)

▶ Watch it solved

Solve x + 3z = 5, 2x - y + 3z = 5, x + 3y + 3z = 11 by Cramer's rule Harder

The question

Question: solve x + 3z = 5, 2x − y + 3z = 5, x + 3y + 3z = 11 by Cramer's rule.

x + 3z = 5①
2x − y + 3z = 5②

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

x + 3y + 3z = 11③

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, 0, 3; 2, −1, 3; 1, 3, 3|1, 0; 2, −1; 1, 3

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

Then every diagonal has three numbers.

Down: 1 × (−1) × 3, 0 × 3 × 1, 3 × 2 × 3.

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

−3, 0, 18.

Together: 15.

= −3 + 0 + 18 = 15

Up: 1 × (−1) × 3, 3 × 3 × 1, 3 × 2 × 0.

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

−3, 9, 0.

Together: 6.

= −3 + 9 + 0 = 6

Down minus up: 15 minus 6 is 9.

So D = 9.

D = 15 − 6 = 9

2 Find Dx

D x: the x column is replaced by the right sides, 5, 5, 11.

Dx = |5, 0, 3; 5, −1, 3; 11, 3, 3|5, 0; 5, −1; 11, 3

Down: 5 × (−1) × 3, 0 × 3 × 11, 3 × 5 × 3.

down = 5 × (−1) × 3 + 0 × 3 × 11 + 3 × 5 × 3

−15, 0, 45.

Together: 30.

= −15 + 0 + 45 = 30

Up: 11 × (−1) × 3, 3 × 3 × 5, 3 × 5 × 0.

up = 11 × (−1) × 3 + 3 × 3 × 5 + 3 × 5 × 0

−33, 45, 0.

Together: 12.

= −33 + 45 + 0 = 12

Down minus up: 30 minus 12 is 18.

So D x = 18.

Dx = 30 − 12 = 18

3 Find Dy

D y: the y column is replaced by the right sides, 5, 5, 11.

Dy = |1, 5, 3; 2, 5, 3; 1, 11, 3|1, 5; 2, 5; 1, 11

Down: 1 × 5 × 3, 5 × 3 × 1, 3 × 2 × 11.

down = 1 × 5 × 3 + 5 × 3 × 1 + 3 × 2 × 11

15, 15, 66.

Together: 96.

= 15 + 15 + 66 = 96

Up: 1 × 5 × 3, 11 × 3 × 1, 3 × 2 × 5.

up = 1 × 5 × 3 + 11 × 3 × 1 + 3 × 2 × 5

15, 33, 30.

Together: 78.

= 15 + 33 + 30 = 78

Down minus up: 96 minus 78 is 18.

So D y = 18.

Dy = 96 − 78 = 18

4 Find Dz

D z: the z column is replaced by the right sides, 5, 5, 11.

Dz = |1, 0, 5; 2, −1, 5; 1, 3, 11|1, 0; 2, −1; 1, 3

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 × (−1) × 11, 0 × 5 × 1, 5 × 2 × 3.

down = 1 × (−1) × 11 + 0 × 5 × 1 + 5 × 2 × 3

−11, 0, 30.

Together: 19.

= −11 + 0 + 30 = 19

Up: 1 × (−1) × 5, 3 × 5 × 1, 11 × 2 × 0.

up = 1 × (−1) × 5 + 3 × 5 × 1 + 11 × 2 × 0

−5, 15, 0.

Together: 10.

= −5 + 15 + 0 = 10

Down minus up: 19 minus 10 is 9.

So D z = 9.

Dz = 19 − 10 = 9

5 Divide

x is D x over D: 18/9 is 2.

x = DxD = 189 = 2

y is D y over D: 18/9 is 2.

y = DyD = 189 = 2

z is D z over D: 9/9 is 1.

z = DzD = 99 = 1

So the answer is (2, 2, 1).

(x, y, z) = (2, 2, 1)

6 Check

Let's check in every equation.

We put x = 2, y = 2, z = 1 into ①.

① 2 + 3(1) = 2 + 3 = 5

We get 5, the same as the right side.

Now ②.

② 2(2) − 2 + 3(1) = 4 − 2 + 3 = 5

We get 5, the same as the right side.

Now ③.

③ 2 + 3(2) + 3(1) = 2 + 6 + 3 = 11

We get 11, 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) = (2, 2, 1)

▶ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. solve x + y + z = 6, 2x - y + z = 3, x + 2y - z = 2 by Cramer's rule
    Answer
    (x, y, z) = (1, 2, 3)
  2. solve x + y + z = 5, 2x + 3y + z = 15, x + 2y - z = 11 by Cramer's rule
    Answer
    (x, y, z) = (2, 4, −1)
  3. solve x + 2y + z = 14, 2x + z = 10, 2x + y + z = 14 by Cramer's rule
    Answer
    (x, y, z) = (4, 4, 2)
  4. solve x + y + 2z = 11, x + 3z = 11, x + 2z = 7 by Cramer's rule
    Answer
    (x, y, z) = (−1, 4, 4)
  5. solve 3x + y = 1, 2x + 2y + z = -1, 3x + 2y + z = 0 by Cramer's rule
    Answer
    (x, y, z) = (1, −2, 1)
  6. solve 2x + 3y + 3z = 13, x + y + 3z = 7, x + y + z = 5 by Cramer's rule
    Answer
    (x, y, z) = (2, 2, 1)
  7. solve x - y + z = 1, 3x - y + z = 5, 2x + y + z = 9 by Cramer's rule
    Answer
    (x, y, z) = (2, 3, 2)

Medium

  1. solve x + 2z = 5, 2x + y + z = 11, x - y + 3z = 2 by Cramer's rule
    Answer
    (x, y, z) = (3, 4, 1)
  2. solve 3x + 2y + z = 11, x + y - z = 2, 2x + y = 5 by Cramer's rule
    Answer
    (x, y, z) = (1, 3, 2)
  3. solve x + y + z = 4, x + 3y = 12, x + y + 3z = 0 by Cramer's rule
    Answer
    (x, y, z) = (3, 3, −2)
  4. solve x + 2y + z = 2, 3x + y = 10, x + y = 2 by Cramer's rule
    Answer
    (x, y, z) = (4, −2, 2)
  5. solve 2x + z = 3, x - y - z = -1, x + y = 6 by Cramer's rule
    Answer
    (x, y, z) = (2, 4, −1)
  6. solve 3x + y + z = 11, x + y - z = 7, 2x + y + 3z = 6 by Cramer's rule
    Answer
    (x, y, z) = (3, 3, −1)
  7. solve x + 2y + 2z = 13, 3x + 3y + z = 10, x + 2y + z = 9 by Cramer's rule
    Answer
    (x, y, z) = (−1, 3, 4)

Harder

  1. solve x + 3z = 5, 2x - y + 3z = 5, x + 3y + 3z = 11 by Cramer's rule
    Answer
    (x, y, z) = (2, 2, 1)
  2. solve x + 2y - z = 2, x + 2y = 1, 3x + 3y - z = 7 by Cramer's rule
    Answer
    (x, y, z) = (3, −1, −1)
  3. solve x + 3y = -3, x + 2y + 3z = -7, 2x + 3y + z = -2 by Cramer's rule
    Answer
    (x, y, z) = (3, −2, −2)
  4. solve x + 2y = -6, 2x + y + 3z = -3, 2x + y + z = -5 by Cramer's rule
    Answer
    (x, y, z) = (−2, −2, 1)
  5. solve x + 2z = 2, x + 2y + z = -2, x - y - z = -3 by Cramer's rule
    Answer
    (x, y, z) = (−2, −1, 2)
  6. solve 2x + 3y = 7, x - y = -4, 2x + 3y + z = 8 by Cramer's rule
    Answer
    (x, y, z) = (−1, 3, 1)