Solve x + 2y + z = 14, 2x + z = 10, 2x + y + z = 14 by Cramer's rule Step by Step

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The question

Question: solve x + 2y + z = 14, 2x + z = 10, 2x + y + z = 14 by Cramer's rule.

x + 2y + z = 14①

2x + z = 10②

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

2x + y + z = 14③

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 = D_xD, y = D_yD, z = D_zD

1 Find D

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

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

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

Then every diagonal has three numbers.

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

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

0, 4, 2.

Together: 6.

= 0 + 4 + 2 = 6

Up: 2 × 0 × 1, 1 × 1 × 1, 1 × 2 × 2.

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

0, 1, 4.

Together: 5.

= 0 + 1 + 4 = 5

Down minus up: 6 minus 5 is 1.

So D = 1.

D = 6 − 5 = 1

2 Find Dx

D x: the x column is replaced by the right sides, 14, 10, 14.

D_x = |14, 2, 1; 10, 0, 1; 14, 1, 1|14, 2; 10, 0; 14, 1

Down: 14 × 0 × 1, 2 × 1 × 14, 1 × 10 × 1.

down = 14 × 0 × 1 + 2 × 1 × 14 + 1 × 10 × 1

0, 28, 10.

Together: 38.

= 0 + 28 + 10 = 38

Up: 14 × 0 × 1, 1 × 1 × 14, 1 × 10 × 2.

up = 14 × 0 × 1 + 1 × 1 × 14 + 1 × 10 × 2

0, 14, 20.

Together: 34.

= 0 + 14 + 20 = 34

Down minus up: 38 minus 34 is 4.

So D x = 4.

D_x = 38 − 34 = 4

3 Find Dy

D y: the y column is replaced by the right sides, 14, 10, 14.

D_y = |1, 14, 1; 2, 10, 1; 2, 14, 1|1, 14; 2, 10; 2, 14

Down: 1 × 10 × 1, 14 × 1 × 2, 1 × 2 × 14.

down = 1 × 10 × 1 + 14 × 1 × 2 + 1 × 2 × 14

10, 28, 28.

Together: 66.

= 10 + 28 + 28 = 66

Up: 2 × 10 × 1, 14 × 1 × 1, 1 × 2 × 14.

up = 2 × 10 × 1 + 14 × 1 × 1 + 1 × 2 × 14

20, 14, 28.

Together: 62.

= 20 + 14 + 28 = 62

Down minus up: 66 minus 62 is 4.

So D y = 4.

D_y = 66 − 62 = 4

4 Find Dz

D z: the z column is replaced by the right sides, 14, 10, 14.

D_z = |1, 2, 14; 2, 0, 10; 2, 1, 14|1, 2; 2, 0; 2, 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 × 0 × 14, 2 × 10 × 2, 14 × 2 × 1.

down = 1 × 0 × 14 + 2 × 10 × 2 + 14 × 2 × 1

0, 40, 28.

Together: 68.

= 0 + 40 + 28 = 68

Up: 2 × 0 × 14, 1 × 10 × 1, 14 × 2 × 2.

up = 2 × 0 × 14 + 1 × 10 × 1 + 14 × 2 × 2

0, 10, 56.

Together: 66.

= 0 + 10 + 56 = 66

Down minus up: 68 minus 66 is 2.

So D z = 2.

D_z = 68 − 66 = 2

5 Divide

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

x = D_xD = 41 = 4

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

y = D_yD = 41 = 4

z is D z over D: 2/1 is 2.

z = D_zD = 21 = 2

So the answer is (4, 4, 2).

(x, y, z) = (4, 4, 2)

6 Check

Let's check in every equation.

We put x = 4, y = 4, z = 2 into ①.

① 4 + 2(4) + 2 = 4 + 8 + 2 = 14

We get 14, the same as the right side.

Now ②.

② 2(4) + 2 = 8 + 2 = 10

We get 10, the same as the right side.

Now ③.

③ 2(4) + 4 + 2 = 8 + 4 + 2 = 14

We get 14, 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