Solve 2x + 3y + 3z = 13, x + y + 3z = 7, x + y + z = 5 by Cramer's… Step by Step

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Question: solve 2x + 3y + 3z = 13, x + y + 3z = 7, x + y + z = 5 by Cramer's rule.

2x + 3y + 3z = 13①

x + y + 3z = 7②

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

x + y + z = 5③

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 = |2, 3, 3; 1, 1, 3; 1, 1, 1|2, 3; 1, 1; 1, 1

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

Then every diagonal has three numbers.

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

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

2, 9, 3.

Together: 14.

= 2 + 9 + 3 = 14

Up: 1 × 1 × 3, 1 × 3 × 2, 1 × 1 × 3.

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

3, 6, 3.

Together: 12.

= 3 + 6 + 3 = 12

Down minus up: 14 minus 12 is 2.

So D = 2.

D = 14 − 12 = 2

2 Find Dx

D x: the x column is replaced by the right sides, 13, 7, 5.

D_x = |13, 3, 3; 7, 1, 3; 5, 1, 1|13, 3; 7, 1; 5, 1

Down: 13 × 1 × 1, 3 × 3 × 5, 3 × 7 × 1.

down = 13 × 1 × 1 + 3 × 3 × 5 + 3 × 7 × 1

13, 45, 21.

Together: 79.

= 13 + 45 + 21 = 79

Up: 5 × 1 × 3, 1 × 3 × 13, 1 × 7 × 3.

up = 5 × 1 × 3 + 1 × 3 × 13 + 1 × 7 × 3

15, 39, 21.

Together: 75.

= 15 + 39 + 21 = 75

Down minus up: 79 minus 75 is 4.

So D x = 4.

D_x = 79 − 75 = 4

3 Find Dy

D y: the y column is replaced by the right sides, 13, 7, 5.

D_y = |2, 13, 3; 1, 7, 3; 1, 5, 1|2, 13; 1, 7; 1, 5

Down: 2 × 7 × 1, 13 × 3 × 1, 3 × 1 × 5.

down = 2 × 7 × 1 + 13 × 3 × 1 + 3 × 1 × 5

14, 39, 15.

Together: 68.

= 14 + 39 + 15 = 68

Up: 1 × 7 × 3, 5 × 3 × 2, 1 × 1 × 13.

up = 1 × 7 × 3 + 5 × 3 × 2 + 1 × 1 × 13

21, 30, 13.

Together: 64.

= 21 + 30 + 13 = 64

Down minus up: 68 minus 64 is 4.

So D y = 4.

D_y = 68 − 64 = 4

4 Find Dz

D z: the z column is replaced by the right sides, 13, 7, 5.

D_z = |2, 3, 13; 1, 1, 7; 1, 1, 5|2, 3; 1, 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: 2 × 1 × 5, 3 × 7 × 1, 13 × 1 × 1.

down = 2 × 1 × 5 + 3 × 7 × 1 + 13 × 1 × 1

10, 21, 13.

Together: 44.

= 10 + 21 + 13 = 44

Up: 1 × 1 × 13, 1 × 7 × 2, 5 × 1 × 3.

up = 1 × 1 × 13 + 1 × 7 × 2 + 5 × 1 × 3

13, 14, 15.

Together: 42.

= 13 + 14 + 15 = 42

Down minus up: 44 minus 42 is 2.

So D z = 2.

D_z = 44 − 42 = 2

5 Divide

x is D x over D: 4/2 is 2.

x = D_xD = 42 = 2

y is D y over D: 4/2 is 2.

y = D_yD = 42 = 2

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

z = D_zD = 22 = 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(2) + 3(2) + 3(1) = 4 + 6 + 3 = 13

We get 13, the same as the right side.

Now ②.

② 2 + 2 + 3(1) = 2 + 2 + 3 = 7

We get 7, the same as the right side.

Now ③.

③ 2 + 2 + 1 = 5

We get 5, 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