Step by step
The question
Question: solve x + y + z = 5, 2x + 3y + z = 15, x + 2y − z = 11 by Cramer's rule.
Three equations, three letters: x, y and z.
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.
1 Find D
D: the numbers in front of x, y and z, in their places.
We write the first two columns again, on the right.
Then every diagonal has three numbers.
Down: 1 × 3 × (−1), 1 × 1 × 1, 1 × 2 × 2.
−3, 1, 4.
Together: 2.
Up: 1 × 3 × 1, 2 × 1 × 1, −1 × 2 × 1.
3, 2, −2.
Together: 3.
Down minus up: 2 minus 3 is −1.
So D = −1.
2 Find Dx
D x: the x column is replaced by the right sides, 5, 15, 11.
Down: 5 × 3 × (−1), 1 × 1 × 11, 1 × 15 × 2.
−15, 11, 30.
Together: 26.
Up: 11 × 3 × 1, 2 × 1 × 5, −1 × 15 × 1.
33, 10, −15.
Together: 28.
Down minus up: 26 minus 28 is −2.
So D x = −2.
3 Find Dy
D y: the y column is replaced by the right sides, 5, 15, 11.
Down: 1 × 15 × (−1), 5 × 1 × 1, 1 × 2 × 11.
−15, 5, 22.
Together: 12.
Up: 1 × 15 × 1, 11 × 1 × 1, −1 × 2 × 5.
15, 11, −10.
Together: 16.
Down minus up: 12 minus 16 is −4.
So D y = −4.
4 Find Dz
D z: the z column is replaced by the right sides, 5, 15, 11.
Careful!
For D z we replace the z column, the third one.
The other two stay.
Down: 1 × 3 × 11, 1 × 15 × 1, 5 × 2 × 2.
33, 15, 20.
Together: 68.
Up: 1 × 3 × 5, 2 × 15 × 1, 11 × 2 × 1.
15, 30, 22.
Together: 67.
Down minus up: 68 minus 67 is 1.
So D z = 1.
5 Divide
x is D x over D: −2 over (−1) is 2.
y is D y over D: −4 over (−1) is 4.
z is D z over D: 1 over (−1) is −1.
So the answer is (2, 4, −1).
6 Check
Let's check in every equation.
We put x = 2, y = 4, z = −1 into ①.
We get 5, the same as the right side.
Now ②.
We get 15, the same as the right side.
Now ③.
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.