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

Step by step

The question

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

3x + 2y + z = 11①

x + y − z = 2②

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

2x + y = 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 = |3, 2, 1; 1, 1, −1; 2, 1, 0|3, 2; 1, 1; 2, 1

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

Then every diagonal has three numbers.

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

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

0, −4, 1.

Together: −3.

= 0 + (−4) + 1 = −3

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

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

2, −3, 0.

Together: −1.

= 2 + (−3) + 0 = −1

Down minus up: −3 minus (−1) is −2.

So D = −2.

D = −3 − (−1) = −2

2 Find Dx

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

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

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

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

0, −10, 2.

Together: −8.

= 0 + (−10) + 2 = −8

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

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

5, −11, 0.

Together: −6.

= 5 + (−11) + 0 = −6

Down minus up: −8 minus (−6) is −2.

So D x = −2.

D_x = −8 − (−6) = −2

3 Find Dy

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

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

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

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

0, −22, 5.

Together: −17.

= 0 + (−22) + 5 = −17

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

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

4, −15, 0.

Together: −11.

= 4 + (−15) + 0 = −11

Down minus up: −17 minus (−11) is −6.

So D y = −6.

D_y = −17 − (−11) = −6

4 Find Dz

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

D_z = |3, 2, 11; 1, 1, 2; 2, 1, 5|3, 2; 1, 1; 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: 3 × 1 × 5, 2 × 2 × 2, 11 × 1 × 1.

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

15, 8, 11.

Together: 34.

= 15 + 8 + 11 = 34

Up: 2 × 1 × 11, 1 × 2 × 3, 5 × 1 × 2.

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

22, 6, 10.

Together: 38.

= 22 + 6 + 10 = 38

Down minus up: 34 minus 38 is −4.

So D z = −4.

D_z = 34 − 38 = −4

5 Divide

x is D x over D: −2 over (−2) is 1.

x = D_xD = −2−2 = 1

y is D y over D: −6 over (−2) is 3.

y = D_yD = −6−2 = 3

z is D z over D: −4 over (−2) is 2.

z = D_zD = −4−2 = 2

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

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

6 Check

Let's check in every equation.

We put x = 1, y = 3, z = 2 into ①.

① 3(1) + 2(3) + 2 = 3 + 6 + 2 = 11

We get 11, the same as the right side.

Now ②.

② 1 + 3 − 2 = 2

We get 2, the same as the right side.

Now ③.

③ 2(1) + 3 = 2 + 3 = 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