Det by row operations [3, 6, 3; 1, 4, 2; 2, 6, 7] Step by Step

Step by step

The question

Question: find the determinant by row operations of this matrix.

D = |3, 6, 3; 1, 4, 2; 2, 6, 7|

We call the rows R 1, R 2 and R 3.

1 The rules

Row moves change D in a known way.

A swap changes its sign.

A number taken out of a row goes in front.

Adding a multiple of one row to another changes nothing.

And with 0s under the diagonal, D is just the diagonal multiplied.

swap two rows: D changes sign

take a number out of a row: it goes in front

add a multiple of a row: D stays the same

0s under the diagonal: D = the diagonal multiplied

2 Get 0s under the diagonal

D = 3 × |1, 2, 1; 1, 4, 2; 2, 6, 7|

D = 3 × |1, 2, 1; 0, 2, 1; 0, 2, 5|

D = 3 × |1, 2, 1; 0, 2, 1; 0, 0, 4|

Careful!

The 3 we took out stays in front.

D is 3 times the new determinant.

Careful: the 3 stays in front

3 Multiply the diagonal

The diagonal: 1, 2 and 4.

Under it, all 0s.

So D is 3 times the diagonal multiplied: 1 × 2 × 4.

D = 3 × (1 × 2 × 4)

The diagonal makes 8.

D = 3 × 8

So D = 24.

D = 24

4 Check

Let's check with the diagonal method, on the first matrix.

D = |3, 6, 3; 1, 4, 2; 2, 6, 7|3, 6; 1, 4; 2, 6

Down: 126.

down: 3 × 4 × 7 + 6 × 2 × 2 + 3 × 1 × 6

= 126

Up: 102.

up: 2 × 4 × 3 + 6 × 2 × 3 + 7 × 1 × 6 = 102

Down minus up: D = 24.

D = 126 − 102 = 24

We get D = 24 again.

So the answer is right!

Let's look back.

First: the rules.

Then: get 0s under the diagonal.

Then: multiply the diagonal.

Last: check.

Answer