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

Step by step

The question

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

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

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 = |1, 2, 3; 0, 0, 0; 0, −1, −2|

D = −|1, 2, 3; 0, −1, −2; 0, 0, 0|

Careful!

The swap changed the sign of D.

Keep the minus in front, or the answer has the wrong sign.

Careful: a swap changes the sign of D

3 Multiply the diagonal

The diagonal: 1, −1 and 0.

Under it, all 0s.

So D is minus the diagonal multiplied: 1 × (−1) × 0.

D = −(1 × (−1) × 0)

The diagonal makes 0.

D = −0

So D = 0.

D = 0

4 Check

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

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

Down: 22.

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

= 22

Up: 22.

up: 1 × 4 × 3 + 1 × 6 × 1 + 1 × 2 × 2 = 22

Down minus up: D = 0.

D = 22 − 22 = 0

We get D = 0 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