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Det by row operations [1, 2, 3; 0, 1, 4; 2, 5, 9]

Step by step, with the reason for every step · Determinant rules · Linear Algebra

Step by step

The question

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

D = |1, 2, 3; 0, 1, 4; 2, 5, 9|

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, 1, 4; 0, 1, 3|
D = |1, 2, 3; 0, 1, 4; 0, 0, −1|

Careful!

Taking away a multiple of a row changes the entries, but not D.

Careful: adding a multiple of a row: D stays the same

3 Multiply the diagonal

The diagonal: 1, 1 and −1.

Under it, all 0s.

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

D = 1 × 1 × (−1)

So D = −1.

D = −1

4 Check

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

D = |1, 2, 3; 0, 1, 4; 2, 5, 9|1, 2; 0, 1; 2, 5

Down: 25.

down: 1 × 1 × 9 + 2 × 4 × 2 + 3 × 0 × 5
= 25

Up: 26.

up: 2 × 1 × 3 + 5 × 4 × 1 + 9 × 0 × 2 = 26

Down minus up: D = −1.

D = 25 − 26 = −1

We get D = −1 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
D = −1

Your turn

Try this one yourself, then check:

By row operations: D = |1, 3, 1; 0, 1, 5; 1, 4, 7|
Show the answer
D = 1

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