Inverse of [3, 1, 1; 1, 0, -1; 2, 0, -1] Step by Step

Step by step

The question

Question: find the inverse of A.

A = [3, 1, 1; 1, 0, −1; 2, 0, −1]

The inverse A⁻¹ is the matrix that times A gives I: 1s down the diagonal, 0s elsewhere.

A × A⁻¹ = I

1 Find the determinant

First we need D, the determinant of A, by the diagonals.

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

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

Then every diagonal has three numbers.

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

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

0, −2, 0.

Together: −2.

= 0 + (−2) + 0 = −2

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

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

0, 0, −1.

Together: −1.

= 0 + 0 + (−1) = −1

Down minus up: −2 minus (−1) is −1.

So D = −1.

D = −2 − (−1) = −1

2 Find the minors

Each entry has a minor: cover its row and its column.

A 2 by 2 is left.

Its determinant, down minus up, is the minor.

minors = [0×(−1) − 0×(−1), 1×(−1) − 2×(−1), 1×0 − 2×0; 1×(−1) − 0×1, 3×(−1) − 2×1, 3×0 − 2×1; 1×(−1) − 0×1, 3×(−1) − 1×1, 3×0 − 1×1]

Work out each one: 0, 1, 0; −1, −5, −2; −1, −4, −1.

= [0, 1, 0; −1, −5, −2; −1, −4, −1]

3 Change the signs

The signs go plus, minus, plus, like a checkerboard.

Where there is a minus, the minor changes sign.

signs: [+, −, +; −, +, −; +, −, +]

So the cofactors are rows 0, −1, 0, 1, −5, 2 and −1, 4, −1.

The purple ones changed sign.

C = [0, −1, 0; 1, −5, 2; −1, 4, −1]

Careful!

Only the minus places change sign: where the row number plus the column number is odd.

Careful: the − places: row + column is odd

4 Swap rows and columns

Row 1 of C becomes column 1, row 2 becomes column 2, row 3 becomes column 3.

Then we divide by D: times 1 over −1.

A⁻¹ = 1−1 × [0, 1, −1; −1, −5, 4; 0, 2, −1]

5 Divide by the determinant

Dividing by −1 changes every sign.

So the answer is rows 0, −1, 1, 1, 5, −4 and 0, −2, 1.

A⁻¹ = [0, −1, 1; 1, 5, −4; 0, −2, 1]

6 Check

Let's check: A times our answer must be I.

Row 1 of A times column 1 of the answer: 1.

row 1 × column 1: 3 × 0 + 1 × 1 + 1 × 0 = 1

Row 1 times column 2: 0.

row 1 × column 2: 3 × (−1) + 1 × 5 + 1 × (−2) = 0

The other entries work the same way: 1s down the diagonal, 0s elsewhere.

So the answer is right!

A × A⁻¹ = I

Let's look back.

First: find the determinant.

Then: find the minors.

Then: change the signs.

Then: swap rows and columns.

Then: divide by the determinant.

Last: check.

Answer