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

Step by step

The question

Question: find the inverse of A.

A = [1, −2, 0; −1, 3, 1; 0, −1, 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 = |1, −2, 0; −1, 3, 1; 0, −1, 1|1, −2; −1, 3; 0, −1

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

Then every diagonal has three numbers.

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

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

3, 0, 0.

Together: 3.

= 3 + 0 + 0 = 3

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

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

0, −1, 2.

Together: 1.

= 0 + (−1) + 2 = 1

Down minus up: 3 minus 1 is 2.

So D = 2.

D = 3 − 1 = 2

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

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

= [4, −1, 1; −2, 1, −1; −2, 1, 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 4, 1, 1, 2, 1, 1 and −2, −1, 1.

The purple ones changed sign.

C = [4, 1, 1; 2, 1, 1; −2, −1, 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 1/2.

A⁻¹ = 12 × [4, 2, −2; 1, 1, −1; 1, 1, 1]

5 Divide by the determinant

Every entry over 2, made simpler: rows 2, 1, −1, 1/2, 1/2, −1/2 and 1/2, 1/2, 1/2.

A⁻¹ = [2, 1, −1; 1 over 2, 1 over 2, −1 over 2; 1 over 2, 1 over 2, 1 over 2]

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: 1 × 2 + −2 × 12 + 0 × 12 = 1

Row 1 times column 2: 0.

row 1 × column 2: 1 × 1 + −2 × 12 + 0 × 12 = 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