Worked examples
The question
Question: find the inverse of A.
A = [1, 2, 0; 0, 1, 0; 0, 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 = |1, 2, 0; 0, 1, 0; 0, 0, 1|1, 2; 0, 1; 0, 0
We write the first two columns again, on the right.
Then every diagonal has three numbers.
Down: 1 × 1 × 1, 2 × 0 × 0, 0 × 0 × 0.
down = 1 × 1 × 1 + 2 × 0 × 0 + 0 × 0 × 0
1, 0, 0.
Together: 1.
= 1 + 0 + 0 = 1
Up: 0 × 1 × 0, 0 × 0 × 1, 1 × 0 × 2.
up = 0 × 1 × 0 + 0 × 0 × 1 + 1 × 0 × 2
0, 0, 0.
Together: 0.
= 0 + 0 + 0 = 0
Down minus up: 1 minus 0 is 1.
So D = 1.
D = 1 − 0 = 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 = [1×1 − 0×0, 0×1 − 0×0, 0×0 − 0×1; 2×1 − 0×0, 1×1 − 0×0, 1×0 − 0×2; 2×0 − 1×0, 1×0 − 0×0, 1×1 − 0×2]
Work out each one: 1, 0, 0; 2, 1, 0; 0, 0, 1.
= [1, 0, 0; 2, 1, 0; 0, 0, 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 1, 0, 0, −2, 1, 0 and 0, 0, 1.
The purple ones changed sign.
C = [1, 0, 0; −2, 1, 0; 0, 0, 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 1/1.
A−1 = 11 × [1, −2, 0; 0, 1, 0; 0, 0, 1]
5 Divide by the determinant
Dividing by 1 changes nothing.
So the answer is rows 1, −2, 0, 0, 1, 0 and 0, 0, 1.
A−1 = [1, −2, 0; 0, 1, 0; 0, 0, 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: 1 × 1 + 2 × 0 + 0 × 0 = 1
Row 1 times column 2: 0.
row 1 × column 2: 1 × (−2) + 2 × 1 + 0 × 0 = 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 A−1 = [1, −2, 0; 0, 1, 0; 0, 0, 1]
▶ Watch it solved
The question
Question: find the inverse of A.
A = [2, 0, 1; 0, 3, 1; 3, −2, 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 = |2, 0, 1; 0, 3, 1; 3, −2, 1|2, 0; 0, 3; 3, −2
We write the first two columns again, on the right.
Then every diagonal has three numbers.
Down: 2 × 3 × 1, 0 × 1 × 3, 1 × 0 × (−2).
down = 2 × 3 × 1 + 0 × 1 × 3 + 1 × 0 × (−2)
6, 0, 0.
Together: 6.
= 6 + 0 + 0 = 6
Up: 3 × 3 × 1, −2 × 1 × 2, 1 × 0 × 0.
up = 3 × 3 × 1 + (−2) × 1 × 2 + 1 × 0 × 0
9, −4, 0.
Together: 5.
= 9 + (−4) + 0 = 5
Down minus up: 6 minus 5 is 1.
So D = 1.
D = 6 − 5 = 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 = [3×1 − −2×1, 0×1 − 3×1, 0×(−2) − 3×3; 0×1 − −2×1, 2×1 − 3×1, 2×(−2) − 3×0; 0×1 − 3×1, 2×1 − 0×1, 2×3 − 0×0]
Work out each one: 5, −3, −9; 2, −1, −4; −3, 2, 6.
= [5, −3, −9; 2, −1, −4; −3, 2, 6]
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 5, 3, −9, −2, −1, 4 and −3, −2, 6.
The purple ones changed sign.
C = [5, 3, −9; −2, −1, 4; −3, −2, 6]
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 1/1.
A−1 = 11 × [5, −2, −3; 3, −1, −2; −9, 4, 6]
5 Divide by the determinant
Dividing by 1 changes nothing.
So the answer is rows 5, −2, −3, 3, −1, −2 and −9, 4, 6.
A−1 = [5, −2, −3; 3, −1, −2; −9, 4, 6]
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: 2 × 5 + 0 × 3 + 1 × (−9) = 1
Row 1 times column 2: 0.
row 1 × column 2: 2 × (−2) + 0 × (−1) + 1 × 4 = 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 A−1 = [5, −2, −3; 3, −1, −2; −9, 4, 6]
▶ Watch it solved
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 A−1 = [2, 1, −1; 1 over 2, 1 over 2, −1 over 2; 1 over 2, 1 over 2, 1 over 2]
▶ Watch it solved
