Worked examples
The question
Question: find the inverse of A.
A = [3, 5; 1, 2]
The inverse of A is written A⁻¹.
A times its inverse gives I: 1s down the diagonal, 0s elsewhere.
Times I changes nothing, like times 1.
A × A−1 = I, I = [1, 0; 0, 1]
1 Find the determinant
First we need D, the determinant of A.
D = |3, 5; 1, 2|
So D = 3 × 2 − 1 × 5.
D = 3 × 2 − 1 × 5
3 times 2 is 6.
1 times 5 is 5.
D = 6 − 5
6 minus 5 is 1.
So D = 1.
D = 1
2 Swap and change signs
The top left and bottom right swap places: 3 and 2.
The other two stay where they are, but change their signs: 5 becomes −5, 1 becomes −1.
Then we divide by D.
Dividing by 1 is the same as times 1/1.
A−1 = 11 × [2, −5; −1, 3]
Careful!
Only the other two change sign.
3 and 2 just swap places.
Careful: 3 and 2 swap, keep their signs
3 Divide by the determinant
Dividing by 1 changes nothing.
So the answer is rows 2, −5 and −1, 3.
A−1 = [2, −5; −1, 3]
4 Check
Let's check: A times the swapped matrix.
It must give 1s down the diagonal, 0s elsewhere.
= [3, 5; 1, 2] × [2, −5; −1, 3]
= [3 × 2 + 5 × (−1), 3 × (−5) + 5 × 3; 1 × 2 + 2 × (−1), 1 × (−5) + 2 × 3]
We work out every times.
= [6 + (−5), −15 + 15; 2 + (−2), −5 + 6]
Then we add: 1 down the diagonal, 0 everywhere else.
= [1, 0; 0, 1]
Divided by 1, that is I.
So A times our answer is I: the answer is right!
A × A−1 = I
Let's look back.
First: find the determinant.
Then: swap and change signs.
Then: divide by the determinant.
Last: check.
Answer A−1 = [2, −5; −1, 3]
▶ Watch it solved
The question
Question: find the inverse of A.
A = [4, 7; 2, 4]
The inverse of A is written A⁻¹.
A times its inverse gives I: 1s down the diagonal, 0s elsewhere.
Times I changes nothing, like times 1.
A × A−1 = I, I = [1, 0; 0, 1]
1 Find the determinant
First we need D, the determinant of A.
D = |4, 7; 2, 4|
So D = 4 × 4 − 2 × 7.
D = 4 × 4 − 2 × 7
4 times 4 is 16.
2 times 7 is 14.
D = 16 − 14
16 minus 14 is 2.
So D = 2.
D = 2
2 Swap and change signs
The top left and bottom right swap places: 4 and 4.
The other two stay where they are, but change their signs: 7 becomes −7, 2 becomes −2.
Then we divide by D.
Dividing by 2 is the same as times 1/2.
A−1 = 12 × [4, −7; −2, 4]
Careful!
Only the other two change sign.
4 and 4 just swap places.
Careful: 4 and 4 swap, keep their signs
3 Divide by the determinant
Every entry over 2.
A−1 = [4 over 2, −7 over 2; −2 over 2, 4 over 2]
Then we make each one simpler.
The answer: rows 2, −7/2 and −1, 2.
A−1 = [2, −7 over 2; −1, 2]
4 Check
Let's check: A times the swapped matrix.
It must give 2s down the diagonal, 0s elsewhere.
= [4, 7; 2, 4] × [4, −7; −2, 4]
= [4 × 4 + 7 × (−2), 4 × (−7) + 7 × 4; 2 × 4 + 4 × (−2), 2 × (−7) + 4 × 4]
We work out every times.
= [16 + (−14), −28 + 28; 8 + (−8), −14 + 16]
Then we add: 2 down the diagonal, 0 everywhere else.
= [2, 0; 0, 2]
Divided by 2, that is I.
So A times our answer is I: the answer is right!
A × A−1 = I
Let's look back.
First: find the determinant.
Then: swap and change signs.
Then: divide by the determinant.
Last: check.
Answer A−1 = [2, −7 over 2; −1, 2]
▶ Watch it solved
The question
Question: find the inverse of A.
A = [3, −2; 4, −2]
The inverse of A is written A⁻¹.
A times its inverse gives I: 1s down the diagonal, 0s elsewhere.
Times I changes nothing, like times 1.
A × A−1 = I, I = [1, 0; 0, 1]
1 Find the determinant
First we need D, the determinant of A.
D = |3, −2; 4, −2|
So D = 3 × (−2) − 4 × (−2).
D = 3 × (−2) − 4 × (−2)
3 times (−2) is −6.
4 times (−2) is −8.
D = −6 − (−8)
Taking away a negative is adding: −6 + 8.
−6 − (−8) = −6 + 8
−6 minus (−8) is 2.
So D = 2.
D = 2
2 Swap and change signs
The top left and bottom right swap places: 3 and −2.
The other two stay where they are, but change their signs: −2 becomes 2, 4 becomes −4.
Then we divide by D.
Dividing by 2 is the same as times 1/2.
A−1 = 12 × [−2, 2; −4, 3]
Careful!
Only the other two change sign.
3 and −2 just swap places.
Careful: 3 and −2 swap, keep their signs
3 Divide by the determinant
Every entry over 2.
A−1 = [−2 over 2, 2 over 2; −4 over 2, 3 over 2]
Then we make each one simpler.
The answer: rows −1, 1 and −2, 3/2.
A−1 = [−1, 1; −2, 3 over 2]
4 Check
Let's check: A times the swapped matrix.
It must give 2s down the diagonal, 0s elsewhere.
= [3, −2; 4, −2] × [−2, 2; −4, 3]
= [3 × (−2) + −2 × (−4), 3 × 2 + −2 × 3; 4 × (−2) + −2 × (−4), 4 × 2 + −2 × 3]
We work out every times.
= [−6 + 8, 6 + (−6); −8 + 8, 8 + (−6)]
Then we add: 2 down the diagonal, 0 everywhere else.
= [2, 0; 0, 2]
Divided by 2, that is I.
So A times our answer is I: the answer is right!
A × A−1 = I
Let's look back.
First: find the determinant.
Then: swap and change signs.
Then: divide by the determinant.
Last: check.
Answer A−1 = [−1, 1; −2, 3 over 2]
▶ Watch it solved
