math111_logo Inverse

3. Inverse of 2 by 2 matrix

A 1 by 1 matrix A = (a) is invertible if and only if a ≠ 0. Moreover, the inverse A-1 = (a-1).

Assume the 2 by 2 matrix

A = [ a b ]
c d

has

B = [ x z ]
y w

as the inverse. Then from AB = I we have

ax + by = 1, cx + dy = 0, az + bw = 0, cz + dw = 1.

The first two is a system of linear equations with A as the coefficient matrix and x, y as variables. Then d[equation 1] - b[equation 2] and - c[equation 1] + a[equation 2] give us

(ad - bc)x = d, (ad - bc)y = - c.

Similarly, from the 3rd and the 4th equations we get

(ad - bc)z = - b, (ad - bc)w = a.

Thus we conclude the ⇐ direction of the following result.

A 2 by 2 matrix [ a b ] is invertible ⇔ ad - bc ≠ 0. Moreover, the inverse is
c d
1 [ d -b ]
ad - bc -c a

For the ⇒ direction, see this exercise.

Example We have

[ 1 3 ]-1 = (1×4 - 3×2)-1[ 4 -3 ]-1 = [ -2 3/2 ]
2 4 -2 1 1 -1/2

We verify our computation as follows.

[ 1 3 ] [ -2 3/2 ] = [ 1×(-2) + 3×1 1×3/2 + 3×(-1/2) ] = [ 1 0 ]
2 4 1 -1/2 2×(-2) + 4×1 2×3/2 + 4×(-1/2) 0 1
[ -2 3/2 ] [ 1 3 ] = [ (-2)×1 + (3/2)×2 (-2)×3 + (3/2)×4 ] = [ 1 0 ]
1 -1/2 2 4 1×1 + (-1/2)×2 1×3 + (-1/2)×4 0 1

The inverse also tells us that the system

x1 + 3x2 = 1
2x1 + 4x2 = -4

has

[ x1 ] = [ 1 3 ]-1 [ 1 ] = [ -2 3/2 ] [ 1 ] = [ 4 ]
x2 2 4 -4 1 -1/2 4 -1

as the unique solution.

Example The following matrices

[ 2 4 ]
-1 -2
[ 1 5 ]
3 15

are not invertible because 2×(-2) - 4×(-1) = 0, 1×15 - 5×3 = 0.

The formula for the inverse of 2 by 2 matrices can be generalized to bigger matrices. However, the amount of computation involved in such a formula will become too big to be practical.


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