The Inverse of a 2 × 2 Matrix

The matrix that undoes A

The inverse of a square matrix A is the matrix that undoes it. It is written $A^{-1}$A⁻¹, and multiplying the two together in either order gives the identity: $A{A}^{-1}=I$A A⁻¹ = I and $A^{-1}A=I$A⁻¹ A = I. It plays the part that $\frac{1}{5}$1/5 plays for 5, since $5\times \frac{1}{5}=1$5 × 1/5 = 1.

The determinant decides whether there is one. A 2 × 2 matrix whose determinant ad − bc is 0 has no inverse, because it sends different points to the same place and nothing can send them back apart. Every other 2 × 2 matrix has exactly one inverse, and three moves find it.

Three moves

Start from A = (a b; c d). First, swap the two entries on the main diagonal, so that a and d change places. Second, change the signs of the other two entries, b and c, and leave them where they are. That gives the matrix (d −b; −c a).

Last, divide every entry of (d −b; −c a) by the determinant of A, which is ad − bc. The result is the inverse.

The main diagonal of A, a and d, swaps places, and the other two entries keep their places with their signs changed. Dividing this matrix by ad − bc gives the inverse.

Why the three moves work

The determinant lesson multiplied A by (d −b; −c a). The top left entry is a × d + b × (−c) = ad − bc, the top right is a × (−b) + b × a = 0, the bottom left is c × d + d × (−c) = 0, and the bottom right is c × (−b) + d × a = ad − bc. So the product is (ad − bc)I, the identity with every 1 replaced by ad − bc.

Dividing every entry of (d −b; −c a) by ad − bc divides that product by ad − bc as well, which leaves I. Multiplying in the other order, (d −b; −c a) times A, gives (ad − bc)I too, so the same matrix undoes A from either side. And the division is only possible when ad − bc is not 0, which is why a matrix with determinant 0 has no inverse.

An inverse found and checked

Take A = (3 5; 1 2). Its determinant is 3 × 2 − 5 × 1 = 6 − 5 = 1. Swapping the 3 and the 2 gives (2 5; 1 3), and changing the signs of the 5 and the 1 gives (2 −5; −1 3). Dividing by 1 changes nothing, so the inverse of A is the matrix (2 −5; −1 3).

Check it by multiplying back, row by column. Row 1 of A with column 1 of the inverse gives 3 × 2 + 5 × (−1) = 6 − 5 = 1. Row 1 with column 2 gives 3 × (−5) + 5 × 3 = −15 + 15 = 0. Row 2 with column 1 gives 1 × 2 + 2 × (−1) = 0, and row 2 with column 2 gives 1 × (−5) + 2 × 3 = 1. The product is (1 0; 0 1), which is I.

Row 1 of A meets column 2 of the inverse: 3 × (−5) + 5 × 3 = 0, the top right entry of I.

When the determinant is not 1

Take B = (4 2; 5 3). Its determinant is 4 × 3 − 2 × 5 = 12 − 10 = 2. Swapping and changing signs gives (3 −2; −5 4), and now the division matters: every entry is divided by 2. The inverse of B has top row $\frac{3}{2}$3/2 and −1, and bottom row $-\frac{5}{2}$−5/2 and 2.

Check the top row of the product: $4\times \frac{3}{2}+2\times \left(-\frac{5}{2}\right)=6-5=1$4 × 3/2 + 2 × (−5/2) = 6 − 5 = 1, and 4 × (−1) + 2 × 2 = −4 + 4 = 0. The bottom row gives $5\times \frac{3}{2}+3\times \left(-\frac{5}{2}\right)=0$5 × 3/2 + 3 × (−5/2) = 0 and 5 × (−1) + 3 × 2 = 1, so B times its inverse is I.

It is often tidier to leave the division outside, writing the inverse as one half times (3 −2; −5 4). The two forms are the same matrix.

Dividing by the determinant 2 is multiplying every entry by one half, so the 3 in the top left becomes $\frac{3}{2}$3/2.

A negative determinant

For C = (2 3; 1 1), the determinant is 2 × 1 − 3 × 1 = −1. Swapping and changing signs gives (1 −3; −1 2), and dividing by −1 changes the sign of every entry, so the inverse of C is the matrix (−1 3; 1 −2). Check the top left entry of the product: 2 × (−1) + 3 × 1 = 1.

Sending a point back

A matrix moves points, and its inverse moves them back. A = (3 5; 1 2) sends the column (1; 1) to (8; 3), since 3 × 1 + 5 × 1 = 8 and 1 × 1 + 2 × 1 = 3. The inverse sends (8; 3) back to (1; 1), the point it started from, since 2 × 8 + (−5) × 3 = 16 − 15 = 1 and (−1) × 8 + 3 × 3 = −8 + 9 = 1.

That is $A^{-1}A=I$A⁻¹ A = I at work: doing A and then its inverse leaves every point where it was.

The inverse takes (8; 3), where A sent (1; 1), back to (1; 1). The top entry is 2 × 8 + (−5) × 3 = 1.

The usual mistakes

Changing the signs without swapping the diagonal. For (3 5; 1 2) that gives (3 −5; −1 2), and the top left entry of the product is 3 × 3 + 5 × (−1) = 4, not 1.

Swapping the diagonal without changing the signs. (2 5; 1 3) times A has top left entry 2 × 3 + 5 × 1 = 11, so it is not the inverse either.

Changing the signs on the main diagonal instead of the other one. The entries that swap keep their signs; only b and c change sign.

Forgetting to divide. For B = (4 2; 5 3), the matrix (3 −2; −5 4) multiplies B to 2I, not I: its top left entry should be $\frac{3}{2}$3/2, not 3.

Looking for an inverse when the determinant is 0. The division by ad − bc cannot be done, and no matrix undoes A.

A message in code

In the application below, two friends code each pair of letters by multiplying it by a matrix E with determinant 1. The inverse of E, found by the three moves and checked against I, turns the numbers that arrive back into the letters that were sent.