Inverting 2×2 matrices

In this note we invert the general 2×2 matrix as in Theorem 1.4.5 of Anton– Rorres. However, we apply only the standard inversion method, with no guesswork or ingenuity needed.

Theorem 1 The 2×2 matrix A = [ \begin{matrix}a & b \\ c & d \end{matrix} ] is invertible if and only if Δ ≠ 0, where

we write Δ = ad − bc. When Δ ≠ 0, the inverse is

A^{−1} = \frac{1}{Δ} [ \begin{matrix}d & −b \\ −c & a \end{matrix} ]

Proof We row reduce the 2×4 partitioned matrix

[A|I] = [ \begin{matrix}a & b & 1 & 0 \\ c & d & 0 & 1 \end{matrix} ]  (2)

to obtain the reduced row echelon matrix [I|A^{−1}]. There are two cases, depending on whether a = 0 or not.

Case a ≠ 0 We multiply row 1 by 1/a to get

[ \begin{matrix}1 & b/a & 1/a & 0 \\ c & d & 0 & 1 \end{matrix} ]

Then we subtract c times row 1 from row 2 to obtain

[ \begin{matrix}1 & b/a & 1/a & 0 \\ 0 & Δ/a & −c/a & 1 \end{matrix} ]

where we write the entry at row 2, column 2 as d − bc/a = Δ/a. If Δ = 0, inversion breaks down at this point, as we will not get a leading 1 in column 2; otherwise, we multiply row 2 by a/Δ to get the row echelon form

[ \begin{matrix}1 & b/a & 1/a & 0 \\ 0 & 1 & −c/Δ & a/Δ \end{matrix} ]

Finally, we subtract b/a times row 2 from row 1 to get the desired reduced row echelon matrix, whose right half we read off as A^{−1}, in the required form,

[ \begin{matrix}1 & 0 & d/Δ & −b/Δ \\ 0 & 1 & −c/Δ & a/Δ \end{matrix} ]

where at row 1, column 3 we write

\frac{1}{a} + \frac{bc}{aΔ} = \frac{ad − bc + bc}{aΔ} = \frac{d}{Δ}

2 Inverting 2×2 matrices

Case a = 0 We must have c ≠ 0 for inversion to progress, otherwise we have a column of zeros and will never get a leading 1 in column 1. First we switch rows 1 and 2 in (2),

[ \begin{matrix}c & d & 0 & 1 \\ 0 & b & 1 & 0 \end{matrix} ]

Now we multiply row 1 by 1/c to get the leading 1 in row 1,

[ \begin{matrix}1 & d/c & 0 & 1/c \\ 0 & b & 1 & 0 \end{matrix} ]

Again, inversion breaks down here unless b ≠ 0 because we need a leading 1 in column 2. We multiply row 2 by 1/b to get the row echelon matrix

[ \begin{matrix}1 & d/c & 0 & 1/c \\ 0 & 1 & 1/b & 0 \end{matrix} ]

Finally, we subtract d/c times row 2 from row 1 to obtain the reduced row echelon matrix

[ \begin{matrix}1 & 0 & −d/bc & 1/c \\ 0 & 1 & 1/b & 0 \end{matrix} ]

Because now Δ = −bc, this is what we want. (The condition Δ ≠ 0 is exactly what we need to guarantee that c ≠ 0 and b ≠ 0.)

Because the expression Δ occurs everywhere, it deserves a name.

Definition 3 The determinant det(A) of the 2×2 matrix A is the expression

det(A) = Δ = ad − bc

The method generalizes in principle to produce a formula for the inverse of a general n×n matrix, so we know a formula exists. Even for n = 3, we need a better way to find it.