How to find a two-by-two matrix inverse and test invertibility

An inverse matrix reverses a matrix transformation. When an inverse exists, multiplying the matrix by its inverse gives the identity. For a two-by-two matrix, the determinant gives us a direct way to find the inverse and decide whether it exists.

Find the determinant

A = [\begin{array}{r} a & b \\ c & d \end{array}]

\det A = a d - b c

Multiply the main-diagonal entries, then subtract the product from the other diagonal.

Apply the two-by-two inverse formula

If $a d - b c \neq 0$ , swap the main-diagonal entries, change the signs of the other two entries, and multiply by the reciprocal of the determinant.

A^{- 1} = \frac{1}{a d - b c} [\begin{array}{r} d & - b \\ - c & a \end{array}]

Find and check an inverse

K = [\begin{array}{r} - 2 & 4 \\ 3 & 0 \end{array}]

\det K = (- 2) (0) - 4 (3) = - 12

K^{- 1} = [\begin{array}{r} 0 & \frac{1}{3} \\ \frac{1}{4} & \frac{1}{6} \end{array}]

To check the inverse, calculate both products. In $K K^{- 1}$ , the first-row entries are $- 2 (0) + 4 (1 / 4) = 1$ and $- 2 (1 / 3) + 4 (1 / 6) = 0$ . The second-row entries are $0$ and $1$ . The reverse product also gives

K K^{- 1} = K^{- 1} K = I_{2}

Distinguish invertible and singular matrices

A square matrix is invertible when its determinant is nonzero. If its determinant is zero, it is singular and has no ordinary two-sided inverse.

M = [\begin{array}{r} 1 & - 3 \\ 3 & 5 \end{array}]

Here $\det M = 1 (5) - (- 3) (3) = 14$ , so $M$ is invertible.

L = [\begin{array}{r} - 6 & 2 \\ 3 & - 1 \end{array}]

Here $\det L = (- 6) (- 1) - 2 (3) = 0$ , so $L$ is singular. The determinant test also works when some entries are zero, without requiring ratios of entries.

Use an inverse in a matrix equation

For $A X = B$ with invertible $A$ , multiply on the left to obtain $X = A^{- 1} B$ . For $X A = B$ , multiply on the right to obtain $X = B A^{- 1}$ . The side matters because matrix multiplication generally depends on order.

Learn mathKrista KingAugust 20, 2021math, learn online, online course, online math, equilibrium solutions