Math · Precalculus · Worked example

Find the inverse of a 2 × 2 matrix

Find the inverse of the matrix A with rows (2, 1) and (5, 3), and use it to solve 2x + y = 4, 5x + 3y = 7.

A = (\begin{array}{cccccccccc} 2 & 1 \\ 5 & 3 \end{array})

Find the determinant

ad − bc = 2 · 3 − 1 · 5.

2 \cdot 3 - 1 \cdot 5 = 1

Apply the 2 × 2 formula

Swap the diagonal entries, change the signs of the other two and divide by the determinant, 1.

A^{- 1} = (\begin{array}{cccccccccc} 3 & - 1 \\ - 5 & 2 \end{array})

Check that A⁻¹A = I

Row 1 of A⁻¹ times the columns of A gives 1 and 0; row 2 gives 0 and 1.

\begin{array}{l} 3 \cdot 2 - 1 \cdot 5 = 1 \\ 3 \cdot 1 - 1 \cdot 3 = 0 \\ - 5 \cdot 2 + 2 \cdot 5 = 0 \\ - 5 \cdot 1 + 2 \cdot 3 = 1 \end{array}

Solve the system

X = A⁻¹B: multiply A⁻¹ by the column of constants (4, 7).

\begin{array}{l} 3 \cdot 4 - 1 \cdot 7 = 5 \\ - 5 \cdot 4 + 2 \cdot 7 = - 6 \end{array}

Result

A⁻¹ has rows (3, −1) and (−5, 2), and the system’s solution is x = 5, y = −6.

Your turn

Find the inverse of the matrix with rows (4, 7) and (2, 6).

Show the answer and explanation

The matrix with rows (0.6, −0.7) and (−0.2, 0.4).

The determinant is 4 · 6 − 7 · 2 = 10. Swap 4 and 6, change the signs of 7 and 2, and divide every entry by 10.

4 \cdot 6 - 7 \cdot 2 = 10

Keep exploring

In Matrix Transformations, this matrix keeps areas the same, because its determinant is 1. Set d to 2.5 and the determinant drops to 0: the plane collapses onto a line, and there is no inverse.

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