Inverse Matrix

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Inverse Matrix: meaning, method and solved example (Precalculus) – Maths PDF

In short: An inverse matrix A⁻¹ is the matrix that undoes A, so that A times A⁻¹ gives the identity matrix; it exists only for square matrices whose determinant is not zero.

PrecalculusAlgebraA-REI.C.9

What is Inverse Matrix?

An inverse matrix plays the role that a reciprocal plays for numbers. Multiplying 5 by \(\frac{1}{5}\) returns 1, and multiplying a matrix by its inverse returns the identity matrix \(I\), which leaves every other matrix unchanged.

For \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) with \(ad – bc \neq 0\), \(A^{-1} = \frac{1}{ad – bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\). Swap the diagonal entries, negate the others, and divide by the determinant. If the determinant is zero, no inverse exists.

How to spot it

  • A matrix equation AX = B must be solved for X, which needs the inverse matrix of A.
  • A system of linear equations is written with a coefficient grid and a column of unknowns.
  • A question asks you to check whether two grids undo each other by multiplying them.
  • The notation A with a superscript -1 appears, or the determinant of a 2 × 2 grid must be tested first.

The method

  1. Compute the determinant \(ad – bc\); if it is zero, stop, because there is no inverse.
  2. Swap \(a\) and \(d\), change the signs of \(b\) and \(c\), and keep the positions.
  3. Divide every entry by the determinant.
  4. Multiply \(A\) by the inverse matrix you found to confirm it gives the identity matrix.
A4 72 6det A = 24 – 1410A inverse0.6 -0.7-0.2 0.4check: A × A inverse = identityfirst row: 4(0.6) + 7(-0.2) = 1, 4(-0.7) + 7(0.4) = 0second row: 2(0.6) + 6(-0.2) = 0, 2(-0.7) + 6(0.4) = 1
For A with determinant 10, the inverse has entries 0.6, -0.7, -0.2, 0.4.

Inverse Matrix: solved example

Problem

Find the inverse of \(A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}\) and verify it.

Solution

The determinant is \(4(6) – 7(2) = 24 – 14 = 10\), which is nonzero. Swap the diagonal, negate the others: \(\frac{1}{10}\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix}\).

So \(A^{-1} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}\). Check: the first row of \(A\) times the first column of \(A^{-1}\) is \(2.4 – 1.4 = 1\), and the other products give \(0, 0, 1\).

Try it

Question 1

Does \(\begin{bmatrix} 6 & 3 \\ 4 & 2 \end{bmatrix}\) have an inverse?

Show the answer

The determinant is \(12 – 12 = 0\), so no inverse exists.

Question 2

Use \(A^{-1}\) from the example to solve \(4x + 7y = 5\) and \(2x + 6y = 4\).

Show the answer

\(\begin{bmatrix} x \\ y \end{bmatrix} = A^{-1}\begin{bmatrix} 5 \\ 4 \end{bmatrix} = \begin{bmatrix} 0.2 \\ 0.6 \end{bmatrix}\), so \(x = 0.2\), \(y = 0.6\).

Inverse Matrix vs. Determinant

The determinant is one number, and it only decides whether an undo-matrix exists. The inverse is a whole grid of numbers of the same size as the original. Be careful: A⁻¹ does not mean one divided by each entry; a matrix has no entrywise reciprocal that undoes it.

Determinant

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