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Inverse Matrix

Inverse Matrix – definition and example, Precalculus math glossary – Mathovore

Key takeaway: An inverse matrix A⁻¹ is the matrix that satisfies AA⁻¹ = A⁻¹A = I, and it exists only when the determinant of the square matrix A is not zero.

Grade 12 – PrecalculusA-REI.C.9Expressions and Equations

Definition of Inverse Matrix

Definition

An inverse matrix reverses the effect of another matrix, the way dividing undoes multiplying. If a matrix \(A\) moves a point somewhere, then \(A^{-1}\) brings it exactly back. Multiplying the two in either order gives the identity matrix \(I\), which leaves everything unchanged.

Only square matrices can have one, and not all of them do. A matrix whose determinant is zero collapses information and cannot be reversed. The inverse is most useful for solving a system of linear equations written as \(A\mathbf{x}=\mathbf{b}\): multiply both sides by \(A^{-1}\) and the unknowns appear on their own.

Formula and notation

For a 2 by 2 matrix with \(ad-bc\neq 0\):

\[\begin{bmatrix}a & b\\ c & d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}.\]

Swap the diagonal entries, negate the other two, then divide by the determinant. A system \(A\mathbf{x}=\mathbf{b}\) has the solution \(\mathbf{x}=A^{-1}\mathbf{b}\).

(2, 1)3x + y = 74x + 2y = 10A = [3 1; 4 2], det A = 2A⁻¹ = [1 −1/2; −2 3/2]A⁻¹ [7; 10] = [2; 1]x = 2, y = 1
The inverse converts the right-hand side 7, 10 straight into the crossing point (2, 1).

Key facts

  • Check an answer by multiplying: the product with the original must be the identity.
  • The inverse of a product reverses the order: (AB)⁻¹ = B⁻¹A⁻¹.
  • A zero determinant means no inverse matrix exists and the system has no unique solution.
  • The exponent -1 here does not mean a reciprocal of each entry.

Inverse Matrix example

Problem

Use an inverse to solve \(3x+y=7\) and \(4x+2y=10\).

  1. Write the coefficient matrix \(A=\begin{bmatrix}3 & 1\\ 4 & 2\end{bmatrix}\); its determinant is \(6-4=2\).

  2. Build the inverse: \(A^{-1}=\frac{1}{2}\begin{bmatrix}2 & -1\\ -4 & 3\end{bmatrix}=\begin{bmatrix}1 & -\frac{1}{2}\\ -2 & \frac{3}{2}\end{bmatrix}\).

  3. Multiply by the right side: \(A^{-1}\begin{bmatrix}7\\ 10\end{bmatrix}=\begin{bmatrix}7-5\\ -14+15\end{bmatrix}=\begin{bmatrix}2\\ 1\end{bmatrix}\).

The solution is \(x=2,\ y=1\); check: \(6+1=7\) and \(8+2=10\).

Common mistake

Watch out

Bea tries to invert \(\begin{bmatrix}3 & 1\\ 4 & 2\end{bmatrix}\) by taking the reciprocal of every entry, getting \(\begin{bmatrix}\frac{1}{3} & 1\\ \frac{1}{4} & \frac{1}{2}\end{bmatrix}\).

Why it is wrong

An inverse matrix is not a list of entrywise reciprocals. Multiplying her answer by the original does not give the identity matrix. The correct recipe swaps the diagonal, negates the off-diagonal entries and divides by the determinant 2.

Related terms and lessons

Browse the glossary: A–Z list · Precalculus vocabulary · Expressions and Equations

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