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Module 1/Matrices

Matrix Inverse

The inverse A−1A^{-1} is the matrix that undoes AA. This lesson shows you when a 2×22\times 2 matrix has an inverse, how to compute it with a one-line formula, and how inverses behave under multiplication.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

Find the inverse of A=[2111]A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}. Enter A−1A^{-1}.

Find the inverse of A=[1211]A = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}. Enter A−1A^{-1}.

What you’ll be able to do

  • Define the inverse of a square matrix and state the defining equation AA−1=A−1A=IA A^{-1} = A^{-1} A = I.
  • Compute the determinant of a 2×22\times 2 matrix and use it to decide whether the matrix is invertible.
  • Apply the 2×22\times 2 inverse formula to compute A−1A^{-1} by hand and verify the result.
  • State and use the product rule (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} and explain why the order reverses.

In your course

· MATH2015 · Linear Algebra & Probability
§4.4 Operations with matrices. Inverse matrix. Transpose.
  • Theorem 4.4When an inverse exists (equivalent conditions)
    For square AA these are equivalent: A−1A^{-1} exists; the map AA is bijective; the columns of AA form a basis; the columns of AA are linearly independent.
  • Proposition 4.1(A−1)−1=A(A^{-1})^{-1}=A and (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1}
1

What an inverse undoes

Think of a square matrix AA as a transformation: it sends a vector x\mathbf{x} to the vector AxA\mathbf{x}. The inverse A−1A^{-1} is the transformation that undoes this — it sends AxA\mathbf{x} back to x\mathbf{x}.

Formally, for a square matrix AA (equal number of rows and columns), its inverse A−1A^{-1} is the matrix satisfying AA−1=A−1A=I,A A^{-1} = A^{-1} A = I, where II is the identity matrix: the square matrix with 11's on the main diagonal and 00's elsewhere. In the 2×22\times 2 case, I=[1001].I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. The identity plays the role of the number 11 for matrices: Ix=xI\mathbf{x} = \mathbf{x} for every vector x\mathbf{x}, and IA=AI=AIA = AI = A for every compatible matrix AA.

Key facts to anchor your intuition:

  • Only square matrices can have an inverse.
  • Not every square matrix has one. A matrix with an inverse is invertible (also called nonsingular); one without is singular.
  • When the inverse exists, it is unique — there is exactly one matrix that undoes AA.
  • Inverses solve equations: if AA is invertible, the system Ax=bA\mathbf{x} = \mathbf{b} has the single solution x=A−1b\mathbf{x} = A^{-1}\mathbf{b}.
2

The determinant decides everything (2x2)

For a 2×22\times 2 matrix A=[abcd],A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant is the single number det⁡A=ad−bc.\det A = ad - bc. Here a,b,c,da, b, c, d are the four entries of AA: aa is top-left, bb top-right, cc bottom-left, dd bottom-right.

The determinant is the gatekeeper for invertibility: A is invertible  ⟺  det⁡A≠0.A \text{ is invertible} \iff \det A \ne 0. When det⁡A≠0\det A \ne 0, the inverse is given by a one-line formula: A−1=1det⁡A[d−b−ca]=1ad−bc[d−b−ca].A^{-1} = \frac{1}{\det A}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

A handy way to remember the matrix part (before dividing): swap the diagonal entries (aa and dd trade places) and negate the off-diagonal entries (bb and cc get minus signs). Then divide every entry by det⁡A\det A.

If det⁡A=0\det A = 0, the formula would require dividing by zero — a signal that no inverse exists. Geometrically, a zero determinant means AA collapses the plane onto a line (or a point), squashing out information that cannot be recovered, so there is nothing to undo it.

3

Inverting products: socks and shoes

Suppose AA and BB are both invertible square matrices of the same size. Their product ABAB is invertible too, and its inverse reverses the order: (AB)−1=B−1A−1.(AB)^{-1} = B^{-1}A^{-1}.

Why the flip? A quick check confirms it: using associativity and A−1A=IA^{-1}A = I, (AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I.(AB)(B^{-1}A^{-1}) = A(B B^{-1})A^{-1} = A I A^{-1} = A A^{-1} = I. The inner factors BB and B−1B^{-1} cancel first, then AA and A−1A^{-1}.

The classic mnemonic is socks and shoes: to get dressed you put on socks, then shoes (AA then BB); to undo it you must take off shoes first, then socks (B−1B^{-1} then A−1A^{-1}). You reverse the order when you undo a sequence.

Two more useful identities in the same spirit:

  • (A−1)−1=A(A^{-1})^{-1} = A — undoing the undo returns the original.
  • AA−1=A−1A=IAA^{-1} = A^{-1}A = I — the defining relation, which also tells you A−1A^{-1} is itself invertible with inverse AA.
Definition of the Matrix Inverse

A square matrix AA is invertible if there exists a square matrix A−1A^{-1} of the same size such that AA−1=A−1A=IA A^{-1} = A^{-1} A = I, where II is the identity matrix. When such a matrix exists it is unique and is called the inverse of AA.

Intuition. A−1A^{-1} is the transformation that exactly reverses what AA does, just as II (the matrix version of the number 11) leaves every vector unchanged.
Invertibility Criterion and Inverse Formula for 2x2 Matrices

For A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} with det⁡A=ad−bc\det A = ad - bc, the matrix AA is invertible if and only if det⁡A≠0\det A \ne 0. In that case A−1=1ad−bc[d−b−ca]A^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

Intuition. The determinant measures how AA scales area; if it is zero, AA flattens the plane and destroys information, so no inverse can exist. When nonzero, swap the diagonal, negate the off-diagonal, and divide by the determinant.
Inverse of a Product (Socks-and-Shoes Rule)

If AA and BB are invertible square matrices of the same size, then ABAB is invertible and (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.

Intuition. To undo doing AA and then BB, you must first undo BB and then undo AA — like removing shoes before socks. The order reverses.

Worked examples

Example 1

Find the inverse of A=[2312]A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, and verify your answer.

  1. 1

    Label the entries. Compare with the template [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}. Reading off: a=2a = 2 (top-left), b=3b = 3 (top-right), c=1c = 1 (bottom-left), d=2d = 2 (bottom-right).

  2. 2

    Compute the determinant. Using det⁡A=ad−bc\det A = ad - bc: det⁡A=(2)(2)−(3)(1)=4−3=1\det A = (2)(2) - (3)(1) = 4 - 3 = 1.

  3. 3

    Check invertibility. Since det⁡A=1≠0\det A = 1 \ne 0, the matrix AA is invertible, so the formula applies.

  4. 4

    Build the adjugate part. Swap the diagonal entries aa and dd, and negate the off-diagonal entries bb and cc: [d−b−ca]=[2−3−12]\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}.

  5. 5

    Divide by the determinant. Here 1det⁡A=11=1\dfrac{1}{\det A} = \dfrac{1}{1} = 1, so A−1=1⋅[2−3−12]=[2−3−12]A^{-1} = 1 \cdot \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}.

  6. 6

    Verify. Multiply AA−1=[2312][2−3−12]A A^{-1} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}. Top-left: 2⋅2+3⋅(−1)=4−3=12\cdot 2 + 3\cdot(-1) = 4 - 3 = 1. Top-right: 2⋅(−3)+3⋅2=−6+6=02\cdot(-3) + 3\cdot 2 = -6 + 6 = 0. Bottom-left: 1⋅2+2⋅(−1)=2−2=01\cdot 2 + 2\cdot(-1) = 2 - 2 = 0. Bottom-right: 1⋅(−3)+2⋅2=−3+4=11\cdot(-3) + 2\cdot 2 = -3 + 4 = 1. The product is II, confirming the answer.

Answer. A−1=[2−3−12]A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}
Example 2

Determine whether B=[1234]B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} is invertible, and if so find B−1B^{-1}.

Example 3

Let A=[1101]A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} and B=[1011]B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}. Verify that (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.