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Mathematics · Ch 2 — Matrices

Inverse of a Matrix

2.2

Inverse of a Matrix

Definition. If AA is a square matrix of order mm and there exists another square matrix BB of the same order such that AB=BA=IAB = BA = I, where II is the identity matrix of order mm, then BB is called the inverse of AA, written A−1A^{-1}. Writing A−1A^{-1} in place of BB, the defining relation becomes AA−1=A−1A=IAA^{-1} = A^{-1}A = I; by the very same definition, AA is then equally the inverse of BB, so B−1=AB^{-1} = A.

Worked illustration. If A=[2312]A = \begin{bmatrix}2&3\\1&2\end{bmatrix} and B=[2−3−12]B = \begin{bmatrix}2&-3\\-1&2\end{bmatrix}, then AB=[2312][2−3−12]=[4−3−6+62−2−3+4]=[1001]=I2AB = \begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&-3\\-1&2\end{bmatrix} = \begin{bmatrix}4-3&-6+6\\2-2&-3+4\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I_2. Likewise BA=[2−3−12][2312]=[4−36−6−2+2−3+4]=[1001]=I2BA = \begin{bmatrix}2&-3\\-1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix} = \begin{bmatrix}4-3&6-6\\-2+2&-3+4\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I_2. Since both products equal the identity, B=A−1B = A^{-1} and A=B−1A = B^{-1}.

When does an inverse exist? Consider A=[1224]A = \begin{bmatrix}1&2\\2&4\end{bmatrix}: no matrix XX can be found with AX=IAX = I. The reason is that ∣A∣=1⋅4−2⋅2=0|A| = 1\cdot4 - 2\cdot2 = 0. This illustrates the necessary condition for a matrix XX with AX=IAX=I to exist: ∣A∣≠0|A| \neq 0, i.e. AA must be a non-singular matrix.

Three standing notes:

  1. Every square matrix AA of order m×mm\times m has a corresponding determinant, det⁡A=∣A∣\det A = |A|.
  2. A matrix is said to be invertible if its inverse exists.
  3. A square matrix AA has an inverse if and only if ∣A∣≠0|A| \neq 0.

Uniqueness of the inverse. It can be proved that if AA is a square matrix with ∣A∣≠0|A|\neq0, its inverse A−1A^{-1} is unique -- there is never more than one matrix that undoes AA.

Theorem. If a square matrix AA's inverse exists, it is unique.

Proof. Let AA be a square matrix of order mm whose inverse exists. Suppose, if possible, that BB and CC are both inverses of AA. By the definition of inverse, AB=BA=IAB=BA=I and AC=CA=IAC=CA=I. Now consider

B=BI=B(AC)B = BI = B(AC)

∴ B=(BA)C=IC\therefore\ B = (BA)C = IC

∴ B=C\therefore\ B = C

Hence B=CB=C, i.e. the inverse -- when it exists -- is unique. …

Misc 2.2aWorked illustration -- verifying two given 2x2 matrices are inverses of each other

Worked out. A specific matrix A and a candidate matrix B are multiplied both ways (AB and BA); both products come out to the 2x2 identity matrix, confirming B is the inverse of A (and A the inverse of B) directly from the defining equation, before the general theory of computing an inverse from scratch is developed. …

Misc 2.2bProof -- uniqueness of the inverse of a matrix

Worked out. A short algebraic proof that if a square matrix A has an inverse, that inverse is the only one: assuming two candidate inverses B and C both satisfy the defining equations, the proof manipulates B = BI = B(AC) = (BA)C = IC = C using associativity of matrix multiplication to conclude B and C must be the same matri …