Inverse Matrices and Adjugate Matrices

Inverse Matrices

The inverse matrix is an important concept in matrix theory, with important applications in solving systems of linear equations, matrix decompositions and so on.

Inverse Matrix

For an n-th order square matrix AA, if there exists an n-th order square matrix BB such that: AB=BA=IAB = BA = I then AA is called an invertible matrix, BB is called the inverse matrix of AA, and is written A−1A^{-1}.

Note:

  • Only square matrices can possibly have an inverse
  • If the matrix AA is invertible, its inverse is unique
  • Not every square matrix has an inverse
The Inverse Matrix Formula

A−1=1∣A∣A∗A^{-1} = \frac{1}{|A|}A^*

Properties of the Inverse Matrix

  1. Uniqueness: if AA is invertible, its inverse is unique
  2. Double inverse: (A−1)−1=A(A^{-1})^{-1} = A
  3. Inverse of a product: (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} (if both AA and BB are invertible)
  4. Inverse of a transpose: (AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T
  5. Inverse of a scalar multiple: (kA)−1=1kA−1(kA)^{-1} = \frac{1}{k}A^{-1} (k≠0k \neq 0)

Methods for Computing the Inverse

Method 1: The Adjugate Matrix Method

Inverting via the Adjugate Matrix

If ∣A∣≠0|A| \neq 0, then:

A−1=1∣A∣A∗A^{-1} = \frac{1}{|A|}A^*

where A∗A^* is the adjugate matrix of AA.

Example: Find the inverse of the matrix A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}.

Solution:

  1. ∣A∣=1×4−2×3=4−6=−2≠0|A| = 1 \times 4 - 2 \times 3 = 4 - 6 = -2 \neq 0, so AA is invertible
  2. A11=4A_{11} = 4, A12=−3A_{12} = -3, A21=−2A_{21} = -2, A22=1A_{22} = 1
  3. A∗=(4−2−31)A^* = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}
  4. A−1=1−2(4−2−31)=(−211.5−0.5)A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}

Method 2: The Elementary Transformation Method

Steps:

  1. Form the augmented matrix [A∣I][A|I]
  2. Apply elementary row operations to the augmented matrix to reduce AA to the identity matrix
  3. At that point the position of II is A−1A^{-1}

Example: Find the inverse of A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} by the elementary transformation method.

Solution: (12∣1034∣01)\begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 3 & 4 & | & 0 & 1 \end{pmatrix}

R2−3R1R_2 - 3R_1: (12∣100−2∣−31)\begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 0 & -2 & | & -3 & 1 \end{pmatrix}

R2×(−12)R_2 \times (-\frac{1}{2}): (12∣1001∣1.5−0.5)\begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 0 & 1 & | & 1.5 & -0.5 \end{pmatrix}

R1−2R2R_1 - 2R_2: (10∣−2101∣1.5−0.5)\begin{pmatrix} 1 & 0 & | & -2 & 1 \\ 0 & 1 & | & 1.5 & -0.5 \end{pmatrix}

so A−1=(−211.5−0.5)A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}

Adjugate Matrices

The adjugate matrix is an important tool for computing the inverse matrix, and it provides an explicit formula for it.

The Definition of the Adjugate Matrix

Let A=(aij)n×nA = (a_{ij})_{n \times n} be an n-th order square matrix, and let AijA_{ij} be the algebraic cofactor of the entry aija_{ij}. Then the matrix

A∗=(A11A21⋯An1A12A22⋯An2⋮⋮⋱⋮A1nA2n⋯Ann)A^* = \begin{pmatrix} A_{11} & A_{21} & \cdots & A_{n1} \\ A_{12} & A_{22} & \cdots & A_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1n} & A_{2n} & \cdots & A_{nn} \end{pmatrix}

is called the adjugate matrix of AA.

Note: the entry in row ii, column jj of the adjugate matrix is AjiA_{ji}, that is, the transpose of the algebraic cofactors.

Properties of the Adjugate Matrix

  1. Basic relation: AA∗=A∗A=∣A∣IAA^* = A^*A = |A|I
  2. Inverse matrix formula: if ∣A∣≠0|A| \neq 0, then A−1=1∣A∣A∗A^{-1} = \frac{1}{|A|}A^*
  3. Transpose property: (A∗)T=(AT)∗(A^*)^T = (A^T)^*
  4. Scalar multiple property: (kA)∗=kn−1A∗(kA)^* = k^{n-1}A^*

Computing the Adjugate Matrix

Steps:

  1. Compute the algebraic cofactor of each entry
  2. Arrange them in the transposed positions

Example: Find the adjugate matrix of the matrix A=(123014560)A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}.

Solution:

  1. Compute the algebraic cofactor of each entry:

    • A11=∣1460∣=−24A_{11} = \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = -24
    • A12=−∣0450∣=20A_{12} = -\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = 20
    • A13=∣0156∣=−5A_{13} = \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} = -5
    • A21=−∣2360∣=18A_{21} = -\begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix} = 18
    • A22=∣1350∣=−15A_{22} = \begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix} = -15
    • A23=−∣1256∣=4A_{23} = -\begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix} = 4
    • A31=∣2314∣=5A_{31} = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 5
    • A32=−∣1304∣=−4A_{32} = -\begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -4
    • A33=∣1201∣=1A_{33} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1
  2. Arrange them in the transposed positions: adj(A)=(−2418520−15−4−541)\text{adj}(A) = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}

Summary

Symbols Used in This Article

SymbolTypePronunciation / NoteMeaning in This Article
A−1A^{-1}Mathematical symbolA inverseThe inverse matrix of the matrix AA, satisfying AA−1=IAA^{-1}=I
A∗A^*Mathematical symbolA starThe adjugate matrix of the matrix AA, that is, adj(A)\text{adj}(A)
AA∗=∣A∣IAA^* = \lvert A\rvert IMathematical symbolA times A star equals determinant times IThe basic relation for the adjugate matrix
AijA_{ij}Mathematical symbolA i jThe algebraic cofactor of the entry aija_{ij}, (−1)i+jMij(-1)^{i+j}M_{ij}
∣A∣\lvert A\rvertMathematical symboldeterminant of AThe determinant of the matrix AA; AA is invertible when ∣A∣≠0\lvert A\rvert \neq 0
[A∣I][A \mid I]Mathematical symbolaugmented matrixThe augmented matrix, used to compute the inverse by the elementary transformation method

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