MatricesTopic #11 of 30

Inverse Matrices

Invertible matrices, finding inverses, properties of inverse, and singular matrices.

Overview

The inverse of a matrix AA, denoted A−1A^{-1}, is the matrix that when multiplied by AA gives the identity matrix. Not all matrices have inverses.

Definition

For a square matrix AA, the inverse A−1A^{-1} satisfies:

AA−1=A−1A=IAA^{-1} = A^{-1}A = I

Invertibility

A matrix AA is invertible (or non-singular) if A−1A^{-1} exists.

Conditions for Invertibility

A square matrix AA is invertible if and only if:

  • det⁡(A)≠0\det(A) \neq 0
  • AA has nn pivot positions (full rank)
  • Ax=0A\mathbf{x} = \mathbf{0} has only the trivial solution
  • Columns of AA are linearly independent
  • Rows of AA are linearly independent

Finding the Inverse

2×2 Matrix Formula

For A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}:

A−1=1det⁡(A)[d−b−ca]A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

where det⁡(A)=ad−bc\det(A) = ad - bc

Example

A=[3121]A = \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix}

det⁡(A)=3(1)−1(2)=1\det(A) = 3(1) - 1(2) = 1

A−1=[1−1−23]A^{-1} = \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix}

Row Reduction Method

For larger matrices, augment AA with II and row reduce:

[A∣I]→[I∣A−1][A \mid I] \to [I \mid A^{-1}]

Example

[12∣1034∣01]→[12∣100−2∣−31]→[10∣−2101∣3/2−1/2]\begin{bmatrix} 1 & 2 & \vert & 1 & 0 \\ 3 & 4 & \vert & 0 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 2 & \vert & 1 & 0 \\ 0 & -2 & \vert & -3 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & \vert & -2 & 1 \\ 0 & 1 & \vert & 3/2 & -1/2 \end{bmatrix}

A−1=[−213/2−1/2]A^{-1} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}

Properties of Inverse

PropertyFormula
UniquenessInverse is unique if it exists
Involution(A−1)−1=A(A^{-1})^{-1} = A
Product(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}
Transpose(AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T
Scalar(cA)−1=(1/c)A−1(cA)^{-1} = (1/c)A^{-1}
Power(An)−1=(A−1)n(A^n)^{-1} = (A^{-1})^n

Product Inverse

Important: The inverse of a product reverses the order:

(ABC)−1=C−1B−1A−1(ABC)^{-1} = C^{-1}B^{-1}A^{-1}

Singular Matrices

A singular matrix has no inverse.

Characteristics

  • det⁡(A)=0\det(A) = 0
  • Ax=0A\mathbf{x} = \mathbf{0} has non-trivial solutions
  • Columns are linearly dependent
  • Not full rank

Example

A=[1224]A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}

det⁡(A)=1(4)−2(2)=0\det(A) = 1(4) - 2(2) = 0

AA is singular (no inverse exists)

Solving Systems with Inverse

For the system Ax=bA\mathbf{x} = \mathbf{b} where AA is invertible:

x=A−1b\mathbf{x} = A^{-1}\mathbf{b}

Example

Solve:

3x+y=53x + y = 5 2x+y=42x + y = 4

A=[3121],b=[54]A = \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}

A−1=[1−1−23]A^{-1} = \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix}

x=A−1b=[1−1−23][54]=[5−4−10+12]=[12]\mathbf{x} = A^{-1}\mathbf{b} = \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix}\begin{bmatrix} 5 \\ 4 \end{bmatrix} = \begin{bmatrix} 5 - 4 \\ -10 + 12 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}

Solution: x=1x = 1, y=2y = 2

Inverse of Special Matrices

Diagonal Matrix

D=[d1000d2000d3]D−1=[1/d10001/d20001/d3]D = \begin{bmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{bmatrix} \quad D^{-1} = \begin{bmatrix} 1/d_1 & 0 & 0 \\ 0 & 1/d_2 & 0 \\ 0 & 0 & 1/d_3 \end{bmatrix}

Orthogonal Matrix

Q−1=QTQ^{-1} = Q^T

Upper Triangular Matrix

Inverse is also upper triangular (use back substitution).