Inverse of a Matrix

Tags: #◇CSMATH2 #◇Linear Algebra #◇Computational Mathematics #◇Matrices #◇Matrix Operations #◇Inverse Matrix #◇Square Matrix #◇Computer Science #◇Vector Spaces #◇Linear Transformations

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◇Overview

The inverse of a square matrix $A$ is a matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. Not all square matrices are invertible.

◇Key Concepts

Only square matrices can have inverses $A$ is invertible (nonsingular) iff $\det(A) \neq 0$ If $A^{-1}$ exists, it is unique Computed via Gauss-Jordan: $[A | I] \to [I | A^{-1}]$

◇Details / Theory

Definition

An $n \times n$ matrix $A$ is invertible (nonsingular) if there exists an $n \times n$ matrix $A^{-1}$ such that:

$AA^{-1} = A^{-1}A = I_n$

If no such matrix exists, $A$ is singular (not invertible).

◇Nonsingular Matrix

A matrix is nonsingular (invertible) iff:
$\det(A) \neq 0$ $A$ row-reduces to $I$ The system $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$

$2 \times 2$ Formula

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with $\det(A) = ad - bc \neq 0$:

$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Gauss-Jordan Method (General)

1. Form the augmented matrix $[A \mid I]$
2. Apply ◇elementary row operations to reduce $A$ to $I$
3. The right side becomes $A^{-1}$: $[A \mid I] \to [I \mid A^{-1}]$

Stepped Calculation Trace: $3 \times 3$ Gauss-Jordan Inversion

Find $A^{-1}$ for:
$A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{bmatrix}$

Step 1: Augment $A$ with $I_3$

$[A \mid I] = \left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 5 & 3 & 0 & 1 & 0 \\ 1 & 0 & 8 & 0 & 0 & 1 \end{array}\right]$

Step 2: Forward Elimination (Create Zeroes Below Pivot 1)

Perform $R_2 \leftarrow R_2 - 2R_1$ and $R_3 \leftarrow R_3 - R_1$:
$\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & -2 & 5 & -1 & 0 & 1 \end{array}\right]$

Step 3: Eliminate Below Pivot 2

Perform $R_3 \leftarrow R_3 + 2R_2$:
$\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & 0 & -1 & -5 & 2 & 1 \end{array}\right]$

Step 4: Normalize Pivot 3

Perform $R_3 \leftarrow -1 \cdot R_3$:
$\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & 0 & 1 & 5 & -2 & -1 \end{array}\right]$

Step 5: Back-Substitution (Clear Column 3 Above Pivot 3)

Perform $R_2 \leftarrow R_2 + 3R_3$ and $R_1 \leftarrow R_1 - 3R_3$:

• New $R_2 = [0, 1, 0 \mid -2 + 15, 1 - 6, 0 - 3] = [0, 1, 0 \mid 13, -5, -3]$
• New $R_1 = [1, 2, 0 \mid 1 - 15, 0 + 6, 0 + 3] = [1, 2, 0 \mid -14, 6, 3]$

$\left[\begin{array}{ccc|ccc} 1 & 2 & 0 & -14 & 6 & 3 \\ 0 & 1 & 0 & 13 & -5 & -3 \\ 0 & 0 & 1 & 5 & -2 & -1 \end{array}\right]$

Step 6: Clear Column 2 Above Pivot 2

Perform $R_1 \leftarrow R_1 - 2R_2$:

• New $R_1 = [1, 0, 0 \mid -14 - 26, 6 - 2(-5), 3 - 2(-3)] = [1, 0, 0 \mid -40, 16, 9]$

$[I \mid A^{-1}] = \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -40 & 16 & 9 \\ 0 & 1 & 0 & 13 & -5 & -3 \\ 0 & 0 & 1 & 5 & -2 & -1 \end{array}\right]$

Thus, the inverse is:
$A^{-1} = \begin{bmatrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{bmatrix}$

Verification Check: $A \cdot A^{-1} = I_3$

$A A^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{bmatrix} \begin{bmatrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{bmatrix} = \begin{bmatrix} -40 + 26 + 15 & 16 - 10 - 6 & 9 - 6 - 3 \\ -80 + 65 + 15 & 32 - 25 - 6 & 18 - 15 - 3 \\ -40 + 0 + 40 & 16 + 0 - 16 & 9 + 0 - 8 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3$

Properties

◇Common Pitfalls

Assuming all square matrices are invertible — check $\det(A) \neq 0$. Forgetting reverse order in $(AB)^{-1} = B^{-1}A^{-1}$. Non-square matrices never have a (two-sided) inverse.

◇Summary

TL;DR — $A^{-1}$ satisfies $AA^{-1} = I$. Exists iff $\det(A) \neq 0$. For $2 \times 2$, use the formula. For larger matrices, use Gauss-Jordan ($[A|I] \to [I|A^{-1}]$). Product inverse reverses order.

◇Review Questions

1. If $A$ is an invertible matrix and $AB = AC$, prove that $B = C$.
   • Proof: Left-multiply both sides by $A^{-1}$: $A^{-1}(AB) = A^{-1}(AC) \implies (A^{-1}A)B = (A^{-1}A)C \implies IB = IC \implies B = C$.
2. Explain why the inverse of an upper triangular matrix must also be an upper triangular matrix.
   • Explanation: During Gauss-Jordan elimination on $[U \mid I]$, all operations performed to reduce $U$ to $I$ involve scaling rows and clearing entries above the diagonal pivots. Applying these row additions strictly from bottom to top preserves all zeroes below the main diagonal in the right-hand identity block, yielding an upper triangular inverse.
3. Why does a matrix with a row of all zeros have no inverse?
   • Explanation: If matrix $A$ has a zero row, then for any matrix $B$, the product $AB$ will also have a corresponding row of all zeros. Since the identity matrix $I$ has $1$s on the diagonal and no zero rows, $AB = I$ can never be satisfied.

◇Links

Related: ◇Nonsingular Matrix — ◇Elementary Row Operation — ◇Elementary Matrices — ◇Matrix Multiplication — ◇Identity Matrix — ◇Special Types of Square Matrices Source / Reference:

◇References & Navigation

• Previous Topic: ◇Important Remarks
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• Related Notes: ◇Scalar Matrix, ◇Matrix Addition, ◇Matrices Detailed