MathIsimple
LA-4.3
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Matrix Inverse

The inverse of a matrix undoes its action. Invertible matrices correspond to bijective linear maps.

Learning Objectives
  • Define the inverse of a square matrix
  • Prove uniqueness of the inverse
  • Apply the fundamental characterizations of invertibility
  • Compute inverses using Gaussian elimination
  • Apply properties of inverse: product, transpose, powers
  • Understand the connection between invertible matrices and isomorphisms
  • Recognize when a matrix is NOT invertible
  • Use the inverse to solve systems of equations
Prerequisites
  • Matrix operations (LA-4.2)
  • Matrix representation (LA-4.1)
  • Linear map composition
  • Gaussian elimination basics

1. Definition and Motivation

Just as a bijective function has an inverse that "undoes" it, an invertible linear map has an inverse map. The matrix inverse is the matrix representation of this inverse map.

Definition 4.13: Inverse Matrix

Let AMn(F)A \in M_n(F) be an n×nn \times n matrix. If there exists a matrix BMn(F)B \in M_n(F) such that:

AB=BA=InAB = BA = I_n

then AA is called invertible (or nonsingular), and BB is called the inverse of AA, denoted A1A^{-1}.

Remark 4.12: Singular vs Nonsingular

A square matrix that is NOT invertible is called singular. The terminology comes from the fact that singular matrices represent "exceptional" or "degenerate" cases.

Theorem 4.18: Uniqueness of the Inverse

If AA is invertible, then its inverse A1A^{-1} is unique.

Proof:

Suppose BB and CC both satisfy AB=BA=IAB = BA = I and AC=CA=IAC = CA = I. Then:

B=BI=B(AC)=(BA)C=IC=CB = BI = B(AC) = (BA)C = IC = C

Therefore B=CB = C, proving uniqueness.

Remark 4.13: Connection to Isomorphisms

A matrix AA is invertible if and only if the linear map it represents is an isomorphism (bijective). The inverse matrix represents the inverse map.

2. One-Sided Inverses for Square Matrices

Theorem 4.19: One-Sided Implies Two-Sided

For square matrices A,BMn(F)A, B \in M_n(F):

AB=I    BA=I    A,B are both invertibleAB = I \iff BA = I \iff A, B \text{ are both invertible}
Proof:

Suppose AB=IAB = I. We show BB is invertible with inverse AA:

Step 1: Show columns of BB are linearly independent.

If Bx=0Bx = 0, then x=Ix=(AB)x=A(Bx)=A0=0x = Ix = (AB)x = A(Bx) = A \cdot 0 = 0.

Step 2: Since BB has independent columns, it's invertible.

Step 3: From AB=IAB = I, right-multiply by B1B^{-1}: A=B1A = B^{-1}.

Therefore BA=BB1=IBA = BB^{-1} = I.

Remark 4.14: Warning: Non-Square Matrices

This fails for non-square matrices! A left inverse doesn't imply a right inverse:

A=(100100),B=(100010)A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}

Then BA=I2BA = I_2 but ABI3AB \neq I_3.

3. Characterizations of Invertibility

Theorem 4.20: Invertibility Equivalences (The Invertible Matrix Theorem)

For an n×nn \times n matrix AA, the following are equivalent (TFAE):

  1. AA is invertible
  2. det(A)0\det(A) \neq 0
  3. rank(A)=n\text{rank}(A) = n
  4. The columns of AA are linearly independent
  5. The rows of AA are linearly independent
  6. Ax=0Ax = 0 has only the trivial solution x=0x = 0
  7. Ax=bAx = b has a unique solution for every bb
  8. AA is a product of elementary matrices
  9. AA is row equivalent to InI_n
  10. The reduced row echelon form of AA is InI_n
  11. ker(A)={0}\ker(A) = \{0\} (trivial null space)
  12. im(A)=Fn\text{im}(A) = F^n (columns span all of FnF^n)
Remark 4.15: Power of the Theorem

This theorem is central to linear algebra. It connects seemingly different concepts: invertibility, determinants, rank, independence, solvability of equations, and elementary matrices.

4. Properties of the Inverse

Theorem 4.21: Inverse Properties

For invertible matrices AA, BB and scalar λ0\lambda \neq 0:

  1. (A1)1=A(A^{-1})^{-1} = A
  2. (AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1} (order reverses!)
  3. (An)1=(A1)n(A^n)^{-1} = (A^{-1})^n
  4. (λA)1=λ1A1(\lambda A)^{-1} = \lambda^{-1} A^{-1}
  5. (AT)1=(A1)T(A^T)^{-1} = (A^{-1})^T
  6. det(A1)=1/det(A)\det(A^{-1}) = 1/\det(A)
Proof:

(2): Verify directly:

(AB)(B1A1)=A(BB1)A1=AIA1=AA1=I(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I

By uniqueness, (AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1}.

(5): From AA1=IAA^{-1} = I, take transpose of both sides:

(A1)TAT=IT=I(A^{-1})^T A^T = I^T = I

This shows (A1)T(A^{-1})^T is a left inverse of ATA^T. For square matrices, this means (AT)1=(A1)T(A^T)^{-1} = (A^{-1})^T.

Corollary 4.7: Product of Inverses

For invertible matrices A1,A2,,AkA_1, A_2, \ldots, A_k:

(A1A2Ak)1=Ak1A21A11(A_1 A_2 \cdots A_k)^{-1} = A_k^{-1} \cdots A_2^{-1} A_1^{-1}
Remark 4.16: Cancellation Law

If AA is invertible and AB=ACAB = AC, then B=CB = C. Multiply both sides by A1A^{-1} on the left.

5. Computing the Inverse

Theorem 4.22: Gaussian Elimination Method

To compute A1A^{-1}:

  1. Form the augmented matrix [AI][A | I]
  2. Row reduce to reduced echelon form
  3. If the left part becomes II, the right part is A1A^{-1}
  4. If AA cannot be reduced to II, then AA is not invertible
Example 4.8: Computing an Inverse

Find the inverse of A=(111012104)A = \begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 1 & 0 & 4 \end{pmatrix}.

Solution: Form [AI][A | I] and row reduce:

(111100012010104001)\begin{pmatrix} 1 & -1 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 2 & | & 0 & 1 & 0 \\ 1 & 0 & 4 & | & 0 & 0 & 1 \end{pmatrix}

After row operations: R3R3R1R_3 \to R_3 - R_1, then eliminate to get:

(100443010232001111)\begin{pmatrix} 1 & 0 & 0 & | & 4 & 4 & -3 \\ 0 & 1 & 0 & | & 2 & 3 & -2 \\ 0 & 0 & 1 & | & -1 & -1 & 1 \end{pmatrix}

Therefore:

A1=(443232111)A^{-1} = \begin{pmatrix} 4 & 4 & -3 \\ 2 & 3 & -2 \\ -1 & -1 & 1 \end{pmatrix}
Remark 4.17: Why This Works

Row reducing [AI][A|I] applies the same sequence of elementary row operations to both AA and II. If AA reduces to II, then II becomes A1A^{-1} because elementary operations are equivalent to left-multiplying by elementary matrices.

6. Solving Systems with Inverses

Theorem 4.23: Inverse Solution Formula

If AA is invertible, the unique solution to Ax=bAx = b is:

x=A1bx = A^{-1}b
Proof:

Multiply both sides of Ax=bAx = b by A1A^{-1} on the left:

A1(Ax)=A1b    (A1A)x=A1b    Ix=A1b    x=A1bA^{-1}(Ax) = A^{-1}b \implies (A^{-1}A)x = A^{-1}b \implies Ix = A^{-1}b \implies x = A^{-1}b
Remark 4.18: Practical Considerations

While the formula x=A1bx = A^{-1}b is elegant, in practice:

  • Computing A1A^{-1} explicitly is often unnecessary
  • Direct solution (e.g., LU decomposition) is usually faster and more stable
  • A1A^{-1} is useful when solving Ax=bAx = b for many different bb vectors

7. Worked Examples

Example 1: 2×2 Inverse Formula

For A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} with adbc0ad - bc \neq 0:

A1=1adbc(dbca)A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Example: A=(3152)A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}

A1=165(2153)=(2153)A^{-1} = \frac{1}{6-5} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}
Example 2: Diagonal Matrix Inverse

If D=diag(d1,,dn)D = \text{diag}(d_1, \ldots, d_n) with all di0d_i \neq 0:

D1=diag(1/d1,,1/dn)D^{-1} = \text{diag}(1/d_1, \ldots, 1/d_n)

Each diagonal entry is replaced by its reciprocal.

Example 3: Inverse of Product

Problem: If A1=(1234)A^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} and B1=(0110)B^{-1} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, find (AB)1(AB)^{-1}.

Solution:

(AB)1=B1A1=(0110)(1234)=(3412)(AB)^{-1} = B^{-1}A^{-1} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}
Example 4: When is a Matrix Singular?

Problem: For which values of kk is A=(1kk4)A = \begin{pmatrix} 1 & k \\ k & 4 \end{pmatrix} singular?

Solution: AA is singular iff det(A)=0\det(A) = 0:

det(A)=4k2=0    k=±2\det(A) = 4 - k^2 = 0 \implies k = \pm 2

AA is singular when k=2k = 2 or k=2k = -2.

8. Common Mistakes

Mistake 1: Wrong Order for Product Inverse

(AB)1A1B1(AB)^{-1} \neq A^{-1}B^{-1}. The correct formula is (AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1}. Remember: order reverses, just like for transpose.

Mistake 2: Assuming Non-Square Matrices Have Inverses

Only square matrices can have (two-sided) inverses. An m×nm \times n matrix with mnm \neq ncannot be invertible, though it may have one-sided inverses or pseudoinverses.

Mistake 3: Assuming Every Square Matrix is Invertible

Many square matrices are singular (not invertible). Check the determinant, rank, or column independence before assuming invertibility.

Mistake 4: Inverse of Sum

(A+B)1A1+B1(A + B)^{-1} \neq A^{-1} + B^{-1}. There is no simple formula for the inverse of a sum.

9. Key Takeaways

Definition

AA1=A1A=IAA^{-1} = A^{-1}A = I. The inverse undoes the matrix.

Invertibility Test

det(A)0\det(A) \neq 0 iff AA invertible iff rank(A)=n\text{rank}(A) = n.

Product Inverse

(AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1}. Order reverses!

Computation

Row reduce [AI][A|I] to [IA1][I|A^{-1}].

10. Quick Reference Summary

PropertyFormula
DefinitionAA1=A1A=IAA^{-1} = A^{-1}A = I
Double Inverse(A1)1=A(A^{-1})^{-1} = A
Product Inverse(AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1}
Transpose-Inverse(AT)1=(A1)T(A^T)^{-1} = (A^{-1})^T
Scalar Inverse(λA)1=λ1A1(\lambda A)^{-1} = \lambda^{-1}A^{-1}
Determinantdet(A1)=1/det(A)\det(A^{-1}) = 1/\det(A)

11. Additional Practice Problems

Problem 1

Compute the inverse of (2153)\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} using the 2×2 formula.

Problem 2

Prove that if A2=IA^2 = I (involutory matrix), then A=A1A = A^{-1}.

Problem 3

If AA is invertible, show that (Ak)1=(A1)k(A^k)^{-1} = (A^{-1})^k for all positive integers kk.

Problem 4

Find the inverse of (121011110)\begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix} using row reduction.

Problem 5

Prove that if AA and BB are invertible and commute (AB=BAAB = BA), then A1A^{-1} and B1B^{-1} also commute.

Problem 6

For which values of aa is (1a0a1a0a1)\begin{pmatrix} 1 & a & 0 \\ a & 1 & a \\ 0 & a & 1 \end{pmatrix} invertible?

12. Special Types of Invertible Matrices

Theorem 4.24: Special Matrix Inverses

For special matrix types:

  • Diagonal: diag(d1,,dn)1=diag(1/d1,,1/dn)\text{diag}(d_1, \ldots, d_n)^{-1} = \text{diag}(1/d_1, \ldots, 1/d_n) if all di0d_i \neq 0
  • Orthogonal: A1=ATA^{-1} = A^T if ATA=IA^T A = I
  • Triangular: Inverse of upper (lower) triangular is upper (lower) triangular
  • Involutory: A1=AA^{-1} = A if A2=IA^2 = I
Example 4.9: Orthogonal Matrix Inverse

The rotation matrix Rθ=(cosθsinθsinθcosθ)R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} is orthogonal:

Rθ1=RθT=(cosθsinθsinθcosθ)=RθR_\theta^{-1} = R_\theta^T = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} = R_{-\theta}

Rotating by θ\theta then by θ-\theta returns to the original position.

Example 4.10: Upper Triangular Inverse

For U=(1203)U = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}:

U1=(12/301/3)U^{-1} = \begin{pmatrix} 1 & -2/3 \\ 0 & 1/3 \end{pmatrix}

The inverse is also upper triangular.

13. Connections to Other Topics

Determinants

AA is invertible iff det(A)0\det(A) \neq 0. Furthermore:

det(A1)=1det(A)\det(A^{-1}) = \frac{1}{\det(A)}

The adjugate formula: A1=1det(A)adj(A)A^{-1} = \frac{1}{\det(A)} \text{adj}(A)

Eigenvalues

AA is invertible iff 0 is NOT an eigenvalue. For invertible AA:

eigenvalues of A1={1/λ:λ is eigenvalue of A}\text{eigenvalues of } A^{-1} = \{1/\lambda : \lambda \text{ is eigenvalue of } A\}

Eigenvectors are the same for AA and A1A^{-1}.

Linear Maps

AA is invertible iff the linear map it represents is an isomorphism:

  • Injective: ker(A)={0}\ker(A) = \{0\}
  • Surjective: im(A)=Fn\text{im}(A) = F^n
Change of Basis

If PP is the change of basis matrix, then:

[v]B=P1[v]B[v]_{B'} = P^{-1}[v]_B

Change of basis matrices are always invertible.

14. Challenge Problems

Challenge 1: Sherman-Morrison Formula

Prove that if AA is invertible and u,vu, v are column vectors, then:

(A+uvT)1=A1A1uvTA11+vTA1u(A + uv^T)^{-1} = A^{-1} - \frac{A^{-1}uv^T A^{-1}}{1 + v^T A^{-1}u}

(provided 1+vTA1u01 + v^T A^{-1}u \neq 0)

Challenge 2: Block Matrix Inverse

Find the inverse of (AB0D)\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} where AA and DD are invertible square matrices.

Challenge 3: Nilpotent Matrices

Prove that if NN is nilpotent (Nk=0N^k = 0 for some kk), then INI - N is invertible with:

(IN)1=I+N+N2++Nk1(I - N)^{-1} = I + N + N^2 + \cdots + N^{k-1}

15. Theoretical Insights

Theorem 4.25: General Linear Group

The set of all invertible n×nn \times n matrices over FF forms a group under multiplication, denoted GLn(F)GL_n(F):

  • Closure: Product of invertible matrices is invertible
  • Identity: InI_n
  • Inverse: A1A^{-1} exists for each AGLn(F)A \in GL_n(F)
  • Associativity: Inherited from matrix multiplication
Remark 4.19: Notation

GLn(F)GL_n(F) stands for "General Linear group". When F=RF = \mathbb{R}, we write GLn(R)GL_n(\mathbb{R}); when F=CF = \mathbb{C}, we write GLn(C)GL_n(\mathbb{C}).

Theorem 4.26: Density of Invertible Matrices

Over R\mathbb{R} or C\mathbb{C}, the set of invertible matrices is "generic":

  • Singular matrices have measure zero
  • "Almost all" matrices are invertible
  • A random matrix with continuous entries is invertible with probability 1
Corollary 4.8: Perturbation

If AA is singular, then for almost any matrix EE, the perturbed matrix A+ϵEA + \epsilon E is invertible for small ϵ0\epsilon \neq 0.

16. Study Tips

Check Dimensions

Only square matrices can be invertible. Before computing an inverse, verify the matrix is square.

Verify Your Answer

After computing A1A^{-1}, always verify by checking AA1=IAA^{-1} = I.

Use 2×2 Formula

For 2×22 \times 2 matrices, the formula 1adbc(dbca)\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} is fastest.

Remember Order Reversal

(AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1}. This is the same reversal that happens with transpose.

17. More Worked Examples

Example: Solving a System

Problem: Solve (2153)(xy)=(38)\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 8 \end{pmatrix}.

Solution: First find A1A^{-1}:

A1=165(3152)=(3152)A^{-1} = \frac{1}{6-5}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

Then:

(xy)=A1b=(3152)(38)=(11)\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}b = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\begin{pmatrix} 3 \\ 8 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}
Example: Matrix Equation

Problem: Solve AXB=CAXB = C for XX where AA, BB, CC are known and AA, BB are invertible.

Solution:

AXB=CAXB = C
A1(AXB)=A1CA^{-1}(AXB) = A^{-1}C
XB=A1CXB = A^{-1}C
(XB)B1=A1CB1(XB)B^{-1} = A^{-1}CB^{-1}
X=A1CB1X = A^{-1}CB^{-1}
Example: Power of Inverse

Problem: If A3=IA^3 = I, find A1A^{-1}.

Solution: From A3=IA^3 = I:

AA2=IA \cdot A^2 = I

This shows A2A^2 is the inverse of AA:

A1=A2A^{-1} = A^2
Example: Inverse of Sum

Problem: If AB=BAAB = BA and both are invertible, show (A+B)1(A + B)^{-1} commutes with AA.

Solution: We have A(A+B)=A2+AB=A2+BA=(A+B)AA(A + B) = A^2 + AB = A^2 + BA = (A + B)A.

If A+BA + B is invertible:

A(A+B)(A+B)1=(A+B)A(A+B)1A(A + B)(A + B)^{-1} = (A + B)A(A + B)^{-1}
A=(A+B)A(A+B)1A = (A + B)A(A + B)^{-1}

Multiply by (A+B)1(A + B)^{-1} on left:

(A+B)1A=A(A+B)1(A + B)^{-1}A = A(A + B)^{-1}

18. Geometric Interpretation

Undoing Transformations

If AA represents a transformation (rotation, scaling, shear, etc.), then A1A^{-1} represents the inverse transformation that "undoes" it:

  • If AA rotates by θ\theta, then A1A^{-1} rotates by θ-\theta
  • If AA scales by kk, then A1A^{-1} scales by 1/k1/k
  • If AA shears, then A1A^{-1} shears in the opposite direction

When Inversion Fails

A matrix is singular (not invertible) when it "collapses" dimension:

  • A projection onto a line in R2\mathbb{R}^2 is singular
  • A shear that maps everything to a single line is singular
  • Any transformation that loses information cannot be inverted

Determinant as Volume

The determinant det(A)\det(A) represents the factor by which AA scales volumes:

  • det(A)=0\det(A) = 0 means AA collapses to lower dimension (not invertible)
  • det(A1)=1/det(A)\det(A^{-1}) = 1/\det(A) restores the original volume

19. Historical Notes

Cayley's Contributions: Arthur Cayley developed the theory of matrix inverses in his 1858 paper. He recognized that invertibility corresponds to non-zero determinants and developed the adjugate formula.

Gaussian Elimination: While named after Carl Friedrich Gauss (early 1800s), the method was known to Chinese mathematicians nearly 2000 years earlier in texts like "The Nine Chapters on the Mathematical Art." It remains the standard algorithm for computing inverses.

General Linear Group: The group GLn(F)GL_n(F) became central to modern mathematics through the work of Sophus Lie, Felix Klein, and others in the late 19th century. It connects linear algebra to group theory and geometry through the Erlangen program.

Numerical Computing: In the computer age, efficient and stable algorithms for matrix inversion became crucial. LU decomposition, developed by Alan Turing and others in the 1940s-50s, is now the standard approach for numerical work.

Modern Applications: Matrix inverses appear throughout science and engineering: solving systems of equations, computer graphics (inverting transformations), machine learning (solving linear regression), quantum mechanics, and control theory.

20. Left and Right Inverses

Definition 4.14: One-Sided Inverses

For a (possibly non-square) matrix AA:

  • Left inverse: LL such that LA=ILA = I
  • Right inverse: RR such that AR=IAR = I
Theorem 4.27: Existence of One-Sided Inverses

For an m×nm \times n matrix AA:

  • AA has a left inverse iff AA is injective (columns independent) iff rank(A)=n\text{rank}(A) = n
  • AA has a right inverse iff AA is surjective (rows span FmF^m) iff rank(A)=m\text{rank}(A) = m
  • AA has both iff m=nm = n and AA is invertible
Example 4.11: Left Inverse Example

Let A=(10)A = \begin{pmatrix} 1 \\ 0 \end{pmatrix}. A left inverse is any L=(1c)L = \begin{pmatrix} 1 & c \end{pmatrix} for any cc:

LA=(1c)(10)=1LA = \begin{pmatrix} 1 & c \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = 1

Left inverses are NOT unique for non-square matrices!

Remark 4.20: Moore-Penrose Pseudoinverse

The Moore-Penrose pseudoinverse A+A^+ generalizes the inverse to all matrices (including non-square and singular). It satisfies:

  • AA+A=AAA^+A = A
  • A+AA+=A+A^+AA^+ = A^+
  • (AA+)T=AA+(AA^+)^T = AA^+
  • (A+A)T=A+A(A^+A)^T = A^+A

Applications

Solving Linear Systems

Ax=bx=A1bAx = b \Rightarrow x = A^{-1}b. Theoretical but LU is faster in practice.

Computer Graphics

Inverse transformations for camera models, view projections, and animation.

Cryptography

Hill cipher uses matrix multiplication for encryption and inverse for decryption.

Statistics

Covariance matrix inverse in multivariate distributions and regression.

Related Topics

Elementary Matrices
Determinants
Matrix Rank
LU Decomposition
Gaussian Elimination
General Linear Group

Key Takeaways

  • Invertible ⟺ det ≠ 0 ⟺ full rank ⟺ columns independent
  • 2×2 formula: swap diagonals, negate off-diagonals, divide by det
  • Gauss-Jordan: row reduce [A|I] to [I|A⁻¹]
  • (AB)⁻¹ = B⁻¹A⁻¹ — order reverses!

What's Next?

Now that you understand matrix inverses, the next topics build on this foundation:

  • Elementary Matrices: The building blocks that make Gaussian elimination work
  • Matrix Rank: The fundamental dimension that determines invertibility
  • Determinants: An algebraic criterion for invertibility

These concepts together form the theoretical backbone of computational linear algebra.

Matrix Inverse Practice
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If A1A^{-1} exists, what is AA1AA^{-1}?
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Is the zero matrix invertible?
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(AB)1=?(AB)^{-1} = ?
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4
If AA is n×nn \times n and Ax=0Ax = 0 has only x=0x = 0 as solution, is AA invertible?
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(A1)T=?(A^{-1})^T = ?
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A 3×43 \times 4 matrix is invertible:
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If det(A)0\det(A) \neq 0, then AA is:
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How do you compute A1A^{-1} using row reduction?
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If AB=IAB = I for square matrices, is BA=IBA = I?
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(A1)1=?(A^{-1})^{-1} = ?
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If AA is invertible, is ATA^T invertible?
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What is det(A1)\det(A^{-1})?
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