The Hook: The Undo Button of Mathematics
Ever wondered how Google can reverse-engineer where a webpage should rank? They use matrix inverses!
The PageRank algorithm creates a giant matrix representing how web pages link to each other. To find rankings, Google needs to solve a massive linear system — which requires finding the inverse of this matrix.
Real-world applications:
- Computer graphics: Undo transformations (rotate back, scale back)
- Cryptography: Decrypt messages (encryption matrix → inverse for decryption)
- Economics: Input-output analysis (production → resource requirements)
- Engineering: Solve circuit equations, structural analysis
- Machine learning: Parameter optimization, neural network training
The magic: If $A$ represents a transformation, $A^{-1}$ undoes it:
$$A \cdot A^{-1} = A^{-1} \cdot A = I$$Why this matters for JEE: Inverse matrices appear in 3-4 questions in JEE Main and are essential for solving linear systems, eigenvalue problems, and transformation geometry in JEE Advanced.
Prerequisites
Before diving into adjoint and inverse, you should be comfortable with:
- Matrix Basics — Square matrices, identity matrix
- Matrix Algebra — Matrix multiplication, transpose
- Determinants — Computing determinants
- Minors and Cofactors — Finding cofactors
Singular vs Non-Singular Matrices
Definitions
A square matrix $A$ is called:
Non-singular (Invertible): if $|A| \neq 0$
- The matrix has an inverse
- Represents a reversible transformation
- System $AX = B$ has a unique solution
Singular (Non-invertible): if $|A| = 0$
- The matrix does NOT have an inverse
- Represents a dimension-collapsing transformation
- System $AX = B$ may have no solution or infinitely many solutions
Examples
Non-singular → Inverse exists ✓
Singular → Inverse does NOT exist ✗
Notice Row 2 = $\frac{3}{2}$ × Row 1 (proportional rows)
Before finding inverse, always check if $|A| = 0$:
- Zero row/column → Singular
- Identical rows/columns → Singular
- Proportional rows/columns → Singular
- Determinant = 0 → Singular
Don’t waste time trying to find inverse of a singular matrix!
Adjoint of a Matrix (Adjugate)
Definition
The adjoint (or adjugate) of a square matrix $A$, denoted $\text{adj}(A)$, is the transpose of the cofactor matrix.
$$\boxed{\text{adj}(A) = [C_{ij}]^T}$$Steps to find adjoint:
- Find all cofactors $C_{ij}$ of matrix $A$
- Form the cofactor matrix $[C_{ij}]$
- Take transpose: $\text{adj}(A) = [C_{ij}]^T$
Example: 2×2 Matrix
$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$Step 1: Find cofactors
$$C_{11} = (+1) \cdot d = d$$ $$C_{12} = (-1) \cdot c = -c$$ $$C_{21} = (-1) \cdot b = -b$$ $$C_{22} = (+1) \cdot a = a$$Step 2: Cofactor matrix
$$[C_{ij}] = \begin{bmatrix} d & -c \\ -b & a \end{bmatrix}$$Step 3: Transpose
$$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$For 2×2 matrices, there’s a shortcut:
$$\boxed{\text{adj}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}}$$Pattern: Swap diagonal elements, negate off-diagonal elements.
Example: 3×3 Matrix
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{bmatrix}$$Step 1: Find all 9 cofactors
$$C_{11} = (+1) \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} = 24$$ $$C_{12} = (-1) \begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} = -(0-5) = 5$$ $$C_{13} = (+1) \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = -4$$ $$C_{21} = (-1) \begin{vmatrix} 2 & 3 \\ 0 & 6 \end{vmatrix} = -12$$ $$C_{22} = (+1) \begin{vmatrix} 1 & 3 \\ 1 & 6 \end{vmatrix} = 3$$ $$C_{23} = (-1) \begin{vmatrix} 1 & 2 \\ 1 & 0 \end{vmatrix} = 2$$ $$C_{31} = (+1) \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = -2$$ $$C_{32} = (-1) \begin{vmatrix} 1 & 3 \\ 0 & 5 \end{vmatrix} = -5$$ $$C_{33} = (+1) \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} = 4$$Step 2: Cofactor matrix
$$[C_{ij}] = \begin{bmatrix} 24 & 5 & -4 \\ -12 & 3 & 2 \\ -2 & -5 & 4 \end{bmatrix}$$Step 3: Transpose
$$\text{adj}(A) = \begin{bmatrix} 24 & -12 & -2 \\ 5 & 3 & -5 \\ -4 & 2 & 4 \end{bmatrix}$$Properties of Adjoint
Property 1: Product with Original Matrix
$$\boxed{A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| \cdot I}$$This is the KEY property connecting adjoint and inverse!
Property 2: Determinant of Adjoint
For an $n \times n$ matrix:
$$\boxed{|\text{adj}(A)| = |A|^{n-1}}$$Examples:
- For $2 \times 2$: $|\text{adj}(A)| = |A|^{2-1} = |A|$
- For $3 \times 3$: $|\text{adj}(A)| = |A|^{3-1} = |A|^2$
Property 3: Adjoint of Transpose
$$\text{adj}(A^T) = [\text{adj}(A)]^T$$Property 4: Adjoint of Product
$$\text{adj}(AB) = \text{adj}(B) \cdot \text{adj}(A)$$Note: Order reverses (just like $(AB)^T = B^T A^T$)!
Property 5: Adjoint of Adjoint
For non-singular $A$ of order $n$:
$$\text{adj}(\text{adj}(A)) = |A|^{n-2} \cdot A$$Property 6: Adjoint of Scalar Multiple
$$\text{adj}(kA) = k^{n-1} \text{adj}(A)$$where $n$ is the order of matrix $A$.
Most important property:
$$A \cdot \text{adj}(A) = |A| \cdot I$$Why it’s magic: This directly gives us the inverse formula!
Divide both sides by $|A|$:
$$A \cdot \frac{\text{adj}(A)}{|A|} = I$$Therefore: $A^{-1} = \frac{\text{adj}(A)}{|A|}$
Inverse of a Matrix
Definition
The inverse of a square matrix $A$, denoted $A^{-1}$, is the unique matrix such that:
$$\boxed{A \cdot A^{-1} = A^{-1} \cdot A = I}$$Existence condition: $A^{-1}$ exists if and only if $|A| \neq 0$
Inverse Formula Using Adjoint
$$\boxed{A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)}$$This is THE formula for JEE!
Steps to Find Inverse
Step 1: Check if $|A| \neq 0$ (if zero, inverse doesn’t exist)
Step 2: Find all cofactors $C_{ij}$
Step 3: Form adjoint: $\text{adj}(A) = [C_{ij}]^T$
Step 4: Apply formula: $A^{-1} = \frac{1}{|A|} \text{adj}(A)$
Example 1: 2×2 Inverse
$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$Step 1: $|A| = 4 - 6 = -2 \neq 0$ ✓
Step 2:
$$\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$Step 3:
$$A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}$$Verify:
$$A \cdot A^{-1} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$✓
Quick Formula for 2×2 Inverse
For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$:
$$\boxed{A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}}$$Pattern:
- Find determinant: $ad - bc$
- Swap diagonal: $a \leftrightarrow d$
- Negate off-diagonal: $b \to -b$, $c \to -c$
- Divide by determinant
Example 2: 3×3 Inverse
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$$Step 1: Calculate $|A|$
$$|A| = 1(0-24) - 2(0-20) + 3(0-5) = -24 + 40 - 15 = 1$$Since $|A| = 1 \neq 0$, inverse exists ✓
Step 2: Find all 9 cofactors
$$C_{11} = \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = -24$$ $$C_{12} = -\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = 0 - (-20) = 20$$ $$C_{13} = \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} = -5$$ $$C_{21} = -\begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix} = -(-18) = 18$$ $$C_{22} = \begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix} = -15$$ $$C_{23} = -\begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix} = -(6-10) = 4$$ $$C_{31} = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 5$$ $$C_{32} = -\begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -4$$ $$C_{33} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1$$Step 3: Form adjoint (transpose of cofactor matrix)
$$\text{adj}(A) = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix}$$Step 4: Apply formula
$$A^{-1} = \frac{1}{1} \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix}$$Properties of Inverse Matrices
Property 1: Uniqueness
If $A^{-1}$ exists, it is unique.
Property 2: Inverse of Inverse
$$\boxed{(A^{-1})^{-1} = A}$$Inverting twice returns the original matrix.
Property 3: Inverse of Product
$$\boxed{(AB)^{-1} = B^{-1} A^{-1}}$$Order reverses! (Just like transpose and adjoint)
Extended: $(ABC)^{-1} = C^{-1} B^{-1} A^{-1}$
Property 4: Inverse of Transpose
$$\boxed{(A^T)^{-1} = (A^{-1})^T}$$Property 5: Determinant of Inverse
$$\boxed{|A^{-1}| = \frac{1}{|A|}}$$Proof: $A \cdot A^{-1} = I$ → $|A| \cdot |A^{-1}| = |I| = 1$
Property 6: Inverse of Scalar Multiple
$$\boxed{(kA)^{-1} = \frac{1}{k} A^{-1}}$$Property 7: Identity Matrix
$$\boxed{I^{-1} = I}$$The identity is its own inverse.
Property 8: Symmetric Matrix
If $A$ is symmetric, then $A^{-1}$ is also symmetric:
$$A = A^T \implies A^{-1} = (A^{-1})^T$$Three operations that REVERSE order:
- Transpose: $(AB)^T = B^T A^T$
- Inverse: $(AB)^{-1} = B^{-1} A^{-1}$
- Adjoint: $\text{adj}(AB) = \text{adj}(B) \cdot \text{adj}(A)$
Mnemonic: “TIA Reverses” (Transpose, Inverse, Adjoint)
Think: Taking off shoes and socks — reverse the order!
Special Matrices and Their Inverses
1. Diagonal Matrix
$$D = \begin{bmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{bmatrix} \implies D^{-1} = \begin{bmatrix} \frac{1}{d_1} & 0 & 0 \\ 0 & \frac{1}{d_2} & 0 \\ 0 & 0 & \frac{1}{d_3} \end{bmatrix}$$Rule: Reciprocal of each diagonal element (if all non-zero).
2. Orthogonal Matrix
A matrix $A$ is orthogonal if:
$$\boxed{A^T A = AA^T = I}$$For orthogonal matrices:
$$\boxed{A^{-1} = A^T}$$Example: Rotation matrices are orthogonal.
3. Involutory Matrix
A matrix $A$ is involutory if:
$$A^2 = I$$For involutory matrices:
$$\boxed{A^{-1} = A}$$The matrix is its own inverse!
Example: $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Elementary Row Operations Method
An alternative method to find inverse is using elementary row operations (Gauss-Jordan elimination):
Method
Write the augmented matrix $[A | I]$ and perform row operations to convert it to $[I | A^{-1}]$.
Example
Find inverse of $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
Step 1: Form augmented matrix
$$\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{array}\right]$$Step 2: $R_2 \to R_2 - 3R_1$
$$\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -2 & -3 & 1 \end{array}\right]$$Step 3: $R_2 \to -\frac{1}{2}R_2$
$$\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{3}{2} & -\frac{1}{2} \end{array}\right]$$Step 4: $R_1 \to R_1 - 2R_2$
$$\left[\begin{array}{cc|cc} 1 & 0 & -2 & 1 \\ 0 & 1 & \frac{3}{2} & -\frac{1}{2} \end{array}\right]$$Result:
$$A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$$Adjoint method:
- ✓ Good for theoretical problems
- ✓ When determinant is simple
- ✓ JEE Advanced symbolic questions
- ✗ Lots of calculations for $3 \times 3$
Row operations method:
- ✓ Faster for numerical $3 \times 3$ and larger
- ✓ Systematic, less prone to errors
- ✓ Good for computational problems
- ✗ Not useful for symbolic matrices
For JEE: Master both! Use adjoint for $2 \times 2$ and theoretical, use row operations for numerical $3 \times 3$.
Common Mistakes to Avoid
Wrong: Finding inverse when $|A| = 0$
Right: Always check $|A| \neq 0$ first!
If $|A| = 0$, the matrix is singular and has no inverse.
Don’t waste time calculating adjoint if determinant is zero!
Wrong: $\text{adj}(A) = [C_{ij}]$ (cofactor matrix)
Right: $\text{adj}(A) = [C_{ij}]^T$ (transpose of cofactor matrix)
Always transpose the cofactor matrix to get adjoint!
Wrong: $(AB)^{-1} = A^{-1} B^{-1}$
Right: $(AB)^{-1} = B^{-1} A^{-1}$ (order reverses!)
Think: Undo operations in reverse order.
Wrong: $A^{-1} = \frac{1}{A}$ (dividing each element by 1)
Right: $A^{-1} = \frac{1}{|A|} \text{adj}(A)$ (multiply adjoint by $\frac{1}{|A|}$)
Matrix inverse is NOT element-wise reciprocal!
Wrong: $(A + B)^{-1} = A^{-1} + B^{-1}$
Right: There is NO simple formula for $(A + B)^{-1}$!
This is a common trick question in JEE. Don’t fall for it!
Practice Problems
Level 1: Foundation (NCERT)
Find the adjoint of $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$
Solution:
For $2 \times 2$: Swap diagonal, negate off-diagonal
$$\text{adj}(A) = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$$Answer: $\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$
Find the inverse of $A = \begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$
Solution:
$$|A| = 7 - 6 = 1$$ $$\text{adj}(A) = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}$$ $$A^{-1} = \frac{1}{1} \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}$$Answer: $\begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}$
Verify that $A \cdot \text{adj}(A) = |A| \cdot I$ for $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
Solution:
$$|A| = -2$$ $$\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$ $$A \cdot \text{adj}(A) = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}$$ $$|A| \cdot I = -2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}$$✓
Verified!
Level 2: JEE Main
If $A = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$, find $(A^{-1})^T$.
Solution:
$$|A| = 14 - 15 = -1$$ $$A^{-1} = \frac{1}{-1} \begin{bmatrix} 7 & -3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 3 \\ 5 & -2 \end{bmatrix}$$ $$(A^{-1})^T = \begin{bmatrix} -7 & 5 \\ 3 & -2 \end{bmatrix}$$Answer: $\begin{bmatrix} -7 & 5 \\ 3 & -2 \end{bmatrix}$
If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{bmatrix}$, find $A^{-1}$.
Solution:
Diagonal matrix! Inverse = reciprocal of diagonal elements
$$A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$$Answer: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.5 & 0 \\ 0 & 0 & 0.2 \end{bmatrix}$
If $|A| = 5$ for a $3 \times 3$ matrix, find $|\text{adj}(A)|$.
Solution:
For $n \times n$ matrix: $|\text{adj}(A)| = |A|^{n-1}$
For $3 \times 3$: $|\text{adj}(A)| = |A|^{3-1} = 5^2 = 25$
Answer: 25
Level 3: JEE Advanced
If $A$ and $B$ are invertible matrices, prove that $(AB)^{-1} = B^{-1} A^{-1}$.
Solution:
We need to show: $(AB) \cdot (B^{-1} A^{-1}) = I$
$$(AB)(B^{-1} A^{-1}) = A(BB^{-1})A^{-1}$$ $$= A \cdot I \cdot A^{-1}$$ $$= AA^{-1} = I$$✓
Similarly: $(B^{-1} A^{-1})(AB) = I$
Therefore: $(AB)^{-1} = B^{-1} A^{-1}$ Proved!
If $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, show that $A^{-1} = A^T$ (orthogonal matrix).
Solution:
$$A^T = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$ $$A \cdot A^T = \begin{bmatrix} \cos^2\theta + \sin^2\theta & 0 \\ 0 & \sin^2\theta + \cos^2\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$Since $AA^T = I$, we have $A^{-1} = A^T$ ✓
Proved! (Rotation matrices are orthogonal)
If $A^2 - 3A + 2I = O$, express $A^{-1}$ in terms of $A$ and $I$.
Solution:
$$A^2 - 3A + 2I = O$$ $$A^2 - 3A = -2I$$ $$A(A - 3I) = -2I$$ $$A \cdot \frac{A - 3I}{-2} = I$$Therefore:
$$A^{-1} = \frac{3I - A}{2}$$Answer: $A^{-1} = \frac{1}{2}(3I - A)$
Quick Revision Box
| Concept | Formula/Property |
|---|---|
| Non-singular | $\|A\| \neq 0$ (inverse exists) |
| Singular | $\|A\| = 0$ (no inverse) |
| Adjoint | $\text{adj}(A) = [C_{ij}]^T$ |
| Inverse formula | $A^{-1} = \frac{1}{\|A\|} \text{adj}(A)$ |
| $2 \times 2$ adjoint | Swap diagonal, negate off-diagonal |
| $A \cdot \text{adj}(A)$ | $\|A\| \cdot I$ |
| $\|\text{adj}(A)\|$ | $\|A\|^{n-1}$ for $n \times n$ |
| $(AB)^{-1}$ | $B^{-1} A^{-1}$ (order reverses!) |
| $(A^T)^{-1}$ | $(A^{-1})^T$ |
| $\|A^{-1}\|$ | $\frac{1}{\|A\|}$ |
| Diagonal inverse | Reciprocal of diagonal elements |
| Orthogonal | $A^{-1} = A^T$ |
Related Topics
Within Matrices & Determinants Chapter
- Matrix Basics — Foundation of matrices
- Matrix Algebra — Operations needed for inverse
- Determinants — Check if inverse exists
- Minors and Cofactors — Building blocks of adjoint
- Linear Equations — Using inverse to solve systems
Math Connections
- Coordinate Geometry — Transformation matrices
- Vector Algebra — Change of basis matrices
- Complex Numbers — Matrices over complex numbers
Real-World Applications
- Cryptography — RSA encryption, Hill cipher
- Computer graphics — Inverse transformations
- Economics — Leontief input-output model
- Engineering — Circuit analysis, structural mechanics
- Machine learning — Parameter optimization
Teacher’s Summary
- Inverse exists ⟺ $|A| \neq 0$ (non-singular matrix)
- Adjoint = Transpose of cofactor matrix: $\text{adj}(A) = [C_{ij}]^T$
- Inverse formula: $A^{-1} = \frac{1}{|A|} \text{adj}(A)$ (THE formula!)
- Magic property: $A \cdot \text{adj}(A) = |A| \cdot I$
- 2×2 shortcut: Swap diagonal, negate off-diagonal, divide by determinant
- Order reversal: $(AB)^{-1} = B^{-1} A^{-1}$ (like transpose!)
- Diagonal matrix: Inverse = reciprocal of each diagonal element
- Orthogonal matrix: $A^{-1} = A^T$ (rotation matrices!)
- Check first: Always verify $|A| \neq 0$ before finding inverse
“First check the determinant — don’t waste time on singular matrices!”
Exam Strategy:
- $2 \times 2$: Use shortcut formula (10 seconds)
- $3 \times 3$ numerical: Row operations OR adjoint method
- $3 \times 3$ symbolic: Adjoint method (shows all work)
- Special matrices: Use properties (diagonal, orthogonal, etc.)