The Hook: The Mathematics Behind Movie Transformations
When you watch Toy Story or Spider-Man: Into the Spider-Verse, every character movement is controlled by matrix operations!
Buzz Lightyear’s rotation? Matrix multiplication. Scaling a character bigger? Scalar multiplication of matrices. Moving across the screen? Matrix addition.
A simple character rotation by 45° uses:
$$R_{45°} = \begin{bmatrix} \cos 45° & -\sin 45° \\ \sin 45° & \cos 45° \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$Combining rotations? Multiply rotation matrices! That’s why matrix multiplication is fundamental — it combines transformations.
Real applications: Game engines (Unity, Unreal), robot arm movements, GPS coordinate transformations, Google Search ranking algorithm (PageRank uses matrix powers!).
Why this matters for JEE: Matrix operations appear in 4-5 questions in JEE Main and form the basis for solving systems of equations, eigenvalue problems, and coordinate geometry transformations in JEE Advanced.
Interactive: Matrix Transformations Visualizer
See how a 2x2 matrix transforms the coordinate plane! Adjust the matrix values or use preset buttons to explore rotations, reflections, scaling, and shearing. Watch how the unit square and basis vectors transform in real-time.
- The determinant tells you the area scaling factor (negative = orientation flip)
- Eigenvectors (dashed orange/cyan lines) are directions that only get scaled, not rotated
- Basis vectors i’ (blue) and j’ (green) show where the original i-hat and j-hat land
- The matrix columns [a, c] and [b, d] directly give you the transformed basis vectors!
Prerequisites
Before diving into matrix algebra, you should be comfortable with:
- Matrix Basics — Matrix notation, types, and order
- Basic algebra — Distributive property, associative laws
- Ordered pairs and functions
Matrix Addition and Subtraction
Definition
Two matrices can be added or subtracted if and only if they have the same order.
For matrices $A = [a_{ij}]_{m \times n}$ and $B = [b_{ij}]_{m \times n}$:
$$\boxed{(A + B)_{ij} = a_{ij} + b_{ij}}$$ $$\boxed{(A - B)_{ij} = a_{ij} - b_{ij}}$$In words: Add (or subtract) corresponding elements.
Example
$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$$ $$A + B = \begin{bmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}$$ $$A - B = \begin{bmatrix} 1-5 & 2-6 \\ 3-7 & 4-8 \end{bmatrix} = \begin{bmatrix} -4 & -4 \\ -4 & -4 \end{bmatrix}$$Properties of Addition
For matrices $A, B, C$ of the same order:
| Property | Formula | Name |
|---|---|---|
| Commutative | $A + B = B + A$ | Order doesn’t matter |
| Associative | $(A + B) + C = A + (B + C)$ | Grouping doesn’t matter |
| Additive Identity | $A + O = O + A = A$ | Zero matrix is identity |
| Additive Inverse | $A + (-A) = O$ | Negative exists |
Matrix addition behaves exactly like adding numbers!
Just remember: Same order required — you can’t add a $2 \times 2$ to a $3 \times 3$.
Think: You can’t add apples to a table of different dimensions!
Scalar Multiplication
Definition
Multiplying a matrix by a scalar (real number) $k$:
$$\boxed{(kA)_{ij} = k \cdot a_{ij}}$$In words: Multiply every element by $k$.
Example
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}, \quad 3A = \begin{bmatrix} 3 & 6 & 9 \\ 12 & 15 & 18 \end{bmatrix}$$Properties of Scalar Multiplication
For scalars $k, l$ and matrices $A, B$:
| Property | Formula |
|---|---|
| Associative with scalars | $k(lA) = (kl)A$ |
| Distributive over matrix addition | $k(A + B) = kA + kB$ |
| Distributive over scalar addition | $(k + l)A = kA + lA$ |
| Identity | $1 \cdot A = A$ |
| Zero | $0 \cdot A = O$ |
| Negation | $(-1)A = -A$ |
Matrix Multiplication
The Rule
Matrices $A$ and $B$ can be multiplied to get $AB$ if and only if:
$$\boxed{\text{Number of columns in } A = \text{Number of rows in } B}$$If $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$.
Visual memory:
$$A_{m \times \color{red}{n}} \cdot B_{\color{red}{n} \times p} = (AB)_{m \times p}$$The middle indices must match!
Definition
For $A = [a_{ij}]_{m \times n}$ and $B = [b_{jk}]_{n \times p}$:
$$\boxed{(AB)_{ik} = \sum_{j=1}^{n} a_{ij} \cdot b_{jk}}$$In words: Element $(i,k)$ of $AB$ = (row $i$ of $A$) $\cdot$ (column $k$ of $B$)
Step-by-Step Process
To find element $(i,k)$ in $AB$:
- Take row $i$ of matrix $A$
- Take column $k$ of matrix $B$
- Multiply corresponding elements
- Add all the products
Example 1: Basic Multiplication
$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$$Find $AB$:
$(AB)_{11}$ = Row 1 of $A$ $\cdot$ Column 1 of $B$ = $(1)(5) + (2)(7) = 5 + 14 = 19$
$(AB)_{12}$ = Row 1 of $A$ $\cdot$ Column 2 of $B$ = $(1)(6) + (2)(8) = 6 + 16 = 22$
$(AB)_{21}$ = Row 2 of $A$ $\cdot$ Column 1 of $B$ = $(3)(5) + (4)(7) = 15 + 28 = 43$
$(AB)_{22}$ = Row 2 of $A$ $\cdot$ Column 2 of $B$ = $(3)(6) + (4)(8) = 18 + 32 = 50$
$$AB = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$$Example 2: Different Orders
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{2 \times 3}, \quad B = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}_{3 \times 1}$$$AB$ is possible (columns of $A$ = 3, rows of $B$ = 3) ✓
Result will be $2 \times 1$:
$$AB = \begin{bmatrix} (1)(1) + (2)(2) + (3)(3) \\ (4)(1) + (5)(2) + (6)(3) \end{bmatrix} = \begin{bmatrix} 14 \\ 32 \end{bmatrix}$$Can we compute $BA$?
$B$ is $3 \times 1$, $A$ is $2 \times 3$
Columns of $B$ = 1, Rows of $A$ = 2 → NO, $BA$ is not defined! ✗
Matrix multiplication is NOT commutative!
$$AB \neq BA \text{ in general}$$In fact, even if $AB$ exists, $BA$ might not exist!
Example: If $A$ is $2 \times 3$ and $B$ is $3 \times 4$:
- $AB$ exists (and is $2 \times 4$)
- $BA$ does NOT exist (columns of $B$ = 4, rows of $A$ = 2)
JEE Trap: Never assume $AB = BA$! Always check orders first.
Properties of Matrix Multiplication
For matrices with appropriate orders:
| Property | Formula | Name |
|---|---|---|
| NOT Commutative | $AB \neq BA$ (generally) | Order matters! |
| Associative | $(AB)C = A(BC)$ | Grouping doesn’t matter |
| Distributive (left) | $A(B + C) = AB + AC$ | Left distribution |
| Distributive (right) | $(A + B)C = AC + BC$ | Right distribution |
| Identity | $AI = IA = A$ | Identity matrix |
| Zero | $AO = OA = O$ | Zero matrix |
| Scalar | $k(AB) = (kA)B = A(kB)$ | Scalar associativity |
Special Cases Where $AB = BA$
Matrices do commute when:
- One matrix is the identity: $AI = IA$
- One matrix is a scalar matrix: $A(\lambda I) = (\lambda I)A$
- Both are diagonal matrices (of same order)
- Special matrix pairs (rare, must verify)
Think of it as “Row meets Column”
Check if multiplication is possible: middle numbers match
- $A_{2 \times \color{red}{3}} \cdot B_{\color{red}{3} \times 4}$ ✓ → Result is $2 \times 4$
To find an element: Row from first, Column from second
- $(AB)_{23}$ = Row 2 of $A$ times Column 3 of $B$
Order matters: $AB \neq BA$ (unlike regular numbers!)
Mnemonic: “Remember Commuting Breaks” (RCB) — Row-Column-Breaks (no commutative property)
Transpose of a Matrix
Definition
The transpose of a matrix $A$, denoted $A^T$ (or $A'$), is obtained by interchanging rows and columns.
If $A = [a_{ij}]_{m \times n}$, then:
$$\boxed{A^T = [a_{ji}]_{n \times m}}$$In words: Row $i$ of $A$ becomes column $i$ of $A^T$.
Examples
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{2 \times 3} \Rightarrow A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}_{3 \times 2}$$ $$B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \Rightarrow B^T = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix}$$Row matrix ↔ Column matrix:
$$\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}^T = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$Properties of Transpose
| Property | Formula | Note |
|---|---|---|
| Double transpose | $(A^T)^T = A$ | Transpose twice = original |
| Addition | $(A + B)^T = A^T + B^T$ | Distributes over addition |
| Scalar multiplication | $(kA)^T = kA^T$ | Scalar comes out |
| Multiplication | $(AB)^T = B^T A^T$ | ORDER REVERSES! |
| Extended multiplication | $(ABC)^T = C^T B^T A^T$ | Reverse order |
NOT $A^T B^T$! The order reverses!
Think of it like taking off shoes and socks:
- You put them on: Socks first, then shoes
- You take them off: Shoes first, then socks (reverse order!)
Common JEE mistake: Writing $(AB)^T = A^T B^T$ ✗
Transpose and Matrix Types
| Matrix Type | Transpose Property |
|---|---|
| Symmetric | $A^T = A$ |
| Skew-symmetric | $A^T = -A$ |
| Diagonal | $D^T = D$ (always symmetric) |
| Identity | $I^T = I$ |
| Orthogonal | $A^T A = AA^T = I$ |
Important Formulas and Results
Symmetric and Skew-Symmetric Decomposition
Every square matrix can be uniquely expressed as:
$$\boxed{A = \frac{A + A^T}{2} + \frac{A - A^T}{2}}$$where:
- $P = \frac{A + A^T}{2}$ is symmetric ($P^T = P$)
- $Q = \frac{A - A^T}{2}$ is skew-symmetric ($Q^T = -Q$)
Proof:
$$P^T = \left(\frac{A + A^T}{2}\right)^T = \frac{A^T + A}{2} = P$$ $$Q^T = \left(\frac{A - A^T}{2}\right)^T = \frac{A^T - A}{2} = -\frac{A - A^T}{2} = -Q$$Example
$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ $$A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$$ $$P = \frac{1}{2}\begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} = \begin{bmatrix} 1 & 2.5 \\ 2.5 & 4 \end{bmatrix}$$(Symmetric)
$$Q = \frac{1}{2}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -0.5 \\ 0.5 & 0 \end{bmatrix}$$(Skew-symmetric)
Verify: $P + Q = A$ ✓
Common Mistakes to Avoid
Wrong: Assuming $AB = BA$
Right: Always check! Matrix multiplication is not commutative.
Even if both $AB$ and $BA$ exist, they’re usually different!
Example:
$$A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$$ $$AB = \begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix}, \quad BA = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$$$AB \neq BA$ !
Wrong: $(AB)^T = A^T B^T$
Right: $(AB)^T = B^T A^T$ (order reverses!)
JEE Example: If $(AB)^T = BA$, what can you conclude? $B^T A^T = BA$
This happens when $A^T = B$ and $B^T = A$, or special symmetric cases.
Wrong: Any two matrices can be multiplied
Right: Check if multiplication is defined!
For $AB$: Columns of $A$ must equal rows of $B$
Example: Can you multiply $A_{3 \times 2}$ and $B_{4 \times 3}$?
NO! Columns of $A$ = 2, Rows of $B$ = 4 → Not compatible ✗
Wrong: If $AB = O$, then $A = O$ or $B = O$
Right: $AB = O$ does NOT imply either matrix is zero!
Counterexample:
$$A = \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}, B = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$$ $$AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$but neither $A$ nor $B$ is zero!
Matrix algebra doesn’t have the “zero product property” of real numbers!
Wrong: If $AB = AC$ and $A \neq O$, then $B = C$
Right: You cannot cancel matrices like numbers!
Cancellation requires $A$ to be invertible (non-singular).
If $|A| \neq 0$, then $AB = AC \Rightarrow B = C$ ✓
But if $|A| = 0$, cancellation fails!
Power of a Matrix
For a square matrix $A$:
$$A^2 = A \cdot A, \quad A^3 = A \cdot A \cdot A, \quad A^n = \underbrace{A \cdot A \cdots A}_{n \text{ times}}$$Properties:
| Formula | Condition |
|---|---|
| $A^m \cdot A^n = A^{m+n}$ | Always |
| $(A^m)^n = A^{mn}$ | Always |
| $A^m \cdot B^m = (AB)^m$ | Only if $AB = BA$ |
Identity matrix: $I^n = I$ for all $n$
Diagonal matrix:
$$D = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \Rightarrow D^n = \begin{bmatrix} a^n & 0 \\ 0 & b^n \end{bmatrix}$$Nilpotent matrix: $A^k = O$ for some positive integer $k$
Example: $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, then $A^2 = O$
Idempotent matrix: $A^2 = A$
Example: Projection matrices
Practice Problems
Level 1: Foundation (NCERT)
Find $A + B$ if $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}$.
Solution:
$$A + B = \begin{bmatrix} 2+1 & 3+(-2) \\ 1+3 & 4+0 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 4 & 4 \end{bmatrix}$$Answer: $\begin{bmatrix} 3 & 1 \\ 4 & 4 \end{bmatrix}$
If $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$, find $A^T$.
Solution:
Interchange rows and columns:
$$A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$$Answer: $\begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$
Compute $AB$ if $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 6 \end{bmatrix}$.
Solution:
Check: $A$ is $2 \times 2$, $B$ is $2 \times 1$ → $AB$ is $2 \times 1$ ✓
$$AB = \begin{bmatrix} 1(5) + 2(6) \\ 3(5) + 4(6) \end{bmatrix} = \begin{bmatrix} 17 \\ 39 \end{bmatrix}$$Answer: $\begin{bmatrix} 17 \\ 39 \end{bmatrix}$
Level 2: JEE Main
If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, verify that $(A^T)^T = A$.
Solution:
$$A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$$ $$(A^T)^T = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = A$$✓
Verified!
Express $A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$ as the sum of a symmetric and skew-symmetric matrix.
Solution:
$$A^T = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}$$Symmetric part:
$$P = \frac{A + A^T}{2} = \frac{1}{2}\begin{bmatrix} 4 & 7 \\ 7 & 10 \end{bmatrix} = \begin{bmatrix} 2 & 3.5 \\ 3.5 & 5 \end{bmatrix}$$Skew-symmetric part:
$$Q = \frac{A - A^T}{2} = \frac{1}{2}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -0.5 \\ 0.5 & 0 \end{bmatrix}$$Verify: $P + Q = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = A$ ✓
Answer: $P = \begin{bmatrix} 2 & 3.5 \\ 3.5 & 5 \end{bmatrix}$, $Q = \begin{bmatrix} 0 & -0.5 \\ 0.5 & 0 \end{bmatrix}$
If $A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 0 & 4 \end{bmatrix}$, find $(AB)^T$ and verify that $(AB)^T = B^T A^T$.
Solution:
$$AB = \begin{bmatrix} 2 & 3 \\ -2 & 25 \end{bmatrix}$$ $$(AB)^T = \begin{bmatrix} 2 & -2 \\ 3 & 25 \end{bmatrix}$$Now:
$$B^T = \begin{bmatrix} 2 & 0 \\ 3 & 4 \end{bmatrix}, \quad A^T = \begin{bmatrix} 1 & -1 \\ 0 & 7 \end{bmatrix}$$ $$B^T A^T = \begin{bmatrix} 2 & -2 \\ 3 & 25 \end{bmatrix}$$$(AB)^T = B^T A^T$ ✓ Verified!
Level 3: JEE Advanced
If $A$ is a skew-symmetric matrix of odd order, prove that $|A| = 0$.
Solution:
For skew-symmetric matrix: $A^T = -A$
Taking determinant both sides:
$$|A^T| = |-A|$$We know: $|A^T| = |A|$ and $|-A| = (-1)^n |A|$ where $n$ is the order.
$$|A| = (-1)^n |A|$$Since $n$ is odd: $(-1)^n = -1$
$$|A| = -|A|$$ $$2|A| = 0$$ $$|A| = 0$$Proved!
Find matrix $A$ such that $A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ but $A \neq \pm I$.
Solution:
Try: $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
$$A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$✓
Also try: $A = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$ works!
Answer: Multiple solutions exist, e.g., $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ or $\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$
If $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, find $A^T A$ and interpret.
Solution:
$$A^T = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$ $$A^T A = \begin{bmatrix} \cos^2\theta + \sin^2\theta & -\cos\theta\sin\theta + \sin\theta\cos\theta \\ \sin\theta\cos\theta - \cos\theta\sin\theta & \sin^2\theta + \cos^2\theta \end{bmatrix}$$ $$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$Interpretation: $A$ is an orthogonal matrix (rotation matrix).
Rotation matrices preserve length and angle, so $A^T A = I$.
Answer: $A^T A = I$ (orthogonal matrix)
Quick Revision Box
| Operation | Rule | Key Point |
|---|---|---|
| Addition | $(A+B)_{ij} = a_{ij} + b_{ij}$ | Same order required |
| Scalar multiplication | $(kA)_{ij} = ka_{ij}$ | Multiply each element |
| Matrix multiplication | $(AB)_{ik} = \sum a_{ij}b_{jk}$ | Columns of $A$ = Rows of $B$ |
| Transpose | $(A^T)_{ji} = a_{ij}$ | Rows ↔ Columns |
| $(AB)^T$ | $B^T A^T$ | Order reverses! |
| Symmetric decomp | $A = \frac{A+A^T}{2} + \frac{A-A^T}{2}$ | Any square matrix |
| Commutative? | $AB \neq BA$ generally | NOT commutative! |
Related Topics
Within Matrices & Determinants Chapter
- Matrix Basics — Foundation of matrix types
- Determinants Evaluation — Finding determinant of products
- Adjoint and Inverse — Using transpose to find inverse
- Linear Equations — Matrix form of equations
Math Connections
- Coordinate Geometry — Rotation and transformation matrices
- Trigonometry — Rotation matrices use trig functions
- Vector Algebra — Dot product as matrix multiplication
- Complex Numbers — Matrix representation of complex operations
Real-World Applications
- Computer Graphics — Combining transformations via matrix multiplication
- Machine Learning — Gradient descent uses matrix operations
- Cryptography — Hill cipher uses matrix multiplication
- Economics — Leontief input-output model
Teacher’s Summary
- Addition/Subtraction: Only possible for matrices of same order
- Scalar multiplication: Multiply every element by the scalar
- Matrix multiplication: Columns of first = Rows of second (middle numbers match!)
- NOT commutative: $AB \neq BA$ in general — order matters!
- Transpose reverses order: $(AB)^T = B^T A^T$ (not $A^T B^T$!)
- Associative: $(AB)C = A(BC)$ — grouping doesn’t matter
- Symmetric decomposition: Any square matrix = symmetric + skew-symmetric
- Key property: $(A^T)^T = A$ (double transpose returns original)
“Master the transpose reversal rule — it’s in 80% of JEE Advanced matrix problems!”
Exam Strategy:
- For multiplication: Always check order compatibility first!
- For transpose of product: Reverse the order! $(ABC)^T = C^T B^T A^T$
- Watch for non-commutative traps: Never assume $AB = BA$