Chapter 3: Matrices

Exam Weightage & Blueprint

Total: ~6-8 Marks

Matrices is one of the easiest chapters in Class 12. It is often combined with Determinants in the Algebra unit (Total 10 marks). Focus on multiplication and symmetric properties.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	Very High	Order, Equality, Skew-Symmetric definition
Short Answer (2M)	2	High	Matrix Multiplication, Construction of Matrix
Short Answer (3M)	3	Medium	Express as Sum of Symmetric & Skew-Symmetric
Long Answer	5	Low	Word problems (rare), usually from Determinants

Last 24-Hour Checklist

Order of Matrix: $m \times n$ (Rows $\times$ Columns)
Multiplication Rule: Cols of A = Rows of B
Transpose Property: $(AB)' = B'A'$ (Reversal Law)
Symmetric: $A' = A$

Skew-Symmetric: $A' = -A$ (Diagonal elements are 0)
Sum Property: $A = \frac{1}{2}(A+A') + \frac{1}{2}(A-A')$
Invertible: $AB = BA = I$

Types of Matrices (The Cast)

Thinking Strategy for Order ($m \times n$): Think of a Cinema Hall.
Rows ($m$): Horizontal lines (Side to side).
Columns ($n$): Vertical lines (Front to back).
Order is always Row $\times$ Column (RC Cola).

Column Matrix

Analogy: "The Pole" (Standing)

Only 1 column ($m \times 1$).

$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Row Matrix

Analogy: "The Snake" (Sleeping)

Only 1 row ($1 \times n$).

$\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$

Square Matrix

Analogy: "The Perfect Box"

Rows equal Columns ($m = n$).

$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Diagonal Matrix

Analogy: "Only the Elements"

Only diagonal elements are non-zero. Others are 0.

$\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$

Identity Matrix ($I$)

Analogy: "The Number 1"

Diagonal is 1, rest 0. Acts like '1' in multiplication ($AI = A$).

$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Zero Matrix ($O$)

Analogy: "The Number 0"

All elements are 0. Acts like '0' in addition ($A+O = A$).

Operations on Matrices

1. Addition & Subtraction (The Carbon Copy Rule)

Rule: They must be exact "Carbon Copies" in size.
$(2 \times 2)$ can only add to $(2 \times 2)$.
Action: Element 1 adds to Element 1. ($a_{11} + b_{11}$).

Example:

$\begin{bmatrix} 2 & 5 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 6 & 2 \end{bmatrix}$

$= \begin{bmatrix} 2+1 & 5+0 \\ 3+6 & 4+2 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 9 & 6 \end{bmatrix}$

2. Multiplication (The Diving Board Rule)

Can we multiply?
Check Inner Numbers: $(m \times \mathbf{n})$ and $(\mathbf{n} \times p)$.
If they match, YES! The result will be Outer Numbers ($m \times p$).

How to Multiply (Row dives into Column):

Select Row 1 of Matrix A.
DIVE it into Column 1 of Matrix B.
Multiply first element with first, second with second, then ADD them.
Repeat for Row 1 into Column 2, etc.

Mantra: "Row 1 attacks Col 1, then Row 1 attacks Col 2..."

Example: Let $A = \begin{bmatrix} 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$

Row 1 $(1, 2)$ dives into Col 1 $(3, 4)$:

$AB = [(1 \times 3) + (2 \times 4)] = [3 + 8] = [11]$.

3. Transpose (Stand Up!)

Concept: Make the sleeping lines (Rows) stand up (Columns).
Key Property: $(AB)' = B'A'$ (Reversal Law - like taking off shoes then socks).

Symmetric & Skew-Symmetric Matrices

Symmetric Matrix

If $A' = A$

i.e., $a_{ij} = a_{ji}$ for all $i, j$.

Example: $\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$

Skew-Symmetric Matrix

If $A' = -A$

i.e., $a_{ij} = -a_{ji}$ for all $i, j$.

Note: All diagonal elements must be ZERO.

Theorem: Decomposition of Matrix

Any square matrix $A$ can be expressed as the sum of a symmetric and a skew-symmetric matrix: $$A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A')$$ Where $P = \frac{1}{2}(A + A')$ is symmetric and $Q = \frac{1}{2}(A - A')$ is skew-symmetric.

Orthogonal Matrix

A square matrix $A$ is said to be orthogonal if: $$A A^T = A^T A = I$$ where $A^T$ is the transpose of $A$ and $I$ is the identity matrix.
Key Property: If $A$ is orthogonal, then $A^T = A^{-1}$ and $|A| = \pm 1$.

Solved Examples (Board Marking Scheme)

Q1. Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $a_{ij} = \frac{(i+2j)^2}{2}$. (2 Marks)

Step 1: Calculate Elements 1 Mark

$a_{11} = \frac{(1+2)^2}{2} = \frac{9}{2}$

$a_{12} = \frac{(1+4)^2}{2} = \frac{25}{2}$

$a_{21} = \frac{(2+2)^2}{2} = \frac{16}{2} = 8$

$a_{22} = \frac{(2+4)^2}{2} = \frac{36}{2} = 18$

Step 2: Write Matrix 1 Mark

$A = \begin{bmatrix} 9/2 & 25/2 \\ 8 & 18 \end{bmatrix}$

Q2. If $A = \begin{bmatrix} 3 & 5 \\ 2 & a \end{bmatrix}$ is a symmetric matrix, find $a$. (1 Mark)

Solution: 1 Mark

For symmetric matrix, $A' = A$, so $a_{ij} = a_{ji}$.

Here $a_{12} = 5$ and $a_{21} = 2$. Wait, for symmetric $a_{12}$ must equal $a_{21}$.

If the question implies finding a variable to make it symmetric, say $B = \begin{bmatrix} 3 & x \\ 2 & 4 \end{bmatrix}$, then $x=2$.

Correction in Question Logic: A matrix is symmetric if $a_{12} = a_{21}$. In the given matrix, $5 \neq 2$, so it can never be symmetric regardless of $a$. Let's assume question was: Find $x$ if $\begin{bmatrix} 3 & x \\ 2 & 4 \end{bmatrix}$ is symmetric. Then $x=2$.

Q3. Express $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix. (3 Marks)

Step 1: Find $A'$ 0.5 Mark

$A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$

Step 2: Calculate P (Symmetric Part) 1 Mark

$P = \frac{1}{2}(A + A') = \frac{1}{2} \left( \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$

Check: $P' = P$. Correct.

Step 3: Calculate Q (Skew-Symmetric Part) 1 Mark

$Q = \frac{1}{2}(A - A') = \frac{1}{2} \left( \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Check: $Q' = -Q$. Correct.

Step 4: Final Expression 0.5 Mark

$A = P + Q = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): If $A$ is a square matrix such that $A^2 = A$, then $(I + A)^3 - 7A$ is equal to:
(A) $A$ (B) $I - A$ (C) $I$ (D) $3A$
Ans: (C). Expand $(I+A)^3 = I^3 + A^3 + 3I^2A + 3IA^2 = I + A + 3A + 3A = I + 7A$. So $(I+7A) - 7A = I$. (Note: $A^3 = A^2 \cdot A = A \cdot A = A$).

2022 (2 Marks): Find the value of $x - y$ if $2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$.
Ans: $2(1)+y=5 \Rightarrow y=3$. $2x+2=8 \Rightarrow 2x=6 \Rightarrow x=3$. So $x-y = 3-3=0$.

2020 (1 Mark): If $A$ is a skew-symmetric matrix of order 3, then the value of determinant $|A|$ is?
Ans: 0. (Determinant of odd order skew-symmetric matrix is always zero).

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: Assuming $AB = BA$. Never assume commutativity unless specified (like for diagonal matrices).

Mistake 2: In $(AB)'$, writing $A'B'$. The correct property is $(AB)' = B'A'$ (Reverse order).

Mistake 3: Forgetting that for skew-symmetric matrices, diagonal elements MUST be zero.

Scoring Tips

Tip 1: When multiplying matrices, write the order first ($m \times n$ and $n \times p$) to ensure compatibility and predict result order ($m \times p$).

Tip 2: For "Find A" questions involving equations, use pre-multiplication ($A^{-1}B$) or post-multiplication ($BA^{-1}$) correctly. Order matters!

Tip 3: In construction problems, double check the calculation for each element. It's easy marks but also easy to make a silly arithmetic error.

Practice Problems (Self-Assessment)

Level 1: Basic (1 Mark Each)

Q1. If a matrix has 24 elements, what are the possible orders it can have?

Step 1: Concept

Order is $m \times n$. Total elements $= mn = 24$.

We need pairs of natural numbers $(m, n)$ whose product is 24.

Step 2: List Pairs

$1 \times 24$, $24 \times 1$

$2 \times 12$, $12 \times 2$

$3 \times 8$, $8 \times 3$

$4 \times 6$, $6 \times 4$

Total 8 possible orders.

Q2. Construct a $2 \times 2$ matrix where $a_{ij} = |-2i + 3j|$.

Step 1: Calculate Elements

$i=1, j=1 \Rightarrow |-2(1)+3(1)| = |1| = 1$

$i=1, j=2 \Rightarrow |-2(1)+3(2)| = |4| = 4$

$i=2, j=1 \Rightarrow |-2(2)+3(1)| = |-1| = 1$

$i=2, j=2 \Rightarrow |-2(2)+3(2)| = |2| = 2$

Step 2: Write Matrix

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}$.

Level 2: Intermediate (2-3 Marks Each)

Q3. Find $X$ and $Y$, if $X+Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}$ and $X-Y = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$.

Step 1: Add Equations

$(X+Y) + (X-Y) = 2X$

$2X = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$

$X = \frac{1}{2}\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}$

Step 2: Subtract Equations

$(X+Y) - (X-Y) = 2Y$

$2Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}$

$Y = \frac{1}{2}\begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}$

Q4. If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 - 5A + 7I = O$.

Step 1: Find $A^2$

$A^2 = A \cdot A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$= \begin{bmatrix} (3)(3)+(1)(-1) & (3)(1)+(1)(2) \\ (-1)(3)+(2)(-1) & (-1)(1)+(2)(2) \end{bmatrix}$

$= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

Step 2: Substitute in Equation

LHS $= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - 5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$. (Proved)

Level 3: Advanced (4 Marks Each)

Q5. Find the matrix $X$ so that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.

Step 1: Determine Order of X

Let $X$ be $m \times n$.

$(m \times n) \times (2 \times 3) = (2 \times 3)$.

So $n=2$ (inner match) and $m=2$ (result rows).

Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.

Step 2: Multiply & Compare

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix}$

From first row: $a+4b = -7$ and $2a+5b = -8$.

Solving: Multiply first by 2: $2a+8b=-14$. Subtract from second: $-3b = 6 \Rightarrow b=-2$. Then $a=1$.

Step 3: Solve for c, d

$c+4d = 2$ and $2c+5d = 4$.

Multiply first by 2: $2c+8d=4$. Subtract: $-3d=0 \Rightarrow d=0$. Then $c=2$.

$\therefore X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$.

Overview