2026?27 Syllabus Focus
The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.
- Notation, order, equality and matrix types, including zero, identity, symmetric and skew-symmetric matrices.
- Transpose and matrix operations: addition, multiplication and scalar multiplication; non-commutativity and zero products of non-zero 2 ? 2 matrices.
- Invertible matrices and uniqueness of the inverse for real entries.
Last 24-Hour Checklist
Types of Matrices (The Cast)
Rows ($m$): Horizontal lines (Side to side).
Columns ($n$): Vertical lines (Front to back).
Order is always Row $×$ Column (RC Cola).
Column Matrix
Analogy: "The Pole" (Standing)
Only 1 column ($m × 1$).
$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$
Row Matrix
Analogy: "The Snake" (Sleeping)
Only 1 row ($1 × n$).
$\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$
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)
$(2 × 2)$ can only add to $(2 × 2)$.
Action: Element 1 adds to Element 1. ($a_{11} + b_{11}$).
$\begin{bmatrix} 2 & 5 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 6 & 2 \end{bmatrix}$
$= \begin{bmatrix} 2+1 & 5+0 \\ 3+6 & 4+2 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 9 & 6 \end{bmatrix}$
2. Multiplication (The Diving Board Rule)
Check Inner Numbers: $(m × \mathbf{n})$ and $(\mathbf{n} × p)$.
If they match, YES! The result will be Outer Numbers ($m × p$).
- 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..."
Row 1 $(1, 2)$ dives into Col 1 $(3, 4)$:
$AB = [(1 × 3) + (2 × 4)] = [3 + 8] = [11]$.
3. Transpose (Stand Up!)
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
Orthogonal Matrix
Key Property: If $A$ is orthogonal, then $A^T = A^{-1}$ and $|A| = \pm 1$.
- Multiplication Non-Commutativity: In general, $AB \neq BA$. Even if both $AB$ and $BA$ exist, they may not have the same order or equal entries.
- Zero Divisor Property: If $AB = O$, it does NOT imply $A = O$ or $B = O$. Product of two non-zero matrices can be a zero matrix.
- Transpose Reversal Law:
• $(AB)^T = B^T A^T$
• $(ABC)^T = C^T B^T A^T$
• $(kA)^T = k A^T$ - Skew-Symmetric Matrix Core Properties:
• $A^T = -A \iff a_{ji} = -a_{ij}$.
• Diagonal Elements are Always ZERO: For diagonal elements $i = j \implies a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies a_{ii} = 0$.
• Determinant of any odd-order skew-symmetric matrix is always $0$. - Unique Sum Decomposition Theorem: Any square matrix $A$ can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix:
$$A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$$ where $\frac{1}{2}(A + A^T)$ is symmetric and $\frac{1}{2}(A - A^T)$ is skew-symmetric. - Product with Transpose: For any square matrix $A$, both $A A^T$ and $A^T A$ are always symmetric matrices.
- Total Number of Matrices Formula: The number of all possible matrices of order $m \times n$ with each entry $0$ or $1$ (or any $k$ choices) is $k^{mn}$. (For $3 \times 3$ with entries $0$ or $1$: $2^{3 \times 3} = 2^9 = 512$).
- 📌 Syllabus Note: Elementary row/column transformations to compute inverses (NCERT Ex 3.4) are removed from CBSE exams. Matrix inverses are calculated via the adjoint method in Chapter 4.
Solved Examples (Board Marking Scheme)
Q1. Construct a $2 × 2$ matrix $A = [a_{ij}]$ whose elements are given by $aij = \frac{(i+2j)^2}{2}$. (2 Marks)
$a_{11} = \frac{(1+2)^2}{2} = \frac{9}{2}$
$a_{12} = \frac{(1+4)^2}{2} = \frac{25}{2}$
$a_{21} = \frac{(2+2)^2}{2} = \frac{16}{2} = 8$
$a_{22} = \frac{(2+4)^2}{2} = \frac{36}{2} = 18$
$A = \begin{bmatrix} 9/2 & 25/2 \\ 8 & 18 \end{bmatrix}$
Q2. If $A = \begin{bmatrix} 3 & 5 \\ 2 & a \end{bmatrix}$ is a symmetric matrix, find $a$. (1 Mark)
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)
$A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$
$P = \frac{1}{2}(A + A') = \frac{1}{2} ( \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} ) = \frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$
Check: $P' = P$. Correct.
$Q = \frac{1}{2}(A - A') = \frac{1}{2} ( \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} ) = \frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$
Check: $Q' = -Q$. Correct.
$A = P + Q = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$
Previous Year Questions (PYQs)
(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: $A^3 = A^2 \cdot A = A \cdot A = A$).
Ans: $2(1)+y=5 \Rightarrow y=3$. $2x+2=8 \Rightarrow 2x=6 \Rightarrow x=3$. So $x-y = 3-3=0$.
Ans: 0. (Determinant of odd order skew-symmetric matrix is always zero).
Exam Strategy & Mistake Bank
Common Mistakes
Scoring Tips
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?
Order is $m × n$. Total elements $= mn = 24$.
We need pairs of natural numbers $(m, n)$ whose product is 24.
$1 × 24$, $24 × 1$
$2 × 12$, $12 × 2$
$3 × 8$, $8 × 3$
$4 × 6$, $6 × 4$
Total 8 possible orders.
Q2. Construct a $2 × 2$ matrix where $a_{ij} = |-2i + 3j|$.
$i=1, j=1 \Rightarrow |-2(1)+3(1)| = |1| = 1$
$i=1, j=2 \Rightarrow |-2(1)+3(2)| = |4| = 4$
$i=2, j=1 \Rightarrow |-2(2)+3(1)| = |-1| = 1$
$i=2, j=2 \Rightarrow |-2(2)+3(2)| = |2| = 2$
$A = \begin{bmatrix} a11 & a12 \\ a21 & a22 \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}$.
Level 2: Intermediate (2-3 Marks Each)
Q3. Find $X$ and $Y$, if $X+Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}$ and $X-Y = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$.
$(X+Y) + (X-Y) = 2X$
$2X = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$
$X = \frac{1}{2}\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}$
$(X+Y) - (X-Y) = 2Y$
$2Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}$
$Y = \frac{1}{2}\begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}$
Q4. If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 - 5A + 7I = O$.
$A^2 = A · A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$
$= \begin{bmatrix} (3)(3)+(1)(-1) & (3)(1)+(1)(2) \\ (-1)(3)+(2)(-1) & (-1)(1)+(2)(2) \end{bmatrix}$
$= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$
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 $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.
Let $X$ be $m × n$.
$(m × n) × (2 × 3) = (2 × 3)$.
So $n=2$ (inner match) and $m=2$ (result rows).
Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix}$
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 $a=1$.
$c+4d = 2$ and $2c+5d = 4$.
Multiply first by 2: $2c+8d=4$. Subtract: $-3d=0 \Rightarrow d=0$. Then $c=2$.
$\therefore X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$.
📝 10-Mark Self-Assessment Test
Time: 25 MinsSolve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.
(a) Skew-symmetric matrix (b) Symmetric matrix (c) Zero matrix (d) Identity matrix