Chapter 3: Matrices - Board Exam Notes

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

Column & Row

Column: Only 1 column ($m \times 1$)

Row: Only 1 row ($1 \times n$)

Square Matrix

Rows = Columns ($m = n$)

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

Diagonal Matrix

Non-diagonal elements are zero.

$A = \text{diag}(d_1, d_2, \dots)$

Scalar Matrix

Diagonal matrix with equal diagonal elements.

$a_{ij} = k$ for $i=j$, else $0$.

Identity Matrix ($I$)

Scalar matrix with diagonal elements = 1.

$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Zero Matrix ($O$)

All elements are zero.

Denoted by $O$.

Operations on Matrices

1. Addition & Subtraction

Defined only if matrices have the same order.
Add/Subtract corresponding elements: $C_{ij} = A_{ij} \pm B_{ij}$.

2. Multiplication of Matrices

Product $AB$ is defined if:
Columns of A = Rows of B

If $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$.

Properties of Multiplication:
• Not Commutative: Generally $AB \neq BA$.
• Associative: $(AB)C = A(BC)$.
• Distributive: $A(B+C) = AB + AC$.
• Identity: $IA = AI = A$.

3. Transpose of a Matrix ($A'$ or $A^T$)

Obtained by interchanging rows and columns.
Properties:
1. $(A')' = A$
2. $(kA)' = kA'$
3. $(A+B)' = A' + B'$
4. $(AB)' = B'A'$ (Very Important!)

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.

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?

Answer: $1\times24, 2\times12, 3\times8, 4\times6, 6\times4, 8\times3, 12\times2, 24\times1$. (Total 8).

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

Answer: $\begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}$.

Level 2: Intermediate (2-3 Marks Each)

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

Hint: Add equations: $2X = (X+Y) + (X-Y)$. Subtract: $2Y = (X+Y) - (X-Y)$.
Ans: $X = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}, Y = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}$.

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

Hint: Calculate $A^2 = A \cdot A$. Then substitute in expression.

Level 3: Advanced (4 Marks Each)

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

Hint: Let $X$ be of order $2 \times 2$ (since result is $2 \times 3$ and second matrix is $2 \times 3$). Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Multiply and compare elements to solve for $a,b,c,d$.
Ans: $X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$.