Chapter 4: Determinants - Board Exam Notes

Exam Weightage & Blueprint

Total: ~8-10 Marks

Determinants is a crucial chapter often combined with Matrices. The "Matrix Method" for solving linear equations is a guaranteed long-answer question in most board exams.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	Very High	Value of $\|kA\|$, Singular Matrix, Area of Triangle
Short Answer (2M)	2	Medium	Area of Triangle, Adjoint Properties
Short Answer (3M)	3	High	Properties of Determinants (if in syllabus), Inverse
Long Answer	5	Very High	Solving System of Linear Equations ($AX=B$)

Last 24-Hour Checklist

Expansion: Along any row or column (Sign pattern $+ - +$)
Scalar Property: $|kA| = k^n |A|$ (where $n$ is order)
Area of Triangle: $\frac{1}{2} |x_1(y_2-y_3) + \dots|$ using det
Singular Matrix: $|A| = 0$ (No Inverse)

Adjoint Property: $A(\text{adj } A) = |A|I$
Inverse Formula: $A^{-1} = \frac{1}{|A|} \text{adj } A$
Linear Equations: $X = A^{-1}B$ (Matrix Method)

Basic Concepts & Properties

1. Determinant of a Matrix

Only square matrices have determinants.
For $2 \times 2$: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.
For $3 \times 3$: Expand along any row/column using signs $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$.

2. Important Properties

Reflection Property

$|A| = |A'|$

(Value remains same if rows and columns are interchanged)

Scalar Multiplication

$|kA| = k^n |A|$

Where $n$ is the order of matrix $A$. (Most asked MCQ!)

Multiplication Property: $|AB| = |A||B|$.

3. Area of Triangle

Area of $\Delta$ with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is: $$\Delta = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$$ Note: Since area is positive, take absolute value of determinant. For collinear points, Area = 0.

Adjoint and Inverse

Minors and Cofactors

Minor ($M_{ij}$): Determinant obtained by deleting $i$-th row and $j$-th column.
Cofactor ($A_{ij}$): $A_{ij} = (-1)^{i+j} M_{ij}$.

Adjoint of a Matrix

$\text{adj } A = [A_{ij}]^T$ (Transpose of Cofactor Matrix).
Property: $A(\text{adj } A) = (\text{adj } A)A = |A|I$.

Inverse of a Matrix

A square matrix $A$ is invertible if and only if $A$ is non-singular ($|A| \neq 0$). $$A^{-1} = \frac{1}{|A|} (\text{adj } A)$$

Consistent System

If $|A| \neq 0$, unique solution exists: $X = A^{-1}B$.

Inconsistent System

If $|A| = 0$ and $(\text{adj } A)B \neq O$, no solution.

Solved Examples (Board Marking Scheme)

Q1. If $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, then show that $|2A| = 4|A|$. (2 Marks)

Step 1: Calculate LHS 1 Mark

$2A = 2\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$.

$|2A| = (2)(4) - (4)(8) = 8 - 32 = -24$.

Step 2: Calculate RHS 1 Mark

$|A| = (1)(2) - (2)(4) = 2 - 8 = -6$.

$4|A| = 4(-6) = -24$.

Hence, LHS = RHS.

Q2. Find the area of the triangle with vertices $(2, 7), (1, 1), (10, 8)$. (2 Marks)

Step 1: Setup Determinant 0.5 Mark

$\Delta = \frac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}$

Step 2: Expand 1 Mark

$= \frac{1}{2} [ 2(1-8) - 7(1-10) + 1(8-10) ]$

$= \frac{1}{2} [ 2(-7) - 7(-9) + 1(-2) ]$

$= \frac{1}{2} [ -14 + 63 - 2 ] = \frac{1}{2} [ 47 ] = 23.5$.

Step 3: Final Answer 0.5 Mark

Area = 23.5 sq units.

Q3. Solve the system of equations: $2x + 5y = 1$, $3x + 2y = 7$ using matrix method. (5 Marks)

Step 1: Write $AX=B$ 1 Mark

$A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}$.

Step 2: Find $|A|$ 1 Mark

$|A| = 4 - 15 = -11 \neq 0$. Unique solution exists.

Step 3: Find adj A 1 Mark

Cofactors: $A_{11}=2, A_{12}=-3, A_{21}=-5, A_{22}=2$.

$\text{adj } A = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$.

Step 4: Find $X = A^{-1}B$ 2 Marks

$X = \frac{1}{-11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix}$

$= \frac{1}{-11} \begin{bmatrix} 2 - 35 \\ -3 + 14 \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$.

So, $x = 3, y = -1$.

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): If $A$ is a square matrix of order 3 and $|A| = 5$, then the value of $|2A'|$ is:
(A) -10 (B) 10 (C) -40 (D) 40
Ans: (D). $|2A'| = |2A|$ (since $|A'|=|A|$). Using $|kA| = k^n|A|$, $|2A| = 2^3 |A| = 8 \times 5 = 40$.

2022 (1 Mark): If for matrix $A$, $|A| = 3$, find $|A (\text{adj } A)|$.
Ans: We know $A(\text{adj } A) = |A|I$. So $|A(\text{adj } A)| = ||A|I|$. Since $|A|I$ is a scalar matrix with diagonal elements $|A|$, its determinant is $|A|^n$. If order is not given (assume 3), answer is $3^3 = 27$. If order 2, $3^2=9$.

2020 (4 Marks): Using properties of determinants, prove that $\begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} = a^3$.
Hint: Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$. Then expand.

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: Confusing matrix notation $[ ]$ with determinant $| |$. They are different!

Mistake 2: In $|kA|$, writing $k|A|$. Correct is $k^n |A|$.

Mistake 3: Forgetting the sign convention ($+ - +$) while expanding, especially for the middle term.

Scoring Tips

Tip 1: For "Find $A^{-1}$" questions, always check $|A|$ first. If 0, stop and write "Inverse does not exist".

Tip 2: In Area of Triangle problems, if area is given (e.g., 4 units), use $\pm 4$ in calculation to find unknown coordinates.

Tip 3: For long answer linear equations, verify your $x, y, z$ values by substituting them back into one of the original equations.

Practice Problems (Self-Assessment)

Level 1: Basic (1 Mark Each)

Q1. Evaluate $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$.

Answer: $\cos^2\theta - (-\sin^2\theta) = \cos^2\theta + \sin^2\theta = 1$.

Q2. Find $x$ if $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$.

Answer: LHS: $2-20 = -18$. RHS: $2x^2 - 24$. So $2x^2 - 24 = -18 \Rightarrow 2x^2 = 6 \Rightarrow x^2 = 3 \Rightarrow x = \pm\sqrt{3}$.

Level 2: Intermediate (2-3 Marks Each)

Q3. Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants.

Hint: Let $P(x, y)$ be on the line. Area of $\Delta ABP = 0$.
$\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0 \Rightarrow 1(3x - y) = 0 \Rightarrow y = 3x$.

Q4. If $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$, verify that $(AB)^{-1} = B^{-1}A^{-1}$.

Hint: Calculate $AB$, then its inverse. Calculate $B^{-1}$ and $A^{-1}$ separately and multiply. Both should match.

Level 3: Advanced (5 Marks Each)

Q5. Solve the following system of equations by matrix method:
$3x - 2y + 3z = 8$
$2x + y - z = 1$
$4x - 3y + 2z = 4$

Answer: $x = 1, y = 2, z = 3$.
Steps: Find $|A| = -17$. Find adj A. Then $X = A^{-1}B$.