Chapter 4: Determinants

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 (The Value)

Only square matrices have determinants.
Denoted by: $|A|$ or $\det(A)$ or $\Delta$.
Note: It is NOT modulus. It can be negative.

2x2 Matrix (Crossfire)

$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = (ad) - (bc)$

Method: Down-Right minus Up-Right.

Example: $\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = (4) - (-4) = 8$.

3x3 Expansion (Traffic Lights)

Signs: $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$

Expand along Row 1: take element, apply sign, multiply by minor.

2. Important Properties (Cheat Sheet)

Reflection: $|A| = |A'|$ (Rows $\leftrightarrow$ Columns doesn't change value).
Scalar Power: $|kA|_{n \times n} = k^n |A|$ (Pulling $k$ out brings power $n$).
Multiplication: $|AB| = |A||B|$ (Break it apart).

3. Area of Triangle (Shoelace Lite)

$$\Delta = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$$
Key Points:
1. Always take Absolute Value (Area is +ve).
2. If Area is given as 4, use $\pm 4$ for calculation.
3. Collinear Points: Area = 0.

Adjoint and Inverse (The Magic Formula)

The Big Picture:
Inverse exists ONLY if Matrix is Non-Singular ($|A| \neq 0$).

1. Key Terms

Cofactor ($A_{ij}$)

Signed Minor.

Use "Traffic Control" signs ($+ - +$).

Adjoint (adj A)

Transpose of Cofactor Matrix.

Imagine flipping the cofactors over the diagonal.

2. The Formula

$$A^{-1} = \frac{1}{|A|} (\text{adj } A)$$

3. Properties of Adjoint

$A(\text{adj } A) = (\text{adj } A)A = |A|I$
$|\text{adj } A| = |A|^{n-1}$
$\text{adj}(\text{adj } A) = |A|^{n-2}A$
$|\text{adj}(\text{adj } A)| = |A|^{(n-1)^2}$
$\text{adj}(AB) = (\text{adj } B)(\text{adj } A)$ (Reversal Law)

Solving Equations ($AX=B$):
1. Find $|A|$. If $\neq 0$, proceed.
2. Find Inverse ($A^{-1}$).
3. Multiply: $X = A^{-1}B$.

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

Step 1: Cross Multiply

$(\cos\theta)(\cos\theta) - (\sin\theta)(-\sin\theta)$

Step 2: Simplify

$= \cos^2\theta - (-\sin^2\theta)$

$= \cos^2\theta + \sin^2\theta = 1$.

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

Step 1: Evaluate LHS

$|LHS| = (2)(1) - (4)(5) = 2 - 20 = -18$.

Step 2: Evaluate RHS

$|RHS| = (2x)(x) - (4)(6) = 2x^2 - 24$.

Step 3: Equate & Solve

$2x^2 - 24 = -18$

$2x^2 = 6 \Rightarrow x^2 = 3$

$x = \pm\sqrt{3}$.

Level 2: Intermediate (2-3 Marks Each)

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

Step 1: Condition for Collinearity

Let $P(x, y)$ be any point on the line. Area of $\Delta ABP$ must be 0.

$\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0$.

Step 2: Expand Determinant

Expand along $R_3$ (since it has two zeros):

$0(...) - 0(...) + 1[x(3) - y(1)] = 0$

$3x - y = 0 \Rightarrow y = 3x$.

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

Step 1: Find LHS $(AB)^{-1}$

$AB = \begin{bmatrix} 2(1)+3(-1) & 2(-2)+3(3) \\ 1(1)+(-4)(-1) & 1(-2)+(-4)(3) \end{bmatrix} = \begin{bmatrix} 2-3 & -4+9 \\ 1+4 & -2-12 \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}$.

$|AB| = (-1)(-14) - (5)(5) = 14 - 25 = -11$.

$(AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB) = \frac{1}{-11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}$.

Step 2: Find RHS $B^{-1}A^{-1}$

$|A| = (2)(-4) - (3)(1) = -8 - 3 = -11$.

$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix}$.

$|B| = (1)(3) - (-2)(-1) = 3 - 2 = 1$.

$B^{-1} = \frac{1}{|B|} \text{adj}(B) = \frac{1}{1} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$.

$B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} 3(-4)+2(-1) & 3(-3)+2(2) \\ 1(-4)+1(-1) & 1(-3)+1(2) \end{bmatrix}$

$= \frac{1}{-11} \begin{bmatrix} -12-2 & -9+4 \\ -4-1 & -3+2 \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}$.

Step 3: Conclusion

Since LHS = $\frac{1}{-11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}$ and RHS = $\frac{1}{-11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}$, we have LHS = RHS. Verified.

Level 3: Advanced (5 Marks Each)

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

Step 1: Write in Matrix Form $AX=B$

$A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$.

Step 2: Find Determinant of A

$|A| = 3(1 \cdot 2 - (-1) \cdot (-3)) - (-2)(2 \cdot 2 - (-1) \cdot 4) + 3(2 \cdot (-3) - 1 \cdot 4)$

$|A| = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4)$

$|A| = 3(-1) + 2(8) + 3(-10)$

$|A| = -3 + 16 - 30 = -17$.

Since $|A| = -17 \neq 0$, a unique solution exists.

Step 3: Find Adjoint of A

Cofactors:

$A_{11} = \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} = 2-3 = -1$

$A_{12} = -\begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = -(4+4) = -8$

$A_{13} = \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} = -6-4 = -10$

$A_{21} = -\begin{vmatrix} -2 & 3 \\ -3 & 2 \end{vmatrix} = -(-4+9) = -5$

$A_{22} = \begin{vmatrix} 3 & 3 \\ 4 & 2 \end{vmatrix} = 6-12 = -6$

$A_{23} = -\begin{vmatrix} 3 & -2 \\ 4 & -3 \end{vmatrix} = -(-9+8) = 1$

$A_{31} = \begin{vmatrix} -2 & 3 \\ 1 & -1 \end{vmatrix} = 2-3 = -1$

$A_{32} = -\begin{vmatrix} 3 & 3 \\ 2 & -1 \end{vmatrix} = -(-3-6) = 9$

$A_{33} = \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = 3+4 = 7$

Cofactor matrix $C = \begin{bmatrix} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7 \end{bmatrix}$.

$\text{adj } A = C^T = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$.

Step 4: Find Inverse of A

$A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$.

Step 5: Solve for X ($X=A^{-1}B$)

$X = \frac{1}{-17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$

$X = \frac{1}{-17} \begin{bmatrix} (-1)(8) + (-5)(1) + (-1)(4) \\ (-8)(8) + (-6)(1) + (9)(4) \\ (-10)(8) + (1)(1) + (7)(4) \end{bmatrix}$

$X = \frac{1}{-17} \begin{bmatrix} -8 - 5 - 4 \\ -64 - 6 + 36 \\ -80 + 1 + 28 \end{bmatrix}$

$X = \frac{1}{-17} \begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix}$

$X = \begin{bmatrix} \frac{-17}{-17} \\ \frac{-34}{-17} \\ \frac{-51}{-17} \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.

Final Answer: So, $x = 1, y = 2, z = 3$.

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