Determinants – Chapter 4 | AAI ATC Exam Prep

📌 Why Determinants for AAI ATC?

Chapter 4 – Determinants is one of the highest-weightage topics in the AAI ATC written exam. Questions appear on evaluating 2×2 and 3×3 determinants, minors & cofactors, area of triangles, adjoint, inverse matrices, and solving 3-variable systems using the matrix method. Mastering this chapter is essential for a high score.

"All Mathematical truths are relative and conditional." — C.P. Steinmetz

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4.1 – What is a Determinant?

To every square matrix A = [a_ij] of order n, we associate a number (real or complex) called the determinant of A, denoted by |A| or det(A) or Δ.

Only square matrices have determinants. |A| is read as "determinant of A", NOT "modulus of A".

Order 1 × 1

If A = [a], then |A| = a

Order 2 × 2


|A| = |a₁₁  a₁₂| = a₁₁·a₂₂ − a₂₁·a₁₂

      |a₂₁  a₂₂|


Example: |2   4| = 2(2) − 4(−1) = 4 + 4 = 8
         |−1  2|

Order 3 × 3 – Expansion along R₁

|A| = a₁₁(a₂₂a₃₃ − a₃₂a₂₃) − a₁₂(a₂₁a₃₃ − a₃₁a₂₃) + a₁₃(a₂₁a₃₂ − a₃₁a₂₂)

Expanding along any row OR column always gives the same value of the determinant.

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[ Image Space ]
Add your diagram showing step-by-step expansion of 3×3 determinant along R₁

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Tip for AAI ATC: Always expand along the row or column with the maximum number of zeros — it saves calculation time!

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If A = kB (square matrices of order n), then |A| = kⁿ|B|. For a 3×3 matrix: |kA| = k³|A|.

🎯 Practice MCQs – Determinants (Evaluation)

1Evaluate: |2 4; −5 −1| = ?

A) −18

B) 18

C) −22

D) 22

|A| = 2(−1) − 4(−5) = −2 + 20 = 18. ✅ Answer: B

2If A = [[1,2],[4,2]], then |2A| = ?

A) −12

B) −24

C) −6

D) 24

|A| = 2−8 = −6. |2A| = 2²|A| = 4×(−6) = −24. ✅ Answer: B

3If |x 2; 18 x| = |6 2; 18 6|, then x = ?

A) 6

B) ±6

C) −6

D) 0

x² − 36 = 36 − 36 = 0 → x² − 36 = 0 → x² = 36 → x = ±6. ✅ Answer: B

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4.2 – Area of a Triangle Using Determinants

The area of a triangle with vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) is given by:

Δ = (1/2) |x₁ y₁ 1|
|x₂ y₂ 1|
|x₃ y₃ 1|

Always take the absolute value since area is positive.

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[ Image Space ]
Add a worked example diagram: triangle with vertices (3,8), (−4,2), (5,1) with area calculation

Area = positive

Always take |determinant| — area can never be negative.

Given area ±

If area is given, use both + and − values of determinant.

Collinear Points

Three points are collinear ⟺ Area of triangle = 0 ⟺ |det| = 0.

Equation of Line

Line through (x₁,y₁) and (x₂,y₂): set the 3×3 det with point (x,y) = 0.

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AAI ATC Tip: Collinearity questions are very common. Three points A, B, C are collinear if and only if the determinant formed by their coordinates = 0.

🎯 Practice MCQs – Area of Triangle

1Area of triangle with vertices (1,0), (6,0), (4,3) is:

A) 6 sq units

B) 7.5 sq units

C) 9 sq units

D) 5 sq units

Δ = (1/2)|1(0−3) − 0(6−4) + 1(18−0)| = (1/2)|−3 + 18| = (1/2)(15) = 7.5 sq units. ✅ Answer: B

2If area of triangle is 35 sq units with vertices (2,−6), (5,4), (k,4), then k = ?

A) 12

B) −2

C) −12, −2

D) 12, −2

Setting up the determinant and equating |det| = 70, we get two values: k = 12 or k = −2. ✅ Answer: D

3Which condition must hold for A(a, b+c), B(b, c+a), C(c, a+b) to be collinear?

A) a + b + c ≠ 0

B) They are always collinear

C) a = b = c

D) abc = 0

Setting up the 3×3 determinant, after R₂→R₂−R₁ and R₃→R₃−R₁, the determinant evaluates to 0 for all values. So they are always collinear. ✅ Answer: B

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4.3 – Minors and Cofactors

Minor M_ij: The determinant obtained by deleting the i^th row and j^th column of the matrix. Minor of order n matrix is a determinant of order (n−1).

Cofactor A_ij: Defined as A_ij = (−1)^i+j M_ij

Sign Pattern for Cofactors


Sign Pattern (−1)^(i+j):
+ − +
− + −
+ − +
(i+j even → +1, i+j odd → −1)

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[ Image Space ]
Add your diagram showing minor M₂₃ of a 3×3 matrix by crossing out R₂ and C₃

Expansion Using Cofactors

|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ (along R₁)

Sum of products of elements of ANY row/column with their cofactors = |A|

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Critical Rule: If elements of a row (or column) are multiplied with cofactors of a DIFFERENT row (or column), the sum = 0.
e.g. a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0

🎯 Practice MCQs – Minors and Cofactors

1In the determinant |1 2 3; 4 5 6; 7 8 9|, the minor M₂₃ = ?

A) 6

B) −6

C) 8

D) −8

M₂₃ = delete R₂ and C₃: |1 2; 7 8| = 8 − 14 = −6. ✅ Answer: B

2For the determinant |a_ij|₃ₓ₃, the value Δ is given by which of the following?

A) a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃

B) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁

C) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃

D) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Expansion along Column 1: Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁ ✅ Answer: D

3The cofactor A₁₂ of the element a₁₂ in matrix [[1,−2],[4,3]] is:

A) −4

B) 4

C) −2

D) 2

M₁₂ = 4. A₁₂ = (−1)^(1+2) × M₁₂ = (−1)³ × 4 = −4. ✅ Answer: A

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4.4 – Adjoint and Inverse of a Matrix

The adjoint of a square matrix A = [a_ij]_n×n is the transpose of the cofactor matrix [A_ij]_n×n. Denoted by adj A.

Steps to Find adj A

Find the cofactor A_ij for every element a_ij of A.

Form the cofactor matrix [A_ij].

Transpose the cofactor matrix → this is adj A.

Shortcut for 2×2 Matrix

If A = [[a, b],[c, d]], then adj A = [[d, −b],[−c, a]]

Swap diagonal elements, change signs of off-diagonal elements.

Key Theorems

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Theorem: A·(adj A) = (adj A)·A = |A|·I

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Inverse Formula: A⁻¹ = (1/|A|) · adj A, provided |A| ≠ 0

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|adj A| = |A|ⁿ⁻¹, where n is the order of A. For 3×3: |adj A| = |A|²

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[ Image Space ]
Add worked example: find adj A and A⁻¹ for a given 3×3 matrix with full steps

Singular vs Non-Singular

Type	Condition	Inverse Exists?
Singular Matrix	\|A\| = 0	❌ No inverse
Non-Singular Matrix	\|A\| ≠ 0	✅ Inverse exists

🎯 Practice MCQs – Adjoint and Inverse

1Let A be a nonsingular square matrix of order 3×3. Then |adj A| is equal to:

A) |A|

B) |A|²

C) |A|³

D) 3|A|

|adj A| = |A|^(n−1) = |A|^(3−1) = |A|². ✅ Answer: B

2If A is an invertible matrix of order 2, then det(A⁻¹) is equal to:

A) det(A)

B) 1/det(A)

C) 1

D) 0

Since A·A⁻¹ = I, taking determinants: |A|·|A⁻¹| = 1 → |A⁻¹| = 1/|A|. ✅ Answer: B

3Find adj A for A = [[2,3],[1,4]]:

A) [[4,−3],[−1,2]]

B) [[4,3],[1,2]]

C) [[2,−3],[−1,4]]

D) [[4,−3],[1,2]]

For 2×2: adj A = swap diagonals, change signs of off-diagonals = [[4,−3],[−1,2]]. ✅ Answer: A

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4.5 – Solving Linear Equations Using Matrix Method

A system of linear equations a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃ can be written as AX = B where:

A = coefficient matrix, X = [x,y,z]ᵀ, B = [d₁,d₂,d₃]ᵀ

Solution: X = A⁻¹B (when |A| ≠ 0)

Consistency Chart

\|A\|	(adj A)B	Nature
≠ 0	—	✅ Unique solution (Consistent)
= 0	≠ O	❌ No solution (Inconsistent)
= 0	= O	∞ Infinite solutions OR no solution

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[ Image Space ]
Add a solved 3-variable system: 3x−2y+3z=8, 2x+y−z=1, 4x−3y+2z=4 with full matrix solution

Step-by-Step: Matrix Method

Write the system as AX = B, identify coefficient matrix A.

Calculate |A|. If |A| ≠ 0, inverse exists → proceed.

Find cofactors → adj A → A⁻¹ = (1/|A|) adj A.

Compute X = A⁻¹B to get values of x, y, z.

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AAI ATC Exam Tip: In word problems, first convert the given information into 3 equations. Set up AX = B and solve using A⁻¹. The matrix method is faster and more reliable for 3-variable systems.

🎯 Practice MCQs – System of Equations (Matrix Method)

1For AX = B: if |A| = 0 and (adj A)B ≠ O, the system is:

A) Consistent with unique solution

B) Consistent with infinitely many solutions

C) Inconsistent (no solution)

D) Cannot be determined

When |A| = 0 and (adj A)B ≠ O, the system is inconsistent and has no solution. ✅ Answer: C

2Solve: 2x + 5y = 1, 3x + 2y = 7. What is x?

A) 1

B) 2

C) 3

D) 4

A = [[2,5],[3,2]], |A| = 4−15 = −11. X = A⁻¹B = −(1/11)[[2,−5],[−3,2]][1,7]ᵀ = [3,−1]ᵀ. x = 3. ✅ Answer: C

3For A = diag(x, y, z), the inverse A⁻¹ is:

A) diag(1/x, 1/y, 1/z)

B) xyz · diag(1/x, 1/y, 1/z)

C) (1/xyz) · diag(x,y,z)

D) diag(y,z,x)

For a diagonal matrix, the inverse is simply the diagonal matrix of reciprocals of each element. A⁻¹ = diag(1/x, 1/y, 1/z). ✅ Answer: A

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Quick Revision – Chapter 4 Summary

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|A| for 2×2: a₁₁a₂₂ − a₁₂a₂₁. For 3×3: expand along row/column with most zeros.

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|kA| = kⁿ|A| for n×n matrix. So |3A|₃ₓ₃ = 27|A|.

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Area of triangle = (1/2)|det|. Collinear points → det = 0.

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Minor M_ij: delete row i, col j. Cofactor A_ij = (−1)^(i+j) M_ij.

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adj A = transpose of cofactor matrix. A⁻¹ = (1/|A|) adj A.

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A(adj A) = |A|I. |adj A| = |A|^(n−1).

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Matrix method: AX = B → X = A⁻¹B (if |A| ≠ 0) = unique solution.

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|A| = 0 and (adj A)B ≠ O → inconsistent. |A| = 0 and (adj A)B = O → may or may not be consistent.

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