ML Aggarwal Class 12 Maths Solutions Section A Chapter 01 Determinants

Chapter 1. Determinants

Introduction

In 1693, Leibnitz created determinants to quickly solve a system of linear equations. The two vertical line notation used today for determinants was introduced by Arthur Cayley in 1841. This chapter covers determinants, their key properties, and how to use them to solve systems of linear equations.

1.1 Determinant

Consider two homogeneous linear equations with two variables x and y:
a₁x + b₁y = 0 ...(1)
a₂x + b₂y = 0 From these equations, we can write:
- a₁/b₁ = y/x = - a₂/b₂ When we remove the variables x and y from system (1), we get:
- a₁/b₁ = - a₂/b₂, which means a₁b₂ - a₂b₁ = 0 This eliminant is written as:
\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = 0 ...(2) The left side of (2) is called a determinant of order 2 (or second order determinant), and a₁b₂ - a₂b₁ is its value. This leads to the definition: A determinant of order 2 is an arrangement of 2² = 4 numbers (or expressions) in square form along two horizontal lines called rows and two vertical lines called columns, enclosed within two vertical lines. Thus, \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} is called a determinant of order 2 and its value is a₁b₂ - a₂b₁. The numbers a₁, b₁, a₂, b₂ are called the elements of the determinant, and a₁b₂ - a₂b₁ is called its expansion. Example: \begin{vmatrix} 2 & -4 \\ 7 & 5 \end{vmatrix} = 2(5) - 7(-4) = 10 + 28 = 38 The elements a₁, b₁ make up the first row and a₂, b₂ make up the second row. The elements a₁, a₂ and b₁, b₂ form the first and second columns respectively. The elements a₁, b₂ lie on the principal diagonal (or simply the diagonal). A determinant of order 3 is an arrangement of 3² = 9 numbers (or expressions) in square form along three horizontal lines called rows and three vertical lines called columns, enclosed within two vertical lines. Thus, \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} is called a determinant of order 3. The numbers a₁, b₁, c₁ etc. are called elements. A determinant of order 3 contains 9 elements. The elements a₁, b₁, c₁; a₂, b₂, c₂; and a₃, b₃, c₃ form the first, second, and third rows respectively. The elements a₁, a₂, a₃; b₁, b₂, b₃; and c₁, c₂, c₃ form the first, second, and third columns respectively. The elements a₁, b₂, c₃ lie on the principal diagonal. A determinant is usually shown by the symbol Δ or D. Rows and columns are denoted by R₁, R₂, R₃, ... and C₁, C₂, C₃, ... respectively. **Value of a determinant of order 3** Let Δ = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}, then Δ = a₁\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b₁\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c₁\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} = a₁(b₂c₃ - b₃c₂) - b₁(a₂c₃ - a₃c₂) + c₁(a₂b₃ - a₃b₂) = a₁b₂c₃ - a₁b₃c₂ - a₂b₁c₃ + a₃b₁c₂ + a₂b₃c₁ - a₃b₂c₁ ...(1) The expression on the right side of (1) is called the expansion of the determinant by the first row. **Working rule:** (i) Write the elements of the first row with alternately positive and negative signs, with the first element always positive. (ii) Multiply each signed element by the determinant of second order obtained by deleting the row and column in which that element appears. Example: \begin{vmatrix} 3 & -2 & 5 \\ 1 & 2 & -1 \\ 0 & 4 & 7 \end{vmatrix} = 3\begin{vmatrix} 2 & -1 \\ 4 & 7 \end{vmatrix} - (-2)\begin{vmatrix} 1 & -1 \\ 0 & 7 \end{vmatrix} + 5\begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} = 3(14 + 4) + 2(7 - 0) + 5(4 - 0) = 3(18) + 2(7) + 5(4) = 54 + 14 + 20 = 88 **Determinant of order one** Let a be any number (or expression), then |a| is a determinant of order one and its value is a itself, so |a| = a. Examples: |5| = 5, |-7| = -7 **Remark:** A determinant of order one should not be confused with the absolute value of a real or complex number. **Determinants of order four and higher** Let Δ = \begin{vmatrix} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \\ a_4 & b_4 & c_4 & d_4 \end{vmatrix}, then Δ is a determinant of order 4. It has 4² = 16 elements arranged in 4 rows and 4 columns, and its value is found in a way similar to a determinant of order 3. Similarly, determinants of order 5 and higher can be defined. However, this chapter mainly deals with determinants of order ≤ 3. **1.1.1 Minors and Cofactors** Let Δ = \begin{vmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn} \end{vmatrix} be a determinant of order n (n ≥ 2). The determinant of order n - 1 obtained by removing the ith row and jth column is called the minor of element a_{ij}. It is usually denoted as M_{ij}. If M_{ij} is the minor of element a_{ij} in determinant Δ, then the number (-1)^{i+j} M_{ij} is called the cofactor of element a_{ij}, usually denoted as A_{ij}. Thus, A_{ij} = (-1)^{i+j} M_{ij}. Note: A_{ij} = M_{ij} if i + j is even, and A_{ij} = -M_{ij} if i + j is odd. **Examples:** (1) Let Δ = \begin{vmatrix} 2 & -3 \\ 4 & 7 \end{vmatrix}, then M₁₁ = |7| = 7, M₁₂ = |4| = 4, M₂₁ = |-3| = -3, M₂₂ = |2| = 2 A₁₁ = (-1)^{1+1} M₁₁ = 7, A₁₂ = (-1)^{1+2} M₁₂ = -4, A₂₁ = (-1)^{2+1} M₂₁ = -(-3) = 3, A₂₂ = (-1)^{2+2} M₂₂ = 2 (2) Let Δ = \begin{vmatrix} 7 & 4 & -1 \\ 2 & 3 & 0 \\ 1 & -5 & 2 \end{vmatrix}, then M₁₁ = \begin{vmatrix} 3 & 0 \\ -5 & 2 \end{vmatrix} = 6, M₂₂ = \begin{vmatrix} 7 & -1 \\ 1 & 2 \end{vmatrix} = 15, M₃₂ = \begin{vmatrix} 7 & -1 \\ 2 & 0 \end{vmatrix} = 2 A₁₁ = (-1)^{1+1} M₁₁ = 6, A₂₂ = (-1)^{2+2} M₂₂ = 15, A₃₂ = (-1)^{3+2} M₃₂ = -2 For quick reference, the sign patterns for cofactors in order 2 and 3 determinants are: Order 2: \begin{vmatrix} + & - \\ - & + \end{vmatrix} Order 3: \begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix} **Expansion of a determinant by any row or any column** Let Δ = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} be a determinant of order 3. The cofactors are: A₁₁ = (-1)^{1+1} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}, A₁₂ = (-1)^{1+2} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}, A₁₃ = (-1)^{1+3} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}, A₂₁ = (-1)^{2+1} \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix}, A₂₂ = (-1)^{2+2} \begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix}, A₂₃ = (-1)^{2+3} \begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{vmatrix}, A₃₁ = (-1)^{3+1} \begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix}, A₃₂ = (-1)^{3+2} \begin{vmatrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{vmatrix}, A₃₃ = (-1)^{3+3} \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} The value of Δ is given by: Δ = a₁₁\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a₁₂\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a₁₃\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} (Expansion by first row) = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ Similarly, we can show that: Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃, Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁, etc. Thus: The sum of the products of elements of any row (or column) of a determinant with their corresponding cofactors equals the value of the determinant. This result holds for every determinant of order ≥ 2. The value of a determinant can be found by expanding it along any row or column. **Remark:** A determinant can be evaluated very quickly by expanding along a row or column containing the most zeros.

Illustrative Examples

Example 1. Evaluate the following determinants:
(i) \( \begin{vmatrix} -3 & 1 \\ 5 & 6 \end{vmatrix} \) (ii) \( \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \) (iii) \( \begin{vmatrix} \cos 15° & \sin 15° \\ \sin 75° & \cos 75° \end{vmatrix} \)
Solution:
(i) \( \begin{vmatrix} -3 & 1 \\ 5 & 6 \end{vmatrix} = (-3)(6) - (5)(1) = -18 - 5 = -23 \)

(ii) \( \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} = \cos \theta \cdot \cos \theta - \sin \theta \cdot (-\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1 \)

(iii) \( \begin{vmatrix} \cos 15° & \sin 15° \\ \sin 75° & \cos 75° \end{vmatrix} = \cos 15° \cos 75° - \sin 15° \sin 75° = \cos(15° + 75°) = \cos 90° = 0 \)

Example 2. If \( \begin{vmatrix} x - 2 & -3 \\ 3x & 2x \end{vmatrix} = 3 \), find the integral value of x.
Solution:
Given \( \begin{vmatrix} x - 2 & -3 \\ 3x & 2x \end{vmatrix} = 3 \)

\( \implies (x - 2)(2x) - (3x)(-3) = 3 \)

\( \implies 2x^2 - 4x + 9x - 3 = 0 \implies 2x^2 + 5x - 3 = 0 \)

\( \implies (2x - 1)(x + 3) = 0 \implies 2x - 1 = 0 \text{ or } x + 3 = 0 \)

\( \implies x = \frac{1}{2} \text{ or } -3 \text{ but x is an integer} \)

\( \implies x = -3 \)

Example 3. Evaluate \( \begin{vmatrix} 2 & 3 & -5 \\ 7 & 1 & -2 \\ -3 & 4 & 1 \end{vmatrix} \).
Solution:
\( \begin{vmatrix} 2 & 3 & -5 \\ 7 & 1 & -2 \\ -3 & 4 & 1 \end{vmatrix} = 2\begin{vmatrix} 1 & -2 \\ 4 & 1 \end{vmatrix} - 3\begin{vmatrix} 7 & -2 \\ -3 & 1 \end{vmatrix} - 5\begin{vmatrix} 7 & 1 \\ -3 & 4 \end{vmatrix} \)

\( = 2(1 - (-8)) - 3(7 - 6) - 5(28 - (-3)) \)

\( = 2(9) - 3(1) - 5(31) = 18 - 3 - 155 = -140 \)

Example 4. Evaluate \( \begin{vmatrix} 3 & 7 & 13 \\ -5 & 0 & 0 \\ 0 & 11 & -2 \end{vmatrix} \).
Solution:
Since the second row contains two zeros, expanding by the 2nd row:

\( \begin{vmatrix} 3 & 7 & 13 \\ -5 & 0 & 0 \\ 0 & 11 & -2 \end{vmatrix} = -(-5)\begin{vmatrix} 7 & 13 \\ 11 & -2 \end{vmatrix} = 5(-14 - 143) = 5(-157) = -785 \)

Example 5. Show that the value of the determinant \( \begin{vmatrix} 0 & \tan x & 1 \\ 1 & -\sec x & 0 \\ \sec x & 0 & \tan x \end{vmatrix} \) is independent of x.
Solution:
Expanding by the first row:

\( \begin{vmatrix} 0 & \tan x & 1 \\ 1 & -\sec x & 0 \\ \sec x & 0 & \tan x \end{vmatrix} = 0(-\sec x \tan x - 0) - \tan x(\tan x - 0) + 1(0 + \sec^2 x) \)

\( = 0 - \tan^2 x + \sec^2 x = \sec^2 x - \tan^2 x = 1 \)

The result is 1, which does not depend on x.

Example 6. Find the minors and cofactors of each element of the second column of the determinant Δ and hence find the value of the determinant Δ where Δ = \( \begin{vmatrix} 3 & -2 & 1 \\ 4 & 6 & 5 \\ 2 & -1 & 7 \end{vmatrix} \).
Solution:
\( M_{12} = \begin{vmatrix} 4 & 5 \\ 2 & 7 \end{vmatrix} = 28 - 10 = 18, \quad M_{22} = \begin{vmatrix} 3 & 1 \\ 2 & 7 \end{vmatrix} = 21 - 2 = 19 \)

\( M_{32} = \begin{vmatrix} 3 & 1 \\ 4 & 5 \end{vmatrix} = 15 - 4 = 11 \)

\( A_{12} = (-1)^{1+2} M_{12} = (-1)(18) = -18 \)

\( A_{22} = (-1)^{2+2} M_{22} = (1)(19) = 19 \)

\( A_{32} = (-1)^{3+2} M_{32} = (-1)(11) = -11 \)

Expanding by the 2nd column:

\( \Delta = (-2)(-18) + 6(19) + (-1)(-11) = 36 + 114 + 11 = 161 \)

Example 7. There are two values of x which make determinant \( \Delta = \begin{vmatrix} 1 & -2 & 5 \\ 2 & x & -1 \\ 0 & 4 & 2x \end{vmatrix} = 86 \). Find the sum of these numbers.
Solution:
Given \( \Delta = \begin{vmatrix} 1 & -2 & 5 \\ 2 & x & -1 \\ 0 & 4 & 2x \end{vmatrix} = 86 \)

Expanding by C₁:

\( 1\begin{vmatrix} x & -1 \\ 4 & 2x \end{vmatrix} - 2\begin{vmatrix} -2 & 5 \\ 4 & 2x \end{vmatrix} + 0 = 86 \)

\( 1(2x^2 + 4) - 2(-4x - 20) = 86 \)

\( 2x^2 + 4 + 8x + 40 = 86 \implies 2x^2 + 8x - 42 = 0 \implies x^2 + 4x - 21 = 0 \)

If α and β are the roots, then \( \alpha + \beta = -\frac{4}{1} = -4 \)

Hence, the sum of the two values of x = -4.

1.2 Properties of Determinants

The properties of determinants are useful tools for calculating determinant values. Proofs of most properties are beyond the scope of this book, so we state these properties and verify them using examples.

Property 1. If each element in a row or in a column of a determinant is zero, then the value of the determinant is zero.Verification: Let Δ = \( \begin{vmatrix} a_1 & b_1 & c_1 \\ 0 & 0 & 0 \\ a_3 & b_3 & c_3 \end{vmatrix} \) be a determinant where each element in the second row is zero. Expanding Δ by the second row gives Δ = 0.Property 2. If each element on one side of the principal diagonal of a determinant is zero, then the value of the determinant is the product of the diagonal elements.Verification: Let Δ = \( \begin{vmatrix} a_1 & b_1 & c_1 \\ 0 & b_2 & c_2 \\ 0 & 0 & c_3 \end{vmatrix} \) be a determinant where all elements on one side of the principal diagonal are zero. Expanding Δ by C₁ gives Δ = a₁b₂c₃.Property 3. The value of a determinant remains unchanged if its rows and columns are interchanged.Verification: Let Δ = \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \) and Δ₁ = \( \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \) be the determinant obtained by interchanging rows and columns. Expanding Δ by R₁: Δ = a₁(b₂c₃ - b₃c₂) - b₁(a₂c₃ - a₃c₂) + c₁(a₂b₃ - a₃b₂) Expanding Δ₁ by R₁: Δ₁ = a₁(b₂c₃ - c₂b₃) - a₂(b₁c₃ - c₁b₃) + a₃(b₁c₂ - c₁b₂) Both give the same result, so Δ = Δ₁.Property 4. If any two rows (or columns) of a determinant are interchanged, then the value of the determinant changes by a minus sign only.Verification: Let Δ = \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \) and Δ₁ = \( \begin{vmatrix} c_1 & b_1 & a_1 \\ c_2 & b_2 & a_2 \\ c_3 & b_3 & a_3 \end{vmatrix} \) be the determinant obtained by interchanging the first and third columns. Through expansion, we find Δ₁ = -Δ.Corollary. If any row (or column) of a determinant Δ is passed over m rows (or columns), then the resulting determinant Δ₁ = (-1)^m Δ.Property 5. If two parallel lines (rows or columns) of a determinant are identical, then the value of the determinant is zero.Verification: Let Δ = \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_1 & b_1 & c_1 \end{vmatrix} \) be a determinant where the first and third rows are identical. Expanding by R₁: Δ = a₁(b₂c₁ - b₁c₂) - b₁(a₂c₁ - a₁c₂) + c₁(a₂b₁ - a₁b₂) = 0Property 6. If each element of a row (or a column) of a determinant is multiplied by the same number k, then the value of the new determinant is k times the value of the original determinant.Verification: Let Δ = \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \) and Δ₁ = \( \begin{vmatrix} a_1 & b_1 & c_1 \\ ka_2 & kb_2 & kc_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \) be the determinant obtained by multiplying the second row by k. Expanding Δ₁ by R₁ gives Δ₁ = kΔ.Corollary 1. If two parallel lines (rows or columns) of a determinant have elements that are equi-multiples of each other, then the value of the determinant is zero. (Using Properties 6 and 5)Corollary 2. If each element of a determinant Δ is multiplied by k and Δ₁ is the new determinant, then: - Δ₁ = kΔ if order of Δ = 1 - Δ₁ = k²Δ if order of Δ = 2 - Δ₁ = k³Δ if order of Δ = 3, etc.Property 7. If each element of a row (or a column) of a determinant consists of sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants whose other rows (or columns) are not altered.Verification: Let Δ = \( \begin{vmatrix} a_1 + d_1 & b_1 & c_1 \\ a_2 + d_2 & b_2 & c_2 \\ a_3 + d_3 & b_3 & c_3 \end{vmatrix} \) where each element in the first column is a sum of two terms. Expanding Δ by C₁: Δ = \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix} \)Property 8. If to each element of a row (or a column) of a determinant be added the equimultiples of the corresponding elements of one or more rows (or columns), the value of the determinant remains unchanged.Verification: Let Δ = \( \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \) and Δ₁ = \( \begin{vmatrix} a_1 + ka_2 & a_2 & a_3 \\ b_1 + kb_2 & b_2 & b_3 \\ c_1 + kc_2 & c_2 & c_3 \end{vmatrix} \) be obtained by adding k times the second column to the first. Using Property 7: Δ₁ = Δ + k(0) = Δ.Property 9. The sum of the products of elements of any row (or column) with the cofactors of the corresponding elements of some other row (or column) is zero.Verification: Let Δ = \( \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \). The sum of products of elements of the first row with cofactors of the third row equals zero: a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃ = 0.

1.2.1 Elementary Operations

Let Δ be a determinant of order n (n ≥ 2). The rows and columns are denoted as R₁, R₂, R₃, ... and C₁, C₂, C₃, ... respectively. (i) Interchanging the ith row and jth row is denoted by Rᵢ ↔ Rⱼ. Interchanging the ith column and jth column is denoted by Cᵢ ↔ Cⱼ. (ii) Multiplying each element of the ith row by k is denoted by Rᵢ → kRᵢ. Multiplying each element of the ith column by k is denoted by Cᵢ → kCᵢ. (iii) Adding k times the jth row (j ≠ i) to the ith row is denoted by Rᵢ → Rᵢ + kRⱼ. Adding k times the jth column (j ≠ i) to the ith column is denoted by Cᵢ → Cᵢ + kCⱼ.

Illustrative Examples (Properties and Elementary Operations)

Example 1. Without expanding, evaluate the following determinants:
(i) \( \begin{vmatrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{vmatrix} \) (ii) \( \begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{vmatrix} \)
Solution:
(i) Operating C₁ → C₁ - 8C₃ (using Property 8):

\( \begin{vmatrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{vmatrix} = \begin{vmatrix} 49 - 48 & 1 & 6 \\ 39 - 32 & 7 & 4 \\ 26 - 24 & 2 & 3 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 6 \\ 7 & 7 & 4 \\ 2 & 2 & 3 \end{vmatrix} = 0 \) (by Property 5)

(ii) Operating C₃ → C₃ + C₂ (using Property 8):

\( \begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{vmatrix} = \begin{vmatrix} 1 & a & a + b + c \\ 1 & b & a + b + c \\ 1 & c & a + b + c \end{vmatrix} = (a + b + c) \begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix} = (a + b + c) \times 0 = 0 \) (by Property 6 and Property 5)

Example 2. Without expanding, show that:
(i) \( \begin{vmatrix} b - c & c - a & a - b \\ c - a & a - b & b - c \\ a - b & b - c & c - a \end{vmatrix} = 0 \) (ii) \( \begin{vmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{vmatrix} = 0 \)
Solution:
(i) Operating C₁ → C₁ + C₂ + C₃ (using Property 8):

\( \begin{vmatrix} b - c & c - a & a - b \\ c - a & a - b & b - c \\ a - b & b - c & c - a \end{vmatrix} = \begin{vmatrix} 0 & c - a & a - b \\ 0 & a - b & b - c \\ 0 & b - c & c - a \end{vmatrix} = 0 \) (by Property 1)

(ii) Taking out (-1) from C₁, (-1) from C₂, and (-1) from C₃ (using Property 6):

Δ = (-1)³ \( \begin{vmatrix} 0 & -x & -y \\ x & 0 & -z \\ y & z & 0 \end{vmatrix} = - \) \( \begin{vmatrix} 0 & -x & -y \\ x & 0 & -z \\ y & z & 0 \end{vmatrix} \)

Since interchanging rows and columns (using Property 3) gives -Δ, we have 2Δ = 0, so Δ = 0.

Example 3. Without expanding, show that:
(i) \( \begin{vmatrix} b^2c^2 & bc & b + c \\ c^2a^2 & ca & c + a \\ a^2b^2 & ab & a + b \end{vmatrix} = 0 \) (ii) \( \begin{vmatrix} 1 & a & a^2 - bc \\ 1 & b & b^2 - ca \\ 1 & c & c^2 - ab \end{vmatrix} = 0 \)
Solution:
(i) Multiply R₁ by a, R₂ by b, R₃ by c (using Property 6), then use column operations to show the determinant equals zero after several steps.

(ii) Using Property 7 (decomposing the third column) and then using Properties 5 and 6 shows this determinant also equals zero.

Example 4. Without expanding, show that \( \begin{vmatrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = 0 \), where a, b, c are in A.P.
Solution:
Since a, b, c are in A.P., we have a + c = 2b, so a + c - 2b = 0. Operating R₁ → R₁ + R₃ - 2R₂:

\( \begin{vmatrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = \begin{vmatrix} 0 & 0 & a + c - 2b \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = \begin{vmatrix} 0 & 0 & 0 \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = 0 \) (by Property 1)

Example 5. By using properties of determinants, prove that the determinant \( \begin{vmatrix} a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a \end{vmatrix} \) is independent of x. (I.S.C. 2010)
Solution:
Operating R₁ → R₁ - cos x R₂ and R₃ → R₃ - a R₂:

After performing these operations and expanding, we get the result -a³, which does not depend on x.