Problem 44 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(a, b, c\) be such that \(b(a+c) \neq 0\) if [2009] \(\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|+\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0\) then the value of \(n\) is : (a) any even integer (b) any odd integer (c) any integer (d) zero

Step-by-Step Solution

Answer

The value of \( n \) is any even integer.

Step 1: Understand the Problem

We have two determinants added together and set equal to zero. We need to determine values of \( n \) that make this true, under the condition \( b(a+c) eq 0 \).

Step 2: Expand the Determinants

We need to expand the two determinants given in the problem. Let's start with the first determinant \( D_1 \):\[\begin{vmatrix} a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1 \end{vmatrix}\]

Step 3: Calculate a Single Term in First Determinant

For the first row and first column of the determinant \( D_1 \), calculate \\[ a \cdot \begin{vmatrix} b+1 & b-1 \ c-1 & c+1 \end{vmatrix} = a((b+1)(c+1) - (b-1)(c-1)) \]

Step 4: Expand First Determinant

Now we continue expanding the first determinant \( D_1 \) by calculating:* For row1, column2: \\[ -(a+1) \cdot \begin{vmatrix} -b & b-1 \ c & c+1 \end{vmatrix} \]* For row1, column3: \\[ +(a-1) \cdot \begin{vmatrix} -b & b+1 \ c & c-1 \end{vmatrix} \]

Step 5: Calculate the Second Determinant

Now expand the second determinant \( D_2 \):\[\begin{vmatrix} a+1 & b+1 & c-1 \ a-1 & b-1 & c+1 \ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^n c \end{vmatrix}\] Break it down in a similar way by expanding it along the first row.

Step 6: Add Both Determinants and Simplify

After expanding both determinants, add the results. The problem statement tells us they sum to zero. Simplify this resulting expression.

Step 7: Identify Value of n By Conditions

The conditions \( b(a+c) eq 0 \) simplify the problem further. After simplifying, inspect the resulting equation to determine for which \( n \) the sum simplifies based on powers of \((-1)^n \).

Key Concepts

Matrix Algebra

Matrix algebra is a branch of mathematics that deals with matrices and the operations that can be performed on them, such as addition, multiplication, and the determination of determinants. Matrices are essentially arrays of numbers or expressions arranged in rows and columns. They are powerful tools used in various fields, such as engineering, physics, computer graphics, and economics.

Understanding matrices and their properties is fundamental for solving linear equations and systems. The size of a matrix is defined by its dimensions (number of rows and columns), and knowing these dimensions is crucial when performing matrix operations.

In this exercise, we use matrix algebra to work with determinants, specifically focusing on their expansion. Recognizing how matrices can represent complex systems in a compact form is a first step in solving algebraic problems involving matrices. This sets the stage for determinant operations which help to determine unique solutions or dependencies within the matrix structure.

Determinant Expansion

Determinant expansion involves breaking down the determinant of a matrix into simpler parts to compute its value more tractably. For a 3x3 matrix, this process is often done using cofactor expansion, also known as Laplace's expansion.

In our problem, we start by expanding the first determinant along its first row. This involves calculating minor determinants for each element of the row, which are obtained by deleting the row and the column of that element.

• Calculate the determinant by focusing on one row or column.
• Multiply each element by the determinant of its minor.
• Consider the alternating signs (+, -, +) when multiplying by minors.

After calculating the determinant of the first matrix, we apply a similar technique to the second matrix. Both determinants are set to sum up to zero, implying important algebraic properties of these expressions.

Once expanded, the determinant's value can help determine the uniqueness or existence of solutions in a linear system represented by the matrix. Zero determinants indicate either infinite solutions or none, based on the context of the problem.

Integer Solutions in Algebra

Integer solutions in algebra pertain to solving equations or systems where the solutions are whole numbers. In this exercise, the solution for the variable \( n \) was asked for, implying that the sum of two determinants must equate to zero for specific integer values of \( n \).

To determine the integers, we consider the properties of powers of \((-1)^n\). This results in the simplification of the second matrix according to whether \( n \) is even or odd, influencing the sign of each element in that row.

• For \((-1)^n\), if \( n \) is even, the term equals 1.
• If \( n \) is odd, the term is -1.

Utilizing these properties, we can simplify the second determinant and find that only certain \( n \) values lead both terms to cancel each other out.

For this particular problem, simplifying leads to discovering which integer values \( n \) could take, grounded in how the alternating signs of powers affect the determinant's total value. Finding integer solutions in such scenarios can reveal symmetry or periodicity in algebraic structures.