Problem 80 - Algebra for JEE ADVANCED

Question

The system of equations $$ \begin{aligned} &\alpha x-y-z=a-1 \\ &x-\alpha y-z=a-1 \\ &x-y-a z=\alpha-1 \end{aligned} $$ has no solution if \(\alpha\) is a. either \(-2\) or 1 b. \(-2\) c. 1 d. not \(-2\)

Step-by-Step Solution

Answer

\(\alpha\) is either \(-2\) or 1 for no solution (Option a).

Step 1: Arrange the System

We have the following system of equations:\1. \( \alpha x - y - z = a - 1 \)\2. \( x - \alpha y - z = a - 1 \)\3. \( x - y - a z = \alpha - 1 \)

Step 2: Express in Matrix Form

Write the given system of equations in a matrix form:\[\begin{bmatrix} \alpha & -1 & -1 \1 & -\alpha & -1 \1 & -1 & -a \end{bmatrix} \begin{bmatrix} x \y \z \end{bmatrix} = \begin{bmatrix} a-1 \a-1 \\alpha-1 \end{bmatrix}\]

Step 3: Determine the Condition for No Solution

For the system to have no solution, the matrix must be singular (its determinant must be zero). Calculate the determinant of the coefficient matrix:\[\Delta = \begin{vmatrix} \alpha & -1 & -1 \1 & -\alpha & -1 \1 & -1 & -a \end{vmatrix} \]

Step 4: Calculate the Determinant

Expand the determinant along the first row:\[\Delta = \alpha((-\alpha)(-a) - (1)(-1)) - (-1)(1(-a) - 1(-1)) - (-1)(1(-1) - (-\alpha)(1))\]Simplify to get: \[\Delta = \alpha(\alpha a - 1) - (a + 1) - (\alpha - 1)\]

Step 5: Set Determinant to Zero

For no solution, set \( \Delta = 0 \): \[\alpha(\alpha a - 1) - (a + 1) - (\alpha - 1) = 0\]Simplify to find conditions on \(\alpha\).

Step 6: Solve for Specific Values of \( \alpha \)

Plug values of \(\alpha\) from the options to check if they satisfy the equation from Step 5. Calculate for \(\alpha = -2\) and \(\alpha = 1\).

Step 7: Evaluate the Solutions

Compute:\1. For \(\alpha = -2\): Check if determinant = 0.\2. For \(\alpha = 1\): Check if determinant = 0.

Step 8: Conclude Based on Calculations

If both \(\alpha = -2\) and \(\alpha = 1\) make the determinant zero, then the correct option is a. If only \(\alpha = -2\) makes it zero, then the correct option is b.

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements. For a 2x2 matrix, it is quite simple to find, but for larger matrices, it requires more calculations. In this exercise, we work with a 3x3 matrix. The determinant helps us understand certain properties of the matrix, like whether it is singular or invertible.

To calculate the determinant, we often use a method called cofactor expansion. This involves breaking down the matrix into smaller parts and calculating their determinants. For our 3x3 matrix:

• We expand along one row or column, usually the first for simplicity.
• Perform calculations involving the minor matrices that result from removing a specific row and column from the matrix.

When you find the determinant, it allows you to determine if a system of equations has a unique solution, infinitely many solutions, or no solution at all.

Matrix Form of Equations

Writing a system of equations in matrix form is a powerful tool in linear algebra. This involves transforming the equations into a format using matrices, which can then be manipulated more easily. For our system:

• The coefficients of the variables form the coefficient matrix.
• The variables themselves form a column matrix.
• The constants on the right side of the equations form another column matrix.

By placing these components into a single equation, we end up with the matrix equation: \[ A\vec{x} = \vec{b} \]This approach simplifies solving the system, as we can apply different matrix operations to find solutions.

Singular Matrix

A singular matrix is one that does not have an inverse. This is crucial when discussing solutions of a system of linear equations because:

• If the matrix of coefficients (the main matrix) is singular, it means the determinant is zero.
• A zero determinant indicates the system of equations does not have a unique solution.
• The solutions might be either infinite or nonexistent, depending on additional properties of the system.

In our exercise, we examined the determinant of the matrix formed by the coefficients of the system. When the determinant equals zero, it signifies that the system might not be solvable, reinforcing the concept of a singular matrix.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear transformations. Systems of linear equations, like the one in our exercise, are a fundamental application of linear algebra concepts. Through linear algebra, we gain tools and techniques to:

• Manipulate and solve systems of equations efficiently.
• Understand the geometric and algebraic properties of vectors and matrices.
• Apply these methods to real-world problems, ranging from computer science to engineering.

Linear algebra emphasizes the power of matrices and determinants in solving equations, optimizing computations, and analyzing data. This exercise, in particular, highlighted the importance of understanding when a matrix is singular and how it affects the solvability of a system.