Question

If the system of linear equations, \( x - ay - z = 0, ax - y - z = 0, x + y - z = 0 \), has infinite number of solutions, then the possible values of a are

• A. 0, 1
• B. -1, 2
• C. -1, 1
• D. 0, -1

Hint:

For a homogeneous system of linear equations (\(A\vec{x} = \vec{0}\)), the key condition to remember is: - \( \det(A) \neq 0 \implies \) Only the unique trivial solution (\(\vec{x} = \vec{0}\)). - \( \det(A) = 0 \implies \) Infinite number of non-trivial solutions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A system of homogeneous linear equations (where the right-hand side is all zeros) has a non-trivial solution (including infinite solutions) if and only if the determinant of the coefficient matrix is equal to zero. The system always has the trivial solution (\(x = y = z = 0\)). To have infinite solutions, the equations must be linearly dependent, which is signaled by a zero determinant.
Step 2: Key Formula or Approach:
1. Write the system of equations in matrix form \( A\vec{x} = \vec{0} \).
2. The coefficient matrix \( A \) is: \[ A = \begin{pmatrix} 1 & -a & -1 \\ a & -1 & -1 \\ 1 & 1 & -1 \end{pmatrix} \] 3. For infinite solutions, the determinant of \( A \) must be zero: \( \det(A) = 0 \).
4. Calculate the determinant and solve the resulting equation for \( a \).
Step 3: Detailed Calculation:
Calculate the determinant of the matrix \( A \): \[ \det(A) = 1 \begin{vmatrix} -1 & -1 \\ 1 & -1 \end{vmatrix} - (-a) \begin{vmatrix} a & -1 \\ 1 & -1 \end{vmatrix} + (-1) \begin{vmatrix} a & -1 \\ 1 & 1 \end{vmatrix} \] \[ \det(A) = 1((-1)(-1) - (-1)(1)) + a((a)(-1) - (-1)(1)) - 1((a)(1) - (-1)(1)) \] \[ \det(A) = 1(1 + 1) + a(-a + 1) - 1(a + 1) \] \[ \det(A) = 2 - a^2 + a - a - 1 \] \[ \det(A) = 1 - a^2 \] Now, set the determinant to zero to find the values of \( a \) that give infinite solutions: \[ 1 - a^2 = 0 \] \[ a^2 = 1 \] \[ a = \pm 1 \] The possible values of \( a \) are 1 and -1.
Step 4: Final Answer:
The possible values of \( a \) are -1 and 1. This corresponds to option (C).