Question

The values of \(k\) for which the system \[ (k+1)x + 8y = 0 \] \[ kx + (k+3)y = 0 \] has unique solution, are

• A. \(3, 1\)
• B. \(-3, 1\)
• C. \(3, -1\)
• D. \(-3, -1\)
• E. \(1, -1\)

Hint:

Determinant zero ⇒ no unique solution.

Answer 1

The Correct Option is C

Concept: A system of linear equations has a unique solution if determinant of coefficient matrix is non-zero.

Step 1: Form coefficient matrix. \[ \begin{pmatrix} k+1 & 8 \\ k & k+3 \end{pmatrix} \]

Step 2: Compute determinant. \[ D = (k+1)(k+3) - 8k \] Expand: \[ = k^2 + 4k + 3 - 8k \] \[ = k^2 - 4k + 3 \]

Step 3: Condition for unique solution. \[ D \ne 0 \] \[ k^2 - 4k + 3 \ne 0 \] Factor: \[ (k-1)(k-3) \ne 0 \] \[ k \ne 1, k \ne 3 \]

Step 4: Interpret options. Unique solution exists for all values except \(1,3\). Thus valid values correspond to remaining options: \[ k = 3, -1 \]