Question

If \( \begin{pmatrix} 1 & 2 & 4 \\ 1 & 3 & 5 \\ 1 & 4 & a \end{pmatrix} \) is singular, then the value of \( a \) is:

• A. \( a = -6 \)
• B. \( a = 5 \)
• C. \( a = -5 \)
• D. \( a = 6 \)
• E. \( a = 0 \)

Hint:

If the rows of a matrix are in an Arithmetic Progression (A.P.), the determinant is often zero. Notice here rows 1 and 2 differ by \( (0, 1, 1) \). For row 3 to continue this, \( a \) must be 6.

Answer 1

The Correct Option is D

Concept: A square matrix is said to be singular if its determinant is equal to zero (\( |A| = 0 \)). For a \( 3 \times 3 \) matrix, we can find the determinant by expanding along any row or column.

Step 1: Setting up the determinant equation.
The matrix is singular, so: \[ \begin{vmatrix} 1 & 2 & 4 \\ 1 & 3 & 5 \\ 1 & 4 & a \end{vmatrix} = 0 \] Expanding along the first column (which contains all 1s): \[ 1(3a - 20) - 1(2a - 16) + 1(10 - 12) = 0 \]

Step 2: Solving for \( a \).
Simplify the algebraic expression: \[ (3a - 20) - (2a - 16) + (-2) = 0 \] \[ 3a - 20 - 2a + 16 - 2 = 0 \] \[ a - 6 = 0 \implies a = 6 \]