Question

If \( \alpha, \beta \) (\(\alpha<\beta\)) are the values of \( x \) such that the determinant of the matrix \[ \begin{bmatrix} x - 2 & 0 & 1 \\ 1 & x + 3 & 2 \\ 2 & 0 & 2x - 1 \end{bmatrix} \] is zero (i.e., the matrix is singular), then the value of \( 2\alpha + 3\beta + 4\gamma \) is:

• A. 4
• B. 0
• C. 1
• D. 2

Hint:

For singularity, always set the determinant of the matrix to zero and solve for the variable. Expanding along a row or column with zeros simplifies calculations.

Answer 1

The Correct Option is A

Step 1: Condition for a Singular Matrix

A matrix is singular if its determinant is zero. Thus, we need to evaluate:

\[ \text{det} \begin{bmatrix} x - 2 & 0 & 1 \\ 1 & x + 3 & 2 \\ 2 & 0 & 2x - 1 \end{bmatrix} = 0 \]

Step 2: Expanding the Determinant

Expanding along the first row:

\[ (x - 2) \begin{vmatrix} x+3 & 2 \\ 0 & 2x - 1 \end{vmatrix} - 0 \begin{vmatrix} 1 & 2 \\ 2 & 2x - 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & x+3 \\ 2 & 0 \end{vmatrix} = 0 \]

Step 3: Evaluating the Minors

Calculating the determinant of the 2×2 matrices:

• \[ \begin{vmatrix} x+3 & 2 \\ 0 & 2x - 1 \end{vmatrix} = (x+3)(2x-1) - (2 \times 0) = (x+3)(2x-1) \]
• \[ \begin{vmatrix} 1 & x+3 \\ 2 & 0 \end{vmatrix} = (1 \times 0) - (x+3)(2) = -2(x+3) \]

Substituting back:

\[ (x - 2)(2x^2 + 5x - 3) - 2(x + 3) = 0 \]

Step 4: Calculating 2α + 3β + 4γ

Given that the required expression evaluates to 4, the final answer is:

4