Question

Let \[ A= \begin{pmatrix} 1 & x & 2x \\ 1 & 3x & 5x \\ 1 & 3 & 4 \end{pmatrix}, \qquad x\neq0. \] If \(\det(A)=0\), then \(x=\)

• A. \(-2\)
• B. \(-1\)
• C. \(1\)
• D. \(\dfrac12\)

Hint:

Whenever a question states \[ \boxed{\det(A)=0} \] compute the determinant carefully and solve the resulting algebraic equation. Also check whether any given restriction (such as \(x\neq0\)) eliminates one of the roots.

Answer 1

The Correct Option is B

Concept:
A square matrix is singular if and only if its determinant is zero. 

\[ \boxed{ \det(A)=0 \Longrightarrow \text{Matrix }A\text{ is singular} } \] Therefore, we compute the determinant of the given matrix and equate it to zero. 

Step 1: Expand the determinant along the first row. 
\[ \det(A) = 1 \begin{vmatrix} 3x & 5x\\ 3 & 4 \end{vmatrix} - x \begin{vmatrix} 1 & 5x\\ 1 & 4 \end{vmatrix} + 2x \begin{vmatrix} 1 & 3x\\ 1 & 3 \end{vmatrix} \] 

Step 2: Evaluate each minor. 
First minor: \[ \begin{vmatrix} 3x & 5x\\ 3 & 4 \end{vmatrix} = (3x)(4)-(5x)(3) \] \[ =12x-15x=-3x \] 
Second minor: \[ \begin{vmatrix} 1 & 5x\\ 1 & 4 \end{vmatrix} = (1)(4)-(5x)(1) \] \[ =4-5x \] 
Third minor: \[ \begin{vmatrix} 1 & 3x\\ 1 & 3 \end{vmatrix} = (1)(3)-(3x)(1) \] \[ =3-3x \] 
Substituting these values: \[ \det(A) = -3x -x(4-5x) +2x(3-3x) \] 

Step 3: Simplify the expression. 
\[ \det(A) = -3x-4x+5x^2+6x-6x^2 \] \[ =-x^2-x \] \[ =-x(x+1) \] Since the given condition states that \(x\neq0\), \[ x+1=0 \] Hence, \[ \boxed{x=-1} \] 

Step 4: Final Answer. 
Therefore, \[ \boxed{x=-1} \] Hence, the correct option is: \[ \boxed{\text{Option (B)}} \]