Question

Suppose \( \alpha, \beta, \gamma \) are the roots of the equation \( x^3 + qx + r = 0 \) (with \( r \ne 0 \)) and they are in A.P. Then the rank of the matrix \[ \begin{pmatrix} \alpha & \beta & \gamma
\beta & \gamma & \alpha
\gamma & \alpha & \beta \end{pmatrix} \] is:

• A. \( 3 \)
• B. \( 2 \)
• C. \( 0 \)
• D. \( 1 \)

Hint:

For rank problems with roots: \begin{itemize} \item Use symmetric root conditions. \item A.P. roots simplify nicely. \item Check row sums for dependence. \end{itemize}

Answer 1

The Correct Option is B

Concept:
Given cubic: \[ x^3 + qx + r = 0 \] Sum of roots: \[ \alpha + \beta + \gamma = 0 \] Also, roots are in A.P., so let: \[ \alpha = a - d, \quad \beta = a, \quad \gamma = a + d \]
Step 1: Use sum of roots.
\[ (a - d) + a + (a + d) = 0 \] \[ 3a = 0 \Rightarrow a = 0 \] So roots: \[ \alpha = -d, \quad \beta = 0, \quad \gamma = d \]
Step 2: Substitute into matrix.
\[ A = \begin{pmatrix} -d & 0 & d \\ 0 & d & -d \\ d & -d & 0 \end{pmatrix} \]
Step 3: Check row dependence.
Observe: \[ R_1 + R_2 + R_3 = 0 \] So rows are linearly dependent ⇒ rank \( < 3 \).

Step 4: Check if rank = 1 or 2.
Take two rows: \[ (-d, 0, d), \quad (0, d, -d) \] They are not scalar multiples ⇒ independent. Thus rank = 2.