Question

If \(x,y,z\) are in A.P. with common difference \(d\) and the rank of the matrix \[ \begin{pmatrix} 4 & 5 & x \\ 5 & 6 & y\\ 6 & k & z \end{pmatrix} \] is \(2\), then the values of \(k,d\) are:

• A. \(6,\; \dfrac{x}{2}\)
• B. \(5,\; x\)
• C. any arbitrary,\; \(x\)
• D. \(7,\) any arbitrary

Hint:

For a \(3\times3\) matrix:
Rank \(=3 \Rightarrow \det \neq 0\)
Rank \(=2 \Rightarrow \det = 0\) (but rows/columns not all proportional) Always convert sequences into algebraic form before evaluating determinants.

Answer 1

The Correct Option is D

Step 1: Express \(y\) and \(z\) using A.P. property. Since \(x,y,z\) are in A.P. with common difference \(d\), \[ y = x + d,\qquad z = x + 2d \]
Step 2: Use the condition for rank \(=2\). For a \(3\times 3\) matrix to have rank \(2\), \[ \det = 0 \] (but not all rows are proportional).
Step 3: Evaluate the determinant. \[ \begin{vmatrix} 4 & 5 & x\\ 5 & 6 & x+d\\ 6 & k & x+2d \end{vmatrix} \] Expanding along the first row: \[ = 4[6(x+2d)-k(x+d)] -5[5(x+2d)-6(x+d)] +x[5k-36] \] Simplifying, \[ = (k-7)(x-4d) \]
Step 4: Apply determinant condition. \[ (k-7)(x-4d)=0 \] This gives: \[ k=7 \quad \text{or} \quad x=4d \]
Step 5: Interpret the options. From the given choices, the valid and general solution is: \[ k=7,\quad d \text{ arbitrary} \]
Hence, the correct answer is \(\boxed{(D)\; 7,\text{ any arbitrary}\).}