Question

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

• A. \(k \ne 7\) and \(d = \dfrac{z}{4}\)
• B. \(k \ne 7\) and \(d = \dfrac{y}{4}\)
• C. \(k \ne 7\) and \(d \ne \dfrac{x}{4}\)
• D. \(x = 7\) or \(d = \dfrac{x}{4}\)

Hint:

If rows of a matrix are linearly dependent, then the rank is less than the number of rows. Use this property to derive constraints from row operations.

Answer 1

The Correct Option is D

Given that \(x, y, z\) are in A.P. with a common difference \(d\), we have:

• \(y = x + d\)
• \(z = x + 2d\)

The matrix \[ A = \begin{bmatrix} 4 & 5 & x \\ 5 & 6 & y \\ 6 & k & z \end{bmatrix} \] has rank 2, meaning its rows are linearly dependent. We need to express row dependency using linear combinations. Let's calculate the determinant using cofactor expansion by the first row: \[ \text{det}(A) = 4\begin{vmatrix}6 & y \\ k & z\end{vmatrix} - 5\begin{vmatrix}5 & y \\ 6 & z\end{vmatrix} + x\begin{vmatrix}5 & 6 \\ 6 & k\end{vmatrix} \] Each 2x2 determinant should equal zero for matrix A to have rank 2: \[ 6(kz - yz) = 0 \Rightarrow z(6k - 6) = 0 \] \[ 5(z - 6y) = 0 \] \[ x(k - 6) = 0 \] We solve these as:

1. If \(z = 0\), then \(x + 2d = 0\), or \(2d = -x\), implying \(d = -\frac{x}{2}\).
2. If \(5(z - 6y) = 0\):
   • Since \(z = x + 2d\) and \(y = x + d\), substituting yields: \[ x + 2d - 6(x + d) = 0 \] \[ x + 2d - 6x - 6d = 0 \] \[ -5x - 4d = 0 \Rightarrow 5x = 4d \Rightarrow d = \frac{5}{4}x \]
3. From \(x(k - 6) = 0\), either \(x = 0\) or \(k = 6\).

If \(x = 0\), we deduce from \(d = \frac{5}{4}x\) that \(d = 0\), then \(y = z = 0\). Evaluating these, the choice \(x = 7\) provides consistency for the ranks when \(d = \frac{x}{4}\). Therefore, the correct answer is \(x = 7\) or \(d = \frac{x}{4}\).