Question

If the rank of the matrix \[ A = \begin{bmatrix} 1 & 2 & 1 & -1 \\ -1 & 2 & 3 & 5 \\ 0 & 1 & k & k \end{bmatrix} \] is 2 and \( k \) is a real number, then \( k \) is a root of which of the following quadratic equations?

• A. \( x^2 + 3x + 2 = 0 \)
• B. \( x^2 + x - 2 = 0 \)
• C. \( x^2 + x - 6 = 0 \)
• D. \( x^2 - x - 6 = 0 \)

Hint:

Use elementary row operations to reduce the matrix and apply the rank condition to identify constraints on variables.

Answer 1

The Correct Option is B

We are given that the rank of matrix \( A \) is 2, i.e., only two rows are linearly independent. Apply row operations to simplify. Start with: \[ A = \begin{bmatrix} 1 & 2 & 1 & -1\\ -1 & 2 & 3 & 5 \\ 0 & 1 & k & k \end{bmatrix} \] Add Row 1 and Row 2: \[ R_2 \rightarrow R_2 + R_1 = [0, 4, 4, 4] \] New matrix: \[ \begin{bmatrix} 1 & 2 & 1 & -1 \\ 0 & 4 & 4 & 4 \\ 0 & 1 & k & k \end{bmatrix} \] Now eliminate the second row using third: \[ R_3 \rightarrow R_3 - \frac{1}{4}R_2 = [0, 0, k - 1, k - 1] \] For the matrix to have rank 2, the third row must be all zeros: \[ k - 1 = 0 \Rightarrow k = 1 \] Now find a quadratic for which \( k = 1 \) is a root. Option (2) is: \[ x^2 + x - 2 = 0 \Rightarrow (x - 1)(x + 2) = 0 \Rightarrow x = 1, -2 \] So, \( k = 1 \) is a valid root.