Question

If \( \left| \begin{matrix} 1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3 \end{matrix} \right| = A\lambda^3 + B\lambda^2 + C\lambda + D \), then \( D+A = \)

• A. 1
• B. -4
• C. -5
• D. 3

Hint:

To find \( D \) (the constant term), you can simply put \( \lambda = 0 \) in the determinant. \( D = \Delta(0) = \left| \begin{matrix} 1 & 2 & 3\\ 0 & -1 & 2 \\ 1 & 1 & 3 \end{matrix} \right| = 1(-3-2) + 1(4 - (-3)) = -5 + 7 = 2 \). To find \( A \), check the coefficient of the highest power \( \lambda^3 \). It comes from the product of the diagonal terms or terms involving \( \lambda \) in every row/column. Here \( A = 1 \). Then \( D+A = 2+1=3 \). This is much faster.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

We need to expand the given determinant to find the polynomial in \( \lambda \). The coefficients \( A, B, C, D \) correspond to the terms \( \lambda^3, \lambda^2, \lambda, \) and the constant term, respectively. We are asked to find \( D+A \).
Step 2: Detailed Explanation:

Let \( \Delta(\lambda) = \left| \begin{matrix} 1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3 \end{matrix} \right| \). Expanding along the first column (C1): \[ \Delta(\lambda) = 1 \cdot \left| \begin{matrix} -1-\lambda & 2 \\ 1 & 3 \end{matrix} \right| - 0 + (1-\lambda) \cdot \left| \begin{matrix} 2 & 3-\lambda \\ -1-\lambda & 2 \end{matrix} \right| \] Calculate the minors: 1. First minor: \[ ((-1-\lambda)(3) - (2)(1)) = -3 - 3\lambda - 2 = -3\lambda - 5 \] 2. Second minor (multiplied by \( 1-\lambda \)): \[ (2)(2) - (3-\lambda)(-1-\lambda) \] \[ = 4 - [ -3 - 3\lambda + \lambda + \lambda^2 ] \] \[ = 4 - [ \lambda^2 - 2\lambda - 3 ] \] \[ = 4 - \lambda^2 + 2\lambda + 3 = -\lambda^2 + 2\lambda + 7 \] Now combine the terms: \[ \Delta(\lambda) = 1(-3\lambda - 5) + (1-\lambda)(-\lambda^2 + 2\lambda + 7) \] Expand the second part: \[ (1-\lambda)(-\lambda^2 + 2\lambda + 7) = 1(-\lambda^2 + 2\lambda + 7) - \lambda(-\lambda^2 + 2\lambda + 7) \] \[ = -\lambda^2 + 2\lambda + 7 + \lambda^3 - 2\lambda^2 - 7\lambda \] \[ = \lambda^3 - 3\lambda^2 - 5\lambda + 7 \] Add the first part (\( -3\lambda - 5 \)): \[ \Delta(\lambda) = (\lambda^3 - 3\lambda^2 - 5\lambda + 7) + (-3\lambda - 5) \] \[ \Delta(\lambda) = \lambda^3 - 3\lambda^2 - 8\lambda + 2 \] Comparing this with \( A\lambda^3 + B\lambda^2 + C\lambda + D \): \[ A = 1 \] \[ B = -3 \] \[ C = -8 \] \[ D = 2 \] We need \( D + A \): \[ D + A = 2 + 1 = 3 \]
Step 4: Final Answer:

The value of \( D+A \) is 3.