Question

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

• A. \( 1 \)
• B. \( -4 \)
• C. \( -5 \)
• D. \( 3 \)

Hint:

For questions involving polynomials equated to determinants, always look to substitute \(\lambda = 0\) first to find the constant term \(D\) instantly. For finding leading coefficients like \(A\), trace only the product pathways that can yield the maximum power of the variable instead of performing a tedious full expansion!

Answer 1

The Correct Option is D

Concept: A polynomial equation identity holds true for all possible values of its variable. Instead of completely expanding a complex determinant containing a variable like \(\lambda\), we can substitute strategic values of \(\lambda\) to find specific coefficients or combinations of coefficients directly. Key observations for substitution:
• Substituting \(\lambda = 0\) isolates the constant term \(D\).
• The coefficient of \(\lambda^3\) (which is \(A\)) can be found by inspecting only the terms that contribute to the highest power of \(\lambda\) during expansion.

Step 1: Finding the value of \(D\) by substituting \(\lambda = 0\).
Since the given matrix equation is an identity in \(\lambda\), substitute \(\lambda = 0\) on both sides: \[ \begin{vmatrix} 1 & 2 & 3 - 0 0 & -1 - 0 & 2 1 - 0 & 1 & 3 \end{vmatrix} = A(0)^3 + B(0)^2 + C(0) + D \] \[ \begin{vmatrix} 1 & 2 & 3 0 & -1 & 2 1 & 1 & 3 \end{vmatrix} = D \] Now, expand the determinant along the first column (\(C_1\)): \[ D = 1 \cdot \begin{vmatrix} -1 & 2 1 & 3 \end{vmatrix} - 0 + 1 \cdot \begin{vmatrix} 2 & 3 -1 & 2 \end{vmatrix} \] \[ D = 1 \cdot [(-1)(3) - (2)(1)] + 1 \cdot [(2)(2) - (3)(-1)] \] \[ D = 1 \cdot [-3 - 2] + 1 \cdot [4 + 3] = -5 + 7 = 2 \quad \cdots (1) \]

Step 2: Finding the coefficient of \(\lambda^3\) (\(A\)) by structural inspection.
Let us examine the elements containing \(\lambda\): \[ \Delta(\lambda) = \begin{vmatrix} 1 & 2 & 3 - \lambda 0 & -1 - \lambda & 2 1 - \lambda & 1 & 3 \end{vmatrix} \] Expanding completely along the first column (\(C_1\)) to see where the highest power of \(\lambda\) comes from: \[ \Delta(\lambda) = 1 \cdot \begin{vmatrix} -1 - \lambda & 2 1 & 3 \end{vmatrix} + (1 - \lambda) \cdot \begin{vmatrix} 2 & 3 - \lambda -1 - \lambda & 2 \end{vmatrix} \] Let's find the leading term in each component:
• First part: \(1 \cdot [(-1-\lambda)(3) - 2] = -3 - 3\lambda - 2 = -5 - 3\lambda\). (No \(\lambda^3\) term here).
• Second part: \((1 - \lambda) \cdot [4 - (3 - \lambda)(-1 - \lambda)]\). Focusing only on the \((3 - \lambda)(-1 - \lambda)\) multiplication inside the bracket: \[ (3 - \lambda)(-1 - \lambda) = -3 - 3\lambda + \lambda + \lambda^2 = \lambda^2 - 2\lambda - 3 \] Thus, the second part becomes: \[ (1 - \lambda) \cdot [4 - (\lambda^2 - 2\lambda - 3)] = (1 - \lambda)(7 + 2\lambda - \lambda^2) \] Multiplying this out to extract the \(\lambda^3\) term: \[ -\lambda \cdot (-\lambda^2) = \lambda^3 \] Therefore, the coefficient of \(\lambda^3\) is \(A = 1\). Wait, let's look closer at the signs: \[ 4 - (3-\lambda)(-1-\lambda) = 4 - (\lambda-3)(\lambda+1) = 4 - (\lambda^2 - 2\lambda - 3) = -\lambda^2 + 2\lambda + 7 \] Multiplying by \((1-\lambda)\): \[ (1-\lambda)(-\lambda^2 + 2\lambda + 7) = -\lambda^2 + 2\lambda + 7 + \lambda^3 - 2\lambda^2 - 7\lambda = \lambda^3 - 3\lambda^2 - 5\lambda + 7 \] So the overall polynomial expression is: \[ \Delta(\lambda) = (-5 - 3\lambda) + (\lambda^3 - 3\lambda^2 - 5\lambda + 7) = \lambda^3 - 3\lambda^2 - 8\lambda + 2 \] Comparing this with \(A\lambda^3 + B\lambda^2 + C\lambda + D\), we get: \[ A = 1 \quad \text{and} \quad D = 2 \] Wait, let's re-verify the expansion step. Let's do the expansion along the third row \(R_3\): \[ (1-\lambda)[4 - (3-\lambda)(-1-\lambda)] - 1[2 - 0] + 3[1(-1-\lambda) - 0] \] \[ = (1-\lambda)[4 - (\lambda^2 - 2\lambda - 3)] - 2 + 3(-1-\lambda) \] \[ = (1-\lambda)(-\lambda^2 + 2\lambda + 7) - 2 - 3 - 3\lambda \] \[ = (\lambda^3 - 3\lambda^2 - 5\lambda + 7) - 5 - 3\lambda = \lambda^3 - 3\lambda^2 - 8\lambda + 2 \] This perfectly confirms \(A = 1\) and \(D = 2\). Let's check the value of \(D + A\): \[ D + A = 2 + 1 = 3 \] Wait, let's re-verify the option choice. The calculation shows \(A = 1\) and \(D = 2\), so \(D + A = 3\), which corresponds to option (d). Let's update the correct answer option label accordingly.

Step 3: Evaluating the final required value \(D + A\).
From our rigorous full polynomial expansion: \[ A = 1, \quad D = 2 \] Therefore, the required sum is: \[ D + A = 2 + 1 = 3 \]