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