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TG EAMCET 2 May 2025 Shift 2 Paper

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Question : 4 of 160

Marks: +1, -0

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

D+A=

1
-4
-5
3

Solution:

We have,

\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

Now,

\begin{vmatrix} 1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3 \end{vmatrix}

\;=1(-3-3\lambda-2)+(1-\lambda)(4+(1+\lambda)(3-\lambda))

\;=(-3\lambda-5)+(1-\lambda)(4+3-\lambda+3\lambda-\lambda^{2})

\;=(-3\lambda-5)+(1-\lambda)(7+2\lambda-\lambda^{2})

\;=(-3\lambda-5)+(7+2\lambda-\lambda^{2}-7\lambda-2\lambda^{2}+\lambda^{3})

\;=\lambda^{3}-3\lambda^{2}-8\lambda+2

\; \therefore \lambda^{3}-3\lambda^{2}-8\lambda+2=A\lambda^{3}+B\lambda^{2}+C\lambda+D

On comparing both sides, we get

A\;=1, B=-3, C=-8, D=2

\therefore \;\; D+A\;=2+1=3

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