If k∈R and detA=a1b1c1a2b2c2a3b3c3=K then detB=a1b1c1a2+2a1b2+2b1c2+2c1a3b3c3 is equal to

AP EAMCET · Maths · Determinants

If $k \in R$ and $det A = |\begin{array}{l} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}| = K$ then $\det B = |\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} + 2 a_{1} & b_{2} + 2 b_{1} & c_{2} + 2 c_{1} \\ a_{3} & b_{3} & c_{3} \end{matrix}|$ is equal to

A $0$
B $2 K$
C $K$
D $K^{2}$

Verified Solution

Answer & Solution

Correct Answer

(C) $K$

Step-by-step Solution

Detailed explanation

detA=a1b1c1a2b2c2a3b3c3=K detB=a1b1c1a2+2a1b2+2b1c2+2c1a3b3c3 R2→R2-2R1 detB=a1b1c1a2+2a1-2a1b2+2b1-2b1c2+2c1-2c1a3b3c3 detB=a1b1c1a2 b2 c2a3b3c3 detB=K

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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