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Question

If ∣ ∣ ∣b2+c2abacabc2+a2bccacba2+b2∣ ∣ ∣=ka2b2c2,then the value of k is

A
2
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B
4
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C
0
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D
None of the above
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Solution

The correct option is B 4
=∣ ∣ ∣b2+c2abacabc2+a2bccacba2+b2∣ ∣ ∣
Applying R1aR1,R2bR2,R3cR3, we get =1abc×∣ ∣ ∣a(b2+c2)a2ba2cab2b(c2+a2)cb2ac2bc2c(a2+b2)∣ ∣ ∣
Now, applying C11aC1,C21bC2,C31cC3, we get =abcabc∣ ∣ ∣b2+c2a2a2b2c2+a2b2c2c2a2+b2∣ ∣ ∣
=∣ ∣ ∣02c22b2b2c2+a2b2c2c2a2+b2∣ ∣ ∣ [R1R1R2R3]

Taking 2 common from R1 and applying R2R2+R1, R3R3+R1
=2∣ ∣ ∣0c2b2b2a20c20a2∣ ∣ ∣

Evaluating along R1, we get
=2[c2(a2b2)b2(a2c2)]
=4a2b2c2
Hence, k=4.

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