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Question

∣ ∣1+a1+b1a1+b2a1+b3a2+b11+a2+b2a2+b3a3+b1a3+b21+a3+b3∣ ∣ . Evaluate the determinant.

A
1+3i=1(ai+bi)+1i<j3(aiaj)(bjbi)
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B
1+3i=1(ai+bi)+1i<j3(ai+aj)(bj+bi)
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C
1+3i=1(aibi)+1i<j3(aiaj)(bjbi)
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D
None of these.
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Solution

The correct option is A 1+3i=1(ai+bi)+1i<j3(aiaj)(bjbi)
Let Δ=∣ ∣1+a1+b1a1+b2a1+b3a2+b11+a2+b2a2+b3a3+b1a3+b21+a3+b3∣ ∣
=∣ ∣1a1+b2a1+b301+a2+b2a2+b30a3+b21+a3+b3∣ ∣+∣ ∣a1a1+b2a1+b3a21+a2+b2a2+b3a3a3+b21+a3+b3∣ ∣+∣ ∣b1a1+b2a1+b3b11+a2+b2a2+b3b1a3+b21+a3+b3∣ ∣
Δ=Δ1+Δ2+Δ3 say (1)
Expand Δ1 with respect to C1, we get
Δ1=1+a2+a3+b2+b3+(a2a3)(b3b2)
and
Δ2=∣ ∣a1a1+b2a1+b3a21+a2+b2a2+b3a3a3+b21+a3+b3∣ ∣
Applying C2C2C1;C3C3C1
=∣ ∣a1b2b3a21+b2b3a3b21+b3∣ ∣
Applying R2R)2R1;R3R3R1
=∣ ∣a1b2b3a2a110a3a101∣ ∣
Δ2=a1+b2(a1a2)+b3(a1a3)
Δ3=∣ ∣b1a1+b2a1+b3b11+a2+b2a2+b3b1a3+b21+a3+b3∣ ∣
Applying R2R2R1;R3R3R1
∣ ∣b1a1+b+2a1+b301+a2a1a2a10a3a11+a3a1∣ ∣
Δ3=b1b1(a1a2)b1(a1a3)
Therefore from (1), we have
Δ=Δ1+Δ2+Δ3
=1+3i=1(ai+bi)+1i<j3(aiaj)(bjbi)

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