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Question

Consider the determinant Δ=∣ ∣a1a2a3b1b2b3c1c2c3∣ ∣
Mij= Minor of the element of ith row & jth column.
Cij= Cofactor of element of ith row & jth column.
Value of b1.C31+b2.C32+b3.C33 is

A
0
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B
Δ
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C
2Δ
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D
Δ2
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Solution

The correct option is A 0
Value of b1.C31+b2.C32+b3.C33

=b1.(1)3+1a2a3b2b3+b3.(1)3+2a1a3b1b3+b3.(1)3+3a1a3b1b2

=b1(b2a3a2b3)b2(b1a3a1b3)+b3(b1a3a1b2)

=0

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