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Question

lf A=⎧⎪⎨⎪⎩acbbaccba⎫⎪⎬⎪⎭ then the cofactor of a32 in A+AT is

A
(2a(b+c)(b+c)2)
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B
acb2
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C
a2bc
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D
2a(a+c)(a+c)2
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Solution

The correct option is B (2a(b+c)(b+c)2)
By property of transpose, we get
AT=abccabbca
So, A+AT=2ab+cb+cb+c2ab+cc+bb+c2a
So, M32= minor of a32=2ab+cb+cc+c=2a(b+c)(b+c)2
=2ab+2acb2c22bc
So, cofactor of a32=M32(1)3+2
=M32
=[2a(b+c)(b+c)2]

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