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Question

Let A=3x216x,B=[abc], and C=⎢ ⎢(x+2)25x22x5x22x(x+2)22x(x+2)25x2⎥ ⎥ be three given matrices, where a, b, c and x R. Given that tr(AB)=tr(C)xR, where tr(A) denotes trace of A. If f(x)=ax2+bx+c, then the value of f(1) is .................

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Solution

AB=3x216x[abc]
AB=3ax23bx23cx2abc6ax6bx6cx
tr(AB)=3ax2+b+6xc
Given C=⎢ ⎢(x+2)25x22x5x22x(x+2)22x(x+2)25x2⎥ ⎥
tr(C)=6x2+6x+4
Since, tr(AB)=tr(C) (given)
3ax2+b+6xc=6x2+6x+4
On comparing, we get
a=2,b=4,c=1
Now, f(x)=ax2+bx+c
So, f(1)=a+b+c=7

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