Let A=⎡⎢⎣3x216x⎤⎥⎦,B=[a,b,c] 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)∨x∈R, where tr.(A) denotes trace of A. Find the value of (a+b+c)
A
6
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B
7
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C
8
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D
9
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Solution
The correct option is C 7 Given A=⎡⎢⎣3x216x⎤⎥⎦,B=[a,b,c] Here, AB=⎡⎢⎣3x216x⎤⎥⎦[a,b,c] ⇒AB=⎡⎢⎣3ax23bx23cx2abc6ax6bx6cx⎤⎥⎦ tr(AB)=3ax2+b+6cx Now, given C=⎡⎢
⎢⎣(x+2)25x22x5x22x(x+2)22x(x+2)25x2⎤⎥
⎥⎦ tr(C)=(x+2)2+2x+5x2 ⇒tr(C)=6x2+6x+4 Since, we have tr(AB)=tr(C) ⇒3ax2+b+6cx=6x2+6x+4 On comparing, we get a=2,b=4,c=1 So, a+b+c=7