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Question

If a+b+c=0, then the solution of the equation ∣ ∣axcbcbxabacx∣ ∣=0 is

A
0
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B
±32(a2+b2+c2)
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C
0,±32(a2+b2+c2)
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D
0,±(a2+b2+c2)
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Solution

The correct option is C 0,±32(a2+b2+c2)
∣ ∣axcbcbxabacx∣ ∣=0∣ ∣a+b+cxcba+b+cxbxaa+b+cxacx∣ ∣=0x=a=0 (by hypothesis)
or 1{(bx)(cx)a2}c{cxa}+b{ab+x}=0 by expanding the determinant.
or x2(a2+b2+c2)+(ab+bc+ca)=0
or x2(a2)12(a2)=0{a+b+c=0(a+b+c)2=0a2+2ab=0ab=12a2}or x=±32a2
The solution is x= 0 or ±32a2
Trick : Put a=1, b=-1 and c=0 so that they satisfy the condition a+b+c=0. Now the determinant becomes ∣ ∣1x0101x111x∣ ∣=0(1x){x(1+x)1}+1(1+x)=0(1x){x2+x1}+x+1=0x(x23)=0
Now putting these in the options, we find that option (c) gives the same values i.e., 0, ±3.




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