1
Question
If u=ax2+2bxy+cy2,u′=a′x2+2b′xy+c′y2, then prove that ∣∣
∣∣y2−xyx2abca′b′c′∣∣
∣∣=∣∣∣ax+bybx+cya′x+b′yb′x+c′y∣∣∣=−1y∣∣∣uu′ax+bya′x+b′y∣∣∣
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Solution
Consider first part ∣∣
∣∣y2−xyx2abca′b′c′∣∣
∣∣
Expanding along the first row, y2∣∣∣bcb′c′∣∣∣+xy∣∣∣aca′c′∣∣∣+x2∣∣∣aba′b′∣∣∣
=y2(bc′−b′c)+xy(ac′−a′c)+x2(ab′−a′b)
Consider second part,∣∣∣ax+bybx+cya′x+b′yb′x+c′y∣∣∣
=(ax+by)(b′x+c′y)−(bx+cy)(a′x+b′y)
=ab′x2+ac′xy+bb′xy+bc′y2−(a′bx2+bb′xy+a′cxy+b′cy2)
=y2(bc′−b′c)+xy(ac′−a′c)+x2(ab′−a′b)
Consider third part,−1y∣∣∣uu′ax+bya′x+b′y∣∣∣
=−1y∣∣∣ax2+2bxy+cy2a′x2+2b′xy+c′y2ax+bya′x+b′y∣∣∣
=−1y[(ax2+2bxy+cy2)(a′x+b′y)−(a′x2+2b′xy+c′y2)(ax+by)]
=−1y[(aa′x3+ab′x2y+2a′bx2y+2bb′xy2+ac′xy2+b′cy3) −(aa′x3+a′byx2+2b′ax2y+2bb′xy2+ac′xy2+bc′y3)]
=−1y[ab′x2y+2a′bx2y+2bb′xy2+ac′xy2+b′cy3−a′byx2−2b′ax2y−2bb′xy2−ac′xy2−bc′y3]
=−[ab′x2+2a′bx2+2bb′xy+ac′xy+b′cy2−a′bx2−2b′ax2−2bb′xy−ac′xy−bc′y2]
=−[−ab′x2+a′bx2+ac′xy+b′cy2−ac′xy−bc′y2]
=y2(bc′−b′c)+xy(ac′−a′c)+x2(ab′−a′b)
Hence, all the three parts are equal
∣∣
∣∣y2−xyx2abca′b′c′∣∣
∣∣
Expanding along the first row,
y2∣∣∣bcb′c′∣∣∣+xy∣∣∣aca′c′∣∣∣+x2∣∣∣aba′b′∣∣∣
=y2(bc′−b′c)+xy(ac′−a′c)+x2(ab′−a′b)
Consider second part,
∣∣∣ax+bybx+cya′x+b′yb′x+c′y∣∣∣
=(ax+by)(b′x+c′y)−(bx+cy)(a′x+b′y)
=ab′x2+ac′xy+bb′xy+bc′y2−(a′bx2+bb′xy+a′cxy+b′cy2)
=y2(bc′−b′c)+xy(ac′−a′c)+x2(ab′−a′b)
Consider third part,
−1y∣∣∣uu′ax+bya′x+b′y∣∣∣
=−1y∣∣∣ax2+2bxy+cy2a′x2+2b′xy+c′y2ax+bya′x+b′y∣∣∣
=−1y[(ax2+2bxy+cy2)(a′x+b′y)−(a′x2+2b′xy+c′y2)(ax+by)]
=−1y[(aa′x3+ab′x2y+2a′bx2y+2bb′xy2+ac′xy2+b′cy3)
−(aa′x3+a′byx2+2b′ax2y+2bb′xy2+ac′xy2+bc′y3)]
=−1y[ab′x2y+2a′bx2y+2bb′xy2+ac′xy2+b′cy3−a′byx2−2b′ax2y−2bb′xy2−ac′xy2−bc′y3]
=−[ab′x2+2a′bx2+2bb′xy+ac′xy+b′cy2−a′bx2−2b′ax2−2bb′xy−ac′xy−bc′y2]
=−[−ab′x2+a′bx2+ac′xy+b′cy2−ac′xy−bc′y2]
=y2(bc′−b′c)+xy(ac′−a′c)+x2(ab′−a′b)
Hence, all the three parts are equal
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