1
Question
Show that:
∣∣
∣
∣∣(a−x)2(a−y)2(a−z)2(b−x)2(b−y)2(b−z)2(c−x)2(c−y)2(c−z)2∣∣
∣
∣∣=2(b−c)(c−a)(a−b)(y−z)(z−x)(x−y).
Show that:
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Solution
Given ∣∣
∣
∣∣(a−x)2(a−y)2(a−z)2(b−x)2(b−y)2(b−z)2(c−x)2(c−y)2(c−z)2∣∣
∣
∣∣
=∣∣
∣
∣∣(a–x+a−y)(y−x)(a–y+a−z)(z−y)(a−z)2(b–x+b−y)(y−x)(b–y+b−z)(z−y)(b−z)2(c–x+c−y)(y−x)(c–y+c−z)(z−y)(c−z)2∣∣
∣
∣∣[C1→C1–C2;C2→C2–C3]
=(x−y)(y−z)∣∣
∣
∣∣2a–x−y2a−y−z(a−z)22b–x−y2b−y−z(b−z)22c−x−y2c−y−z(c−z)2∣∣
∣
∣∣
=(x−y)(y−z)∣∣
∣
∣∣2a–x−y2a−y−z(a−z)22(b−a)2(b−a)(b−a)(a+b−2z)2(c−a)2(c−a)(c−a)(a+c−2z)∣∣
∣
∣∣[R2→R2–R1;R3→R2–R1]
=(x−y)(y−z)(b−a)(c−a)(b−c)∣∣
∣
∣∣2a–x−y2a−y−z(a−z)200122a+c−2z∣∣
∣
∣∣[R2→R2–R3]
=(x−y)(y−z)(b−a)(c−a)(b−c)∣∣
∣
∣∣2a–x−y2a−y−z(a−z)200122a+c−2z∣∣
∣
∣∣[R2→R2–R3]
=(x−y)(y−z)(b−a)(c−a)(b−c){−1(4a−2x−2y−4a+2y+2z)}
=2(b−c)(c−a)(a−b)(y−z)(z−x)(x−y) [henceproved]
Given ∣∣ ∣ ∣∣(a−x)2(a−y)2(a−z)2(b−x)2(b−y)2(b−z)2(c−x)2(c−y)2(c−z)2∣∣ ∣ ∣∣
=∣∣ ∣ ∣∣(a–x+a−y)(y−x)(a–y+a−z)(z−y)(a−z)2(b–x+b−y)(y−x)(b–y+b−z)(z−y)(b−z)2(c–x+c−y)(y−x)(c–y+c−z)(z−y)(c−z)2∣∣ ∣ ∣∣[C1→C1–C2;C2→C2–C3]
=(x−y)(y−z)∣∣ ∣ ∣∣2a–x−y2a−y−z(a−z)22b–x−y2b−y−z(b−z)22c−x−y2c−y−z(c−z)2∣∣ ∣ ∣∣
=(x−y)(y−z)∣∣ ∣ ∣∣2a–x−y2a−y−z(a−z)22(b−a)2(b−a)(b−a)(a+b−2z)2(c−a)2(c−a)(c−a)(a+c−2z)∣∣ ∣ ∣∣[R2→R2–R1;R3→R2–R1]
=(x−y)(y−z)(b−a)(c−a)(b−c)∣∣ ∣ ∣∣2a–x−y2a−y−z(a−z)200122a+c−2z∣∣ ∣ ∣∣[R2→R2–R3]
=(x−y)(y−z)(b−a)(c−a)(b−c)∣∣ ∣ ∣∣2a–x−y2a−y−z(a−z)200122a+c−2z∣∣ ∣ ∣∣[R2→R2–R3]
=(x−y)(y−z)(b−a)(c−a)(b−c){−1(4a−2x−2y−4a+2y+2z)}
=2(b−c)(c−a)(a−b)(y−z)(z−x)(x−y) [henceproved]
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