|

−a2	ab	ac
ba	−b2	bc
ca	cb	−c2

|
= abc |

−a	b	c
a	−b	c
a	b	−c

|
[Taking out a, b, and c common from R1,R2, and R3 respectively]
= a2b2c2 |

−1	1	1
1	−1	1
1	1	−1

|
[Taking out a, b, and c common from C1,C2, and C3 respectively]
= a2b2c2 |

−1	1	1
0	0	2
0	2	0

| [Applying R2 → R2+R1 and R3 → R3+R1]
= a2b2c2 [(-1) (0 × 0 – 2 × 2)]
= a2b2c2 [- (0 – 4)] = 4 a2b2c2
Hence proved.