<Dl> <Dd> <Table> <Tr> <Th> (show) Proof by substitution </Th> </Tr> <Tr> <Td> Evaluation of the cross product gives <Dl> <Dd> a = c × d = (c 2 d 3 − c 3 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1) (\ displaystyle \ mathbf (a) = \ mathbf (c) \ times \ mathbf (d) = (\ begin (pmatrix) c_ (2) d_ (3) - c_ (3) d_ (2) \ \ c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) \ end (pmatrix))) </Dd> </Dl> <P> Hence, the left hand side equals </P> <Dl> <Dd> (a) × = (0 c 2 d 1 − c 1 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1 0 c 3 d 2 − c 2 d 3 c 1 d 3 − c 3 d 1 c 2 d 3 − c 3 d 2 0) (\ displaystyle (\ mathbf (a)) _ (\ times) = (\ begin (bmatrix) 0&c_ (2) d_ (1) - c_ (1) d_ (2) &c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) &0&c_ (3) d_ (2) - c_ (2) d_ (3) \ \ c_ (1) d_ (3) - c_ (3) d_ (1) &c_ (2) d_ (3) - c_ (3) d_ (2) &0 \ end (bmatrix))) </Dd> </Dl> <P> Now, for the right hand side, </P> <Dl> <Dd> c d T = (c 1 d 1 c 1 d 2 c 1 d 3 c 2 d 1 c 2 d 2 c 2 d 3 c 3 d 1 c 3 d 2 c 3 d 3) (\ displaystyle \ mathbf (c) \ mathbf (d) ^ (\ mathrm (T)) = (\ begin (bmatrix) c_ (1) d_ (1) &c_ (1) d_ (2) &c_ (1) d_ (3) \ \ c_ (2) d_ (1) &c_ (2) d_ (2) &c_ (2) d_ (3) \ \ c_ (3) d_ (1) &c_ (3) d_ (2) &c_ (3) d_ (3) \ end (bmatrix))) </Dd> </Dl> <P> And its transpose is </P> <Dl> <Dd> d c T = (c 1 d 1 c 2 d 1 c 3 d 1 c 1 d 2 c 2 d 2 c 3 d 2 c 1 d 3 c 2 d 3 c 3 d 3) (\ displaystyle \ mathbf (d) \ mathbf (c) ^ (\ mathrm (T)) = (\ begin (bmatrix) c_ (1) d_ (1) &c_ (2) d_ (1) &c_ (3) d_ (1) \ \ c_ (1) d_ (2) &c_ (2) d_ (2) &c_ (3) d_ (2) \ \ c_ (1) d_ (3) &c_ (2) d_ (3) &c_ (3) d_ (3) \ end (bmatrix))) </Dd> </Dl> <P> Evaluation of the right hand side gives </P> <Dl> <Dd> d c T − c d T = (0 c 2 d 1 − c 1 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1 0 c 3 d 2 − c 2 d 3 c 1 d 3 − c 3 d 1 c 2 d 3 − c 3 d 2 0) (\ displaystyle \ mathbf (d) \ mathbf (c) ^ (\ mathrm (T)) - \ mathbf (c) \ mathbf (d) ^ (\ mathrm (T)) = (\ begin (bmatrix) 0&c_ (2) d_ (1) - c_ (1) d_ (2) &c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) &0&c_ (3) d_ (2) - c_ (2) d_ (3) \ \ c_ (1) d_ (3) - c_ (3) d_ (1) &c_ (2) d_ (3) - c_ (3) d_ (2) &0 \ end (bmatrix))) </Dd> </Dl> <P> Comparison shows that the left hand side equals the right hand side . </P> </Td> </Tr> </Table> </Dd> </Dl> <Dd> <Table> <Tr> <Th> (show) Proof by substitution </Th> </Tr> <Tr> <Td> Evaluation of the cross product gives <Dl> <Dd> a = c × d = (c 2 d 3 − c 3 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1) (\ displaystyle \ mathbf (a) = \ mathbf (c) \ times \ mathbf (d) = (\ begin (pmatrix) c_ (2) d_ (3) - c_ (3) d_ (2) \ \ c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) \ end (pmatrix))) </Dd> </Dl> <P> Hence, the left hand side equals </P> <Dl> <Dd> (a) × = (0 c 2 d 1 − c 1 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1 0 c 3 d 2 − c 2 d 3 c 1 d 3 − c 3 d 1 c 2 d 3 − c 3 d 2 0) (\ displaystyle (\ mathbf (a)) _ (\ times) = (\ begin (bmatrix) 0&c_ (2) d_ (1) - c_ (1) d_ (2) &c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) &0&c_ (3) d_ (2) - c_ (2) d_ (3) \ \ c_ (1) d_ (3) - c_ (3) d_ (1) &c_ (2) d_ (3) - c_ (3) d_ (2) &0 \ end (bmatrix))) </Dd> </Dl> <P> Now, for the right hand side, </P> <Dl> <Dd> c d T = (c 1 d 1 c 1 d 2 c 1 d 3 c 2 d 1 c 2 d 2 c 2 d 3 c 3 d 1 c 3 d 2 c 3 d 3) (\ displaystyle \ mathbf (c) \ mathbf (d) ^ (\ mathrm (T)) = (\ begin (bmatrix) c_ (1) d_ (1) &c_ (1) d_ (2) &c_ (1) d_ (3) \ \ c_ (2) d_ (1) &c_ (2) d_ (2) &c_ (2) d_ (3) \ \ c_ (3) d_ (1) &c_ (3) d_ (2) &c_ (3) d_ (3) \ end (bmatrix))) </Dd> </Dl> <P> And its transpose is </P> <Dl> <Dd> d c T = (c 1 d 1 c 2 d 1 c 3 d 1 c 1 d 2 c 2 d 2 c 3 d 2 c 1 d 3 c 2 d 3 c 3 d 3) (\ displaystyle \ mathbf (d) \ mathbf (c) ^ (\ mathrm (T)) = (\ begin (bmatrix) c_ (1) d_ (1) &c_ (2) d_ (1) &c_ (3) d_ (1) \ \ c_ (1) d_ (2) &c_ (2) d_ (2) &c_ (3) d_ (2) \ \ c_ (1) d_ (3) &c_ (2) d_ (3) &c_ (3) d_ (3) \ end (bmatrix))) </Dd> </Dl> <P> Evaluation of the right hand side gives </P> <Dl> <Dd> d c T − c d T = (0 c 2 d 1 − c 1 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1 0 c 3 d 2 − c 2 d 3 c 1 d 3 − c 3 d 1 c 2 d 3 − c 3 d 2 0) (\ displaystyle \ mathbf (d) \ mathbf (c) ^ (\ mathrm (T)) - \ mathbf (c) \ mathbf (d) ^ (\ mathrm (T)) = (\ begin (bmatrix) 0&c_ (2) d_ (1) - c_ (1) d_ (2) &c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) &0&c_ (3) d_ (2) - c_ (2) d_ (3) \ \ c_ (1) d_ (3) - c_ (3) d_ (1) &c_ (2) d_ (3) - c_ (3) d_ (2) &0 \ end (bmatrix))) </Dd> </Dl> <P> Comparison shows that the left hand side equals the right hand side . </P> </Td> </Tr> </Table> </Dd> <Table> <Tr> <Th> (show) Proof by substitution </Th> </Tr> <Tr> <Td> Evaluation of the cross product gives <Dl> <Dd> a = c × d = (c 2 d 3 − c 3 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1) (\ displaystyle \ mathbf (a) = \ mathbf (c) \ times \ mathbf (d) = (\ begin (pmatrix) c_ (2) d_ (3) - c_ (3) d_ (2) \ \ c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) \ end (pmatrix))) </Dd> </Dl> <P> Hence, the left hand side equals </P> <Dl> <Dd> (a) × = (0 c 2 d 1 − c 1 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1 0 c 3 d 2 − c 2 d 3 c 1 d 3 − c 3 d 1 c 2 d 3 − c 3 d 2 0) (\ displaystyle (\ mathbf (a)) _ (\ times) = (\ begin (bmatrix) 0&c_ (2) d_ (1) - c_ (1) d_ (2) &c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) &0&c_ (3) d_ (2) - c_ (2) d_ (3) \ \ c_ (1) d_ (3) - c_ (3) d_ (1) &c_ (2) d_ (3) - c_ (3) d_ (2) &0 \ end (bmatrix))) </Dd> </Dl> <P> Now, for the right hand side, </P> <Dl> <Dd> c d T = (c 1 d 1 c 1 d 2 c 1 d 3 c 2 d 1 c 2 d 2 c 2 d 3 c 3 d 1 c 3 d 2 c 3 d 3) (\ displaystyle \ mathbf (c) \ mathbf (d) ^ (\ mathrm (T)) = (\ begin (bmatrix) c_ (1) d_ (1) &c_ (1) d_ (2) &c_ (1) d_ (3) \ \ c_ (2) d_ (1) &c_ (2) d_ (2) &c_ (2) d_ (3) \ \ c_ (3) d_ (1) &c_ (3) d_ (2) &c_ (3) d_ (3) \ end (bmatrix))) </Dd> </Dl> <P> And its transpose is </P> <Dl> <Dd> d c T = (c 1 d 1 c 2 d 1 c 3 d 1 c 1 d 2 c 2 d 2 c 3 d 2 c 1 d 3 c 2 d 3 c 3 d 3) (\ displaystyle \ mathbf (d) \ mathbf (c) ^ (\ mathrm (T)) = (\ begin (bmatrix) c_ (1) d_ (1) &c_ (2) d_ (1) &c_ (3) d_ (1) \ \ c_ (1) d_ (2) &c_ (2) d_ (2) &c_ (3) d_ (2) \ \ c_ (1) d_ (3) &c_ (2) d_ (3) &c_ (3) d_ (3) \ end (bmatrix))) </Dd> </Dl> <P> Evaluation of the right hand side gives </P> <Dl> <Dd> d c T − c d T = (0 c 2 d 1 − c 1 d 2 c 3 d 1 − c 1 d 3 c 1 d 2 − c 2 d 1 0 c 3 d 2 − c 2 d 3 c 1 d 3 − c 3 d 1 c 2 d 3 − c 3 d 2 0) (\ displaystyle \ mathbf (d) \ mathbf (c) ^ (\ mathrm (T)) - \ mathbf (c) \ mathbf (d) ^ (\ mathrm (T)) = (\ begin (bmatrix) 0&c_ (2) d_ (1) - c_ (1) d_ (2) &c_ (3) d_ (1) - c_ (1) d_ (3) \ \ c_ (1) d_ (2) - c_ (2) d_ (1) &0&c_ (3) d_ (2) - c_ (2) d_ (3) \ \ c_ (1) d_ (3) - c_ (3) d_ (1) &c_ (2) d_ (3) - c_ (3) d_ (2) &0 \ end (bmatrix))) </Dd> </Dl> <P> Comparison shows that the left hand side equals the right hand side . </P> </Td> </Tr> </Table> <Tr> <Th> (show) Proof by substitution </Th> </Tr>

If a cross b equals a cross c does b equal c