Answer to Question #132434 in Algebra for Evans Malepe

Question

Suppose u, v and w are vectors in 3–space. Which of the following is/are defined? Explain
A. (u x v).w
B. u .(v x w)

Suppose u and v are vectors in 3–space where u = (u1;u2;u3) and v = (v1;v2;v3). Evaluate u x v x u
and v x u x u.


Suppose A and B are 3 x 3 matrices. Prove that
(i) (A + B)T = BT + AT
(ii) det(AB)T = det(A) : det(B)

Answer 1

(u x v).w

Here (u x v) is a vector. The cross product of 2 vectors results in a vector.

Let (u x v) = p where p is a vector

(u x v).w= p.w .This gives a numerical value or a scalar as output. Since dot product of 2 vectors results in a scalar.

Thus (u x v).w is defined in the vector space.

u .(v x w)

The cross product of 2 vectors results in a vector. Hence (v x w) is a vector.

Let v x w = q . Where q is a vector

u .(v x w)= u.q Which results in a scalar output.

Thus u .(v x w) is also defined in a vector space

u = (u₁;u₂;u₃) and v = (v₁;v₂;v₃)

The triple cross product of 3 vectors A B C are given as

Ax(BxC)=(A.C)B - (A.B)C

u x v x u =(u.u)v - (u.v)u .......... (u.u=|u||u|cos0=|u|² )

=|u|² v - (u.v)u

Now

v x u x u = v x(u x u)= v x 0 =0 .......(uxu=|u||u|sin0=0, angle between u and u is 0 and sin0=0)

= v x 0

=0

Prove (A+B)^T=B^T+A^T

A=[a_ij]

B=[b_ij]

A+B=[a_ij+b_ij]

(A+B)^T=[a_ji+b_ji] ..........(1)

B^T=[b_ji]

A^T=[a_ji]

B^T+A^T=[b_ji + a_ji] =[a_ji + b_ji] ............(2)

Right hand side of equation (1) and (2) are equal. So Left hand side will also be equal

(A+B)^T=B^T+A^T

We know Determinant of a matrix and its transpose are equal

det(AB)^T=det(AB)

Case 1

If A is not invertible, then AB is not invertible, then the theorem holds,

because

det(A)=0

Thus det(A)det(B)=0

And Rank of AB will be less than 3

Thus det(AB)=0

det(AB) = det(A) det(B)

Case 2

If A is invertible,

Then there exist elementary row matrices E_k, · · · , E₁ such that

A = E_k · · · E₁. That is A could be represented as product of Elementary row matrices

Then,

det(AB) = det(E_k · · · E₁B), (we use the property det(EA) = det(E) det(A) Where E is an elementary row matrix)

= det(E_k) det(E_k−1 · · · E₁B),

= det(E_k)· · · det(E₁) det(B),

= det(E_k · · · E₁) det(B),

= det(A) det(B).

Hence

det(AB)^T=det(A)det(B)