Answer to Question #91473 in Abstract Algebra for Sajid

Question

Q. Find the dimension of the subspace of R4 that is span of the vectors
(█(1¦(-1)@0@1)), (█(2¦1@1@1)),(█(0¦0@0@0)),(█(1¦1@-2@-5))

Q. Choose the correct answer.
Q. Let b and c are elements in a group G and e is identity element of G. If b5=c3=e,then inverse of bcb2 is
a. b2cb
b. b3c2b4
d. b2c2b4

Answer 1

1) Build the matrix of the coordinates:

$$\begin{pmatrix} 1 & -1 & 0 &1 \\ 2 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & -2 &-5 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 0 &1 \\ 0 & 3 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 2 & -2 &-6 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 0 &1 \\ 0 & 3 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -\frac{8}{3} &-\frac{16}{3} \end{pmatrix}$$

where we added the first line to the others to make the first column contain zero, and analogously with the second line. The rank of this matrix is 3, for finally we obtain only 3 nonzero lines.

2)

$$b^5 = e \rightarrow (b^n)^{-1} = b^{5-n}$$ $$c^3 = e \rightarrow (c^n)^{-1} = c^{3-n}$$

hence

$$b \, c \, b^2 \times b^3 \, c^2 \, b^4 = b \, c \times c^2 \, b^4 = b \times b^4 = e$$

The answer is b) b³c²b⁴