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Let $F_2$ denote the field of scalars {0,1} with operations done using mod 2 arithmetic.

Consider the matrix: $$\begin{bmatrix} 1&1&1&0 \\ 1&1&0&1 \\ 0&0&1&1 \end{bmatrix} \in M_{3,4}(F_2)$$

(a) Determine a basis for the column space of A, and list all the elements in the column space.

(b) Determine a basis for the row space of A, and list all the elements in the row space.

(c) Determine a basis for the solution space of A, and list all the elements in the solution space.

From my understanding, the first step is calculating the RREF, here is the RREF: $$\begin{bmatrix}1&1&0&1\\0&0&1&1\\0&0&0&0\end{bmatrix}$$

Part a:

Leading entries are in column 1, 3 therefore the column space has a basis $\{(1,1,0), (1,0,1)\}$.

Hence the span is: $\{\alpha_1(1,1,0) + \alpha_2(1,0,1)\}$

There are four points in the space:

$\alpha_1 = \alpha_2 = 0 : (0,0,0)$

$\alpha_1 = 1, \alpha_2 = 0 : (1,1,0)$

$\alpha_1 = 0,\alpha_2 = 1 : (1,0,1)$

$\alpha_1 = 1, \alpha_2 = 1 : (0,1,1)$

Part b:

Basis of row space of A is non zero rows in RREF, therefore the basis is $\{(1,1,1,0), (1,1,0,1)\}$. Hence, the span is : $\{\alpha_1(1,1,1,0) + \alpha_2(1,1,0,1)\}$

There are four points in the space:

$\alpha_1 = \alpha_2 = 0 : (0,0,0)$

$\alpha_1 = 1, \alpha_2 = 0 : (1,1,1,0)$

$\alpha_1 = 0,\alpha_2 = 1 : (1,1,0,1)$

$\alpha_1 = 1, \alpha_2 = 1 : (0,0,1,1)$

Part c:

A$\begin{bmatrix}x_1 \\ x_2 \\x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix}1&1&0&1\\0&0&1&1\\0&0&0&0\end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$

$x_2 = s, x_4 = t$

$x_1+x_2+x_4=0 => x_1 +s+t+s+t = s+t => x_1 = s+t$ $x_3 + t = 0 => x_3 +t+t= t => x_3=t$

Can someone please verify this solution?

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  • $\begingroup$ @Bernard the operations are only in mod 2 arithmetic, thus the values can only be 1 or 0. $\endgroup$ – jessica Oct 5 at 10:10
  • $\begingroup$ Sorry, I didn't see that. I'll remove my comment. $\endgroup$ – Bernard Oct 5 at 10:12
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Yes, it's all correct.
You still need to list the elements of the solution space: just substitute all 4 possible $(s,t)$ pairs in your solution.

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  • $\begingroup$ So I just need to add elements to part c like s=0,t=0 : (0,0,0,0), s=1, t=1 : (0,1,1,1), s=1,t=0: (1,1,0,0) and s=0, t=1: (1,0,1,1)? $\endgroup$ – jessica Oct 5 at 11:01
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    $\begingroup$ Exactly. You can verify them $\endgroup$ – Berci Oct 5 at 11:04

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