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))

Answer 1

$$A=\begin{pmatrix} 1 & -1 & 0 & 1\\ 2 & 1 & 1 & 1\\ 0 & 0 & 0 & 0\\ 1 & 1 & -2 & -5 \end{pmatrix}$$

2 line minus 1 line and 4 line plus 1 line multiplied by 5:

$$A \sim \begin{pmatrix} 1 & -1 & 0 & 1\\ 1 & 2 & 1 & 0\\ 0 & 0 & 0 & 0\\ 6 & -4 & -2 & 0 \end{pmatrix}$$

4 line plus 2 line multiplied by 2:

$$A \sim \begin{pmatrix} 1 & -1 & 0 & 1\\ 1 & 2 & 1 & 0\\ 0 & 0 & 0 & 0\\ 8 & 0 & 0 & 0 \end{pmatrix}$$

Hence, we have three linearly independent vectors $\begin{pmatrix} 1\\1\\0\\8 \end{pmatrix}, \begin{pmatrix} 0\\1\\0\\0 \end{pmatrix},\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix},$ and

$$\operatorname{rank}A=3$$

So, we have 3-dimension subspace of $\mathbb{R}^4$