# Linearly independent amount

Statement: If $$v_1=[2;1;11;2]$$, $$v_2=[1;0;4;-1]$$, $$v_3=[11;4;56;5]$$ and $$v_4=[2;-1;5;-6]$$, then $$(v_1,v_2,v_3,v_4)$$ is a linearly independent set.

My answer: I think that they are linearly independent. Am I correct?

• Put them into a 4x4 matrix, calculate the determinant and you'll know for sure. – freakish Dec 13 '18 at 15:32
• okey. i will try – Jacob Andreasson Dec 13 '18 at 15:35
• I've calculated it and it turns out to be $0$. So these vectors are linearly dependent unless I've made a mistake. – freakish Dec 13 '18 at 15:37
• you are correct. I did write wrong. it is linearly dependent – Jacob Andreasson Dec 13 '18 at 15:42
• The English term for Swedish “mängd” is “set” rather than “amount”. – egreg Dec 13 '18 at 16:23

## 1 Answer

You are not correct; you can consider the matrix having those vectors as columns and do Gaussian elimination on it \begin{align} \begin{bmatrix} 2 & 1 & 11 & 2 \\ 1 & 0 & 4 & -1 \\ 11 & 4 & 56 & 5 \\ 2 & -1 & 5 & -6 \end{bmatrix} &\to \begin{bmatrix} 1 & 0 & 4 & -1 \\ 2 & 1 & 11 & 2 \\ 11 & 4 & 56 & 5 \\ 2 & -1 & 5 & -6 \end{bmatrix} &&R_1\leftrightarrow R_2 \\[6px] &\to \begin{bmatrix} 1 & 0 & 4 & -1 \\ 0 & 1 & 3 & 4 \\ 0 & 4 & 12 & 16 \\ 0 & -1 & -3 & -4 \end{bmatrix} && \begin{aligned} R_2 &\gets R_2-2R_1 \\ R_3&\gets R_3-11R_1 \\ R_4 &\gets R_4-2R_1 \end{aligned} \\[6px] &\to \begin{bmatrix} 1 & 0 & 4 & -1 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} && \begin{aligned} R_3&\gets R_3-4R_2 \\ R_4 &\gets R_4+R_2 \end{aligned} \end{align} From the last matrix, which is the reduced row echelon form, it follows that $$v_3=4v_1+3v_2,\qquad v_4=-v_1+4v_2$$ This method is more efficient than computing the determinant, which only tells you that the set is linearly dependent, but doesn't show the linear relations between the vectors.