# Find the vector equation such that set is linearly dependent

Please, only hints and suggestions

I've tried many things with all fails.

I knew that this set of vectors $\{v_1, v_2, v_3 \}$ is Linearly Independent. So I have to find some vector $v$ such that

$c_1v_1 + c_2v_2 + c_3v_3 + c_4v = 0$ but one of them isn't 0.

• Why do you think that v1,v2,v3 is linear dependent? – miracle173 Jan 15 '17 at 23:57
• Can you solve the equation for $v$? – benji Jan 15 '17 at 23:59
• @benji, like $v = \frac{c_1v_1 + c_2v_2 + c_3v_3}{c_4}$? – Amad27 Jan 16 '17 at 0:07

## 1 Answer

Let $v=[a\;b\;c\;d]^T$ and consider Gaussian elimination on the matrix \begin{align} \begin{bmatrix} 1 & 0 & 1 & a\\ -1 & 1 & -2 & b\\ 0 & 1 & -2 & c\\ 2 & 0 & 2 & d \end{bmatrix} &\to \begin{bmatrix} 1 & 0 & 1 & a\\ 0 & 1 & -1 & b+a\\ 0 & 1 & -2 & c\\ 0 & 0 & 0 & d-2a \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 & 1 & a\\ 0 & 1 & -1 & b+a\\ 0 & 0 & -1 & c-b-a\\ 0 & 0 & 0 & d-2a \end{bmatrix} \end{align} Thus you see that the vector should satisfy $d=2a$, which is the condition for the linear system $$c_1v_1+c_2v_2+c_3v_3+c_4v=0$$ to have a non trivial solution.

• very fast. keep it going. – SAJW Jan 16 '17 at 0:14
• I am having some trouble. We solve $c_1v_2 + c_2v_2 + c_3v_3 + c_4v_4$? Using Augmented matrices? – Amad27 Jan 17 '17 at 4:45
• Also, you didnt complete the RREF? Is it valid still? – Amad27 Jan 17 '17 at 4:53
• @Amad27 At that stage, we already know what the pivot columns are, the RREF wold just tell us how to write the vector $v$ as a linear combination of the three vectors (provided $d=2a$). We're essentially solving $v=c_1v_1+c_2v_2+c_3v_3$. – egreg Jan 17 '17 at 9:03
• @egreg, also is this matrix augmented? Is the augmented second part $[a, b, c, d]$? – Amad27 Jan 23 '17 at 1:52