In Gilbert Strang's Linear Algebra Book 4th Edition, the question asks to find the zero vector with the combination of three vectors: $w_1$ = (1,2,3), $w_2$ = (4,5,6), $w_3$ = (7,8,9).
Working this out to reduced row echelon form, I get to a certain point where it's obvious that there is no solution since the bottom row are 0's.
\begin{bmatrix}
1 & 4 & 7 & 0 \\
2 & 5 & 8 & 0 \\
3 & 6 & 9 & 0
\end{bmatrix}
becomes:
\begin{bmatrix}
1 & 4 & 7 & 0 \\
0 & 3 & 6 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
In the solutions, though, it's pointed out that $w_1$ – 2$w_2$ + $w_3$ = 0 and thus, no solution. Plugging the original vectors in this equation, it's obviously true. However, how does one get to $w_1$ – 2$w_2$ + $w_3$ = 0 from the original vectors if RREF can't get you there? Is it just a matter of simply looking at the vectors, I should have seen the dependence, or is there another way of working it out besides RREF?
Best Answer
You wish to find $a_1, a_2, a_3$ such that $a_1 w_1 + a_2 w_2 + a_3 w_3 = 0$. The fact that the third row of the reduced matrix is zeroes means that any number may be chosen for $a_3$ and you will be able to find $a_1, a_2$ satisfying the above equation.
You may choose $a_3 = 1$. Then from the second row you have $3a_2 + 6a_3 = 0 \implies a_2 = -2$. Finally, from the first row you have $a_1 + 4a_2 + 7a_3 = 0 \implies a_1 - 8 + 7 = 0 \implies a_1 = 1$.