For example, write (-3,4,1) as a linear combination of (1,2,1) (8,1,2) and (4,3,2)
Is this possible considering these vectors are not linearly independent?
I ask this because when constructing these into a matrix and running a reduced row echelon form, I do not have a clear answer, and I figure this is the reason.
Thank you for the help.
Best Answer
In order to determine whether it is possible to write $(-3,4,1)$ as a linear combination of $(1,2,1), (8,1,2),$ and $(4,3,2),$ the standard approach is to put these vectors together as the columns of a matrix, then row reduce. In particular, we have $$ \left(\begin{array}{cccc} 1 & 8 & 4 & -3\\ 2 & 1 & 3 & 4\\ 1 & 2 & 2 & 1 \end{array}\right) \leadsto \left(\begin{array}{cccc} 1 & 0 & \frac{4}{3} & \frac{7}{3}\\ 0 & 1 & \frac{1}{3} & -\frac{2}{3}\\ 0 & 0 & 0 & 0 \end{array}\right). $$ Because the last column of the row-reduced matrix is not a pivot column, it is possible to write $(-3,4,1)$ as a linear combination of the other three vectors. In particular: we see that in the row-reduced matrix, the last column can be written as $\frac 73$ of the first column added to $- \frac 23$ of the second column. Because this holds for the row-reduced matrix, it also holds for the original matrix. That is, we have $$ (-3,4,1) = \frac 73 \cdot (1,2,1) - \frac 23 \cdot (8,1,2). $$