Question

Express (3a + b + 3c, − a + 6b − c, 2a + b + 2c) as a linear combination of (3, − 1, 2) and (1, 6, 1).

Answer 1

\[\langle 3a+b+3c,~-a+6b-c,~2a+b+2c\rangle=m\langle3,-1,2\rangle+n\langle1,6,1\rangle\]
\[\begin{cases}
3a+b+3c=3m+n\\
-a+6b-c=-m+6n\\
2a+b+2c=2m+n
\end{cases}\]
You want to solve for \(m\) and \(n\). Might I suggest a matrix method? I assume this is linear algebra, so you should be familiar with this.
\[\begin{pmatrix}
3&1&3\\-1&6&-1\\2&1&2
\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}3m+n\\-m+6n\\2m+n\end{pmatrix}\]
or in augmented matrix form,
\[\begin{pmatrix}
3&1&3&|&3m+n\\-1&6&-1&|&-m+6n\\2&1&2&|&2m+n\end{pmatrix}\]
Row reduce to echelon form, then you have
\[\begin{pmatrix}
1&0&1&|&m\\0&1&0&|&n\\0&0&0&|&0\end{pmatrix}\]
So you have \(m=a+c\) and \(n=b\).

Answer 2

Hey thanks a lot! Can I ask another question? Not a problem, just a question. :)

Answer 3

Please do!