Linear Algebra: linear combinations problem So I have problem number 4:


So what I tried doing is that I set the vector from a) equal to c1*(first vector from S) and c2*(second vector from S), and from that I got the 4x3 matrix:
[6,4,-42;-7,6,13;8,-4,-112;6,1,-60]
And this is impossible to solve since this is basically 4 equations with 3 unknowns. Does this lead to the conclusion that a b c and d cannot be written as a linear combination of S ?
 A: Let 
$A=
\begin{bmatrix}
6 & 4\\
-7 & 6\\
8 & -4\\
6 & 1
\end{bmatrix}
$
be a matrix with column vectors from $S$. Let $\textbf{b}$ be the given vectors which you want to solve for. To find the possible coefficients of these two vectors, you want to solve a system $\textbf{A}x=\textbf{b}$. In augmented matrix form, for (a), this is
$
\begin{bmatrix}
6 & 4 & -42\\
-7 & 6 & 113\\
8 & -4 & -112\\
6 & 1 & -60
\end{bmatrix}.
$
Row reducing gives 
$
\begin{bmatrix}
1 & 0 & -11\\
0 & 1 & 6\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix}
$
so $x=\begin{bmatrix}-11 \\ 6\end{bmatrix}$, and you see 
$$
-11\begin{bmatrix}
6\\
-7\\
8\\
6\\
\end{bmatrix}
+6
\begin{bmatrix}
4\\ 
6\\
-4\\
1\\
\end{bmatrix}
=
\begin{bmatrix}
-42\\
113\\
-112\\
-60\\
\end{bmatrix}.$$
 The others follow similarly. If you can find some solution vector $x$, then you can write the given vector $\textbf{b}$ as a linear combination of the vectors in $S$, otherwise you cannot.
A: I don't understand what you've been doing. Let's call $u$ and $v$ the vectors from $S$ and $w$ the one from (a). What do you want to know is if there exists real numbers $x$ and $y$ such that
$$
w = xu + y v \ .
$$
Right?
But this is a system of linear equations, with four equations (right) and just two unknowns ($x$ and $y$). This one:
$$
\begin{pmatrix}
-42 \\ 113 \\ -112 \\ -60
\end{pmatrix}
=
x
\begin{pmatrix}
6 \\ -7 \\ 8 \\ 6
\end{pmatrix}
+
y
\begin{pmatrix}
4 \\ 6 \\ -4 \\ 1
\end{pmatrix}
$$
Which you can write as
$$
\begin{pmatrix}
6 & 4 & \vert & -42 \\
-7 & 6 &\vert & 113 \\
8 & -4 &\vert & -112 \\
6 & 1 & \vert & -60
\end{pmatrix}
$$
If I'm not wrong, this time this system has no solution at all (and so, $w$ is NOT a linear combination of $u$ and $v$), but you don't have to think that, just because a system of linear equations has more unknowns than equations this necessarily means that it has no solutions. For instance, this one
$$
\begin{align}
x + y &=& 1  \\
x + y &=& 1  \\
x + y &=& 1
\end{align}
$$
Has two unknowns and three equations, but an infinite number of solutions. "Oh, but it's always the same equation, so it doesn't count". -Is that what you're thinking? Well, try to solve this one:
$$
\begin{align}
x + y &=& 1  \\
x - y &=& 0  \\
2x    &=& 1
\end{align}
$$
