Show that no choice of numbers $a$ and $b$ can make $ax + by = (3,0,0)$

Question

Show that no choice of numbers $a$ and $b$ can make $ax + by = (3,0,0)$ when $x = (3,-1,0)$ and $y = (0,1,5)$.

The only materials in the chapter talked about are:

• Vector Space Operations
• Standard Basis
• Coordinates of a vector $x$
• Components of a vector $x$

I don't think that the vector $v=(3,0,0)$ has a unique linear combination of coordinates with respect to the standard basis with how $x$ and $y$ are defined. I also don't think stating this is "rigorous" enough of an answer. Is there a better (clearer) way I can show that the above statement is true?

How can I show that the statement is true from a geometric point of view?

Answer 1

$ax+by = a(3, -1, 0) + b(0, 1, 5) =(3,0,0)$ gives you the system of equations $$ 3a = 3 , \, -a+b = 0, \, 5b = 0. $$ Can you go on from here?

Answer 2

You may not have learned this yet, but $n$ vectors in $ \mathbb {R} ^{n}$ are linearly independent if and only if

the determinant of the matrix formed by taking the vectors as its columns is non-zero.

In your particular case, that matrix has a form (upper triangular) that makes it very easy to compute

the determinant (simply multiply the diagonal entries):

$\begin{vmatrix} 3 & 3 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 5 \end{vmatrix}=-15\ne0,$ so they're linearly independent.