1. ## Scalar/Vector problems

I got stuck on a few questions for a linear algebra assignment...

1. Show that there exists scalars c1, c2, and c3, not all 0, such that c1(-2,9,6) + c2(-3,2,1) + c3(1,7,5) = (0,0,0)

2. a) If v = (3,7) is a vector in the xy-coordinate system, what are the components of v in the x1y1-coordinate system?

b) If v = (v1v2) is a vector in the xy-coordinate system, what are the components of v in the x1y1-coordinate system?

This kind of math is new to me this year

2. Hello chrisks

Welcome to Math Help Forum!
Originally Posted by chrisks
I got stuck on a few questions for a linear algebra assignment...

1. Show that there exists scalars c1, c2, and c3, not all 0, such that c1(-2,9,6) + c2(-3,2,1) + c3(1,7,5) = (0,0,0)

2. a) If v = (3,7) is a vector in the xy-coordinate system, what are the components of v in the x1y1-coordinate system?

b) If v = (v1v2) is a vector in the xy-coordinate system, what are the components of v in the x1y1-coordinate system?

This kind of math is new to me this year
For question 1, you can prove that the determinant $\begin{vmatrix}
-2&-3&1\\
9&2&7\\
6&1&5\\
\end{vmatrix}$
is zero. This will prove that its rows are linearly dependent, and hence the scalars $c_1, c_2, c_3$ exist.

Or you can do it directly with the following system of equations:
$\left\{\begin{array}{l l}
-2c_1-3c_2+c_3=0&(1)\\
9c_1+2c_2+7c_3=0&(2)\\
6c_1+c_2+5c_3=0&(3)\\
\end{array}\right.$

Multiply $(1)$ by $3$ and add to $(3)$:
$-8c_2+8c_3=0$

$\Rightarrow c_2=c_3$
Substitute into $(1)$:
$-2c_1 -3c_2+c_2=0$

$c_1=-c_2$
So if we put $c_1 = 1$ and $c_2 = -1 = c_3$, then:
$1\begin{pmatrix}-2\\9\\6\end{pmatrix}-1\begin{pmatrix}-3\\2\\1\end{pmatrix}-1\begin{pmatrix}1\\7\\5\end{pmatrix}=\begin{pmatri x}0\\0\\0\end{pmatrix}$
For questions 2 and 3, I'm afraid I don't understand what
x1y1-coordinate system
means.