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 $\displaystyle \begin{vmatrix} -2&-3&1\\ 9&2&7\\ 6&1&5\\ \end{vmatrix}$ is zero. This will prove that its rows are linearly dependent, and hence the scalars $\displaystyle c_1, c_2, c_3$ exist.

Or you can do it directly with the following system of equations:
$\displaystyle \left\{\begin{array}{l l} -2c_1-3c_2+c_3=0&(1)\\ 9c_1+2c_2+7c_3=0&(2)\\ 6c_1+c_2+5c_3=0&(3)\\ \end{array}\right.$
Multiply $\displaystyle (1)$ by $\displaystyle 3$ and add to $\displaystyle (3)$:
$\displaystyle -8c_2+8c_3=0$

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

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