# vector algebra

##### Member
Given that
$$\displaystyle r1=2a-3b+c$$
$$\displaystyle r2=3a-5b+2c$$
$$\displaystyle r3=4a-5b+c$$

where $$\displaystyle a, b, c$$ are non-zero and non coplannar vectors

How to prove that $$\displaystyle r1, r2 , r3$$ are linearly dependent?

I have moved with $$\displaystyle c1*r1+c2*r2+c3*r3=0$$
but confused how to show that at leat one of $$\displaystyle c1, c2, c3$$ is non-zero. We only have the information $$\displaystyle a,b,c \neq 0$$ and $$\displaystyle [a b c]\neq 0$$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
suvadip;42909I have moved with [MATH said:
c1*r1+c2*r2+c3*r3=0[/MATH]
I'll denote the coefficients by $x_1,x_2,x_3$ instead of $c_1,c_2,c_3$ because the letter $c$ already denotes a vector. Usually the convention is to use, for example, English letters from the beginning of the alphabet possibly followed by subscripts or primes to denote vectors, Greek letters possibly with subscripts or primes to denote real numbers and so on.

You can rearrange the equation
$x_1r_1+x_2r_2+x_3r_3=0$
to have the form
$y_1a+y_2b+y_3c=0\qquad(1)$
where $y_1,y_2,y_3$ are some numbers expressed through $x_1,x_2,x_3$. Since $a,b,c$ are non-coplanar and hence linearly independent, (1) happens iff
$y_1=y_2=y_3=0.\qquad(2)$
Thus you have three equations and three variables $x_1,x_2,x_3$. Since this system is homogeneous (the right-hand side is 0), it has a solution $x_1=x_2=x_3=0$. If there are no other solutions, then no nontrivial combination of $r_1,r_2,r_3$ is 0 and thus the vectors are linearly independent. If there is a nonzero solution to (2), then there exists a nontrivial linear combination of $r_1,r_2,r_3$ that equals zero and so the vectors are linearly dependent.

#### soroban

##### Well-known member

Given that: .$$\begin{array}{ccc}r_1&=&2a-3b+c \\ r_2&=&3a-5b+2c \\ r_3&=&4a-5b+c \end{array}$$

where $$a, b, c$$ are non-zero, non-coplannar vectors

How to prove that $$r_1, r_2 , r_3$$ are linearly dependent?

Show that one of them is a linear combination of the other two.

We will show that: .$$Pr_1 + Qr_2 \:=\:r_3$$ for some integers $$P,Q.$$

. . $$P(2a-3b+c) + Q(3a-5b+2c) \:=\:4a-5a + c$$

. . $$2Pa - 3Pb + PC + 3Qa - 5Qb + 2Qc \:=\:4a-5b+c$$

. . $$(2P+3Q)a - (3P+5Q)b + (P+2Q)c \:=\:4a-5b+c$$

Equate coefficients: .$$\begin{Bmatrix}2P + 3Q &=& 4 \\ 3P + 5Q &=& 5 \\ P+2Q &=& 1 \end{Bmatrix}$$

Solve the system: .$$P = 5,\;Q = \text{-}2$$

. . Note: these values must satisfy all three equations.

Therefore, $$r_1,r_2,r_3$$ are linearly dependent.