MHB Proving Linear Dependence of r1,r2,r3 Given a,b,c ≠ 0

[Suvadip]

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

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

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

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

 

[Evgeny.Makarov]

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]

Hello, suvadip!

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.