# Math Help - Linear Algebra: Sub-Space!

1. ## Linear Algebra: Sub-Space!

For the vectors in R^3, determine whether the first vector can be expressed as a linear combination of the other two.

a) (-2,0,3), (1,3,0), (2,4,-1)
b) (1,2,-3), (-3,2,1), (2,-1,-1)

Thanks a lot!

2. ## Vectors

Hello Vedicmaths
Originally Posted by Vedicmaths
For the vectors in R^3, determine whether the first vector can be expressed as a linear combination of the other two.

a) (-2,0,3), (1,3,0), (2,4,-1)
b) (1,2,-3), (-3,2,1), (2,-1,-1)

Thanks a lot!
(a) If the first vector can be expressed as a linear combination of the other two, then you'll be able to find values of $a$ and $b$ for which

$-2\vec{i} + 0 \vec{j} + 3\vec{k} = a(1\vec{i} + 3\vec{j} + 0\vec{k}) + b(2\vec{i} + 4 \vec{j} -1\vec{k})$

Comparing coefficients of $\vec{i}$, $\vec{j}$ and $\vec{k}$:

$-2 = a + 2b$ (1)

$0 = 3a + 4b$ (2)

$3 = -b$ (3)

So you've now got three equations in 2 unknowns. Use 2 of them to find values of $a$ and $b$. Then plug these values into the third. If it works then, yes, you can do it; if it doesn't, you can't.

So in this case:

From (3): $b = -3$

Subst into (2): $0 = 3a -12$

$\Rightarrow a = 4$

And test in (1): $a + 2b = 4 -6 = -2$, which is correct.

So, in this case, yes the first vector can be expressed as a linear combination of the other two. It is:

$-2\vec{i} + 0 \vec{j} + 3\vec{k} = 4(1\vec{i} + 3\vec{j} + 0\vec{k}) -3(2\vec{i} + 4 \vec{j} -1\vec{k})$

Use the same method in part (b).