Find scalars $a$ and $b$ so that $au + bv = (1, −4, 9, 18)$

Question

Alright so I am running out of ideas about how to approach this problem, I tried setting $(1,-1,3,5)$ to $w$ and solving for $a$ and $b$ algebraically but just ended up with

$a = (w-bv)/u$

and

$b = (w-au)/v$

i tried solving for $a$ and $b$ algebraically by plugging the second equation into the first but just ended up getting $a = a$ ... and this is a problem from the first section so I havent learned anything after 3.1 in this picture below: so I have no idea what I am even supposed to do to get a numerical scale value for a and b as in the back of the book...

Answer 1

Try going to the simplest way, Just treat it like a system of ecuations do something like this

$$a\vec{u} +b\vec{v}=(1, -4, 9, 18)$$ $$a(1, -1, 3 ,5)+b(2, 1, 0, -3)=(1, -4, 9, 18)$$ Now using the fact that if $\vec{x}=(x_{1}, x_{2})$ and $\vec{y}=(y_{1}, y_{2})$ then $\vec{x}+\vec{y}=(x_{1}+y_{1},x_{2}+y_{2})$ $$(a, -a, 3a, 5a)+(2b, b, 0, -3b)=(1, -4, 9, 18)$$ $$(a +2b, -a+b, 3a, 5a-3b)=(1, -4, 9, 18)$$ Resulting an easy-solving equation system $$a+2b=1$$ $$-a+b=-4$$ $$3a=9$$ $$5a-3b=18$$

Answer 2

$$a\cdot \mathbf{u}+b\cdot\mathbf{v}=a(1, -1, 3, 5)+b(2,1,0,-3)$$ $$=(1,-4,9,18)$$ Therefore we get four equations, $$a+2b=1 \space \space (1)$$ $$-a+b=-4 \space \space (2)$$ $$3a+0\cdot b=9 \space \space (3)$$ $$5a-3b=18 \space \space (4)$$

From equation $(3)$ we see that $a=3$ and from $1$ we get $b=-1$. You can see that these values satisfy equations $(2)$ and $(4)$. Therefore $$3(1, -1, 3, 5)-(2,1,0,-3)=(1,-4,9,18).$$

Edit. Alternatively, you can set up an augmented matrix for the system, such as

$$ \left[ \begin{array}{cc|c} 1&2&1\\ -1&1&-4\\ 3&0&9\\ 5&-3&18\\ \end{array} \right] $$ and then reducing to row echelon or reduced row echelon form, yielding

$$ \left[ \begin{array}{cc|c} 1&0&3\\ 0&1&-1\\ 0&0&0\\ 0&0&0\\ \end{array} \right] $$

which also gives $a=3$ and $b=-1$.

Answer 3

Your formula for $a$ (and also $b$) express it as a ratio of two vectors. In general there is no concept of dividing a vector by another vector unless we know they are already proportional.

The trick is to rename them symbols $a,b$ to $x,y$ and the given condition $xu + yb = (1,-4,9,18)$ after substituting for $u$ and $v$ will lead to 4 equations in two unknowns $x,y$ which you can attempt to solve. SOlution is not always guaranteed it depends on the RHS.