1. ## Linear algebra equation

Hi guys, I was hoping someone could walk me through how to find conditions on 'a' and 'b' such that the system has i) no solution ii) one solution and iii) infinite solutions.

x - 2y = 1
ax + by = 5

The solutions are provided below but I really have no idea how I am supposed to arrive at these conclusions.

Thanks for your time and help!

Solutions
i) If b 6=/= 2a, unique solution x = (b+10)/(b+2a) , y = (5-a)/(b+2a)
ii) If b = -2a : no solution if a =/= 5;
iii) If a = 5 the solutions are x = 1 + 2t, y = t

2. Originally Posted by dems
Hi guys, I was hoping someone could walk me through how to find conditions on 'a' and 'b' such that the system has i) no solution ii) one solution and iii) infinite solutions.

x - 2y = 1
ax + by = 5

The solutions are provided below but I really have no idea how I am supposed to arrive at these conclusions.

Thanks for your time and help!

Solutions
i) If b 6=/= 2a, unique solution x = (b+10)/(b+2a) , y = (5-a)/(b+2a)
ii) If b = -2a : no solution if a =/= 5;
iii) If a = 5 the solutions are x = 1 + 2t, y = t
These are "linear" equations, the equations of straight lines.
You need to understand what linear equations are first.

Then, the equations have a unique solution if their slopes are different,
because they will therefore intersect at a single point.

To find their slopes, write the equations in the form y=mx+c, as "m" is the slope, the multiplier of x in this form.

$\displaystyle x-2y=1\Rightarrow\ x=1+2y\Rightarrow\ x-1=2y\Rightarrow\ y=\frac{1}{2}x-\frac{1}{2}$

$\displaystyle ax+by=5\Rightarrow\ by=5-ax\Rightarrow\ y=-\frac{a}{b}x+\frac{5}{b}$

These slopes are different if $\displaystyle -\frac{a}{b}\ \ne\ \frac{1}{2}$

which gives a unique solution for $\displaystyle -b\ \ne\ 2a$

To obtain the co-ordinates of the point of intersection of the lines...

$\displaystyle x=1+2y$

$\displaystyle x=\frac{5-by}{a}$

$\displaystyle \Rightarrow\ (1+2y)a=5-by\Rightarrow\ 2ay+by=5-a$

from which the values of x and y follow.

There is no solution if the lines are parallel and do not touch.

$\displaystyle 5(x-2y)=5(1)=5$

This is still the equation of the first line

$\displaystyle 5x-10y=5\Rightarrow\ y=\frac{5}{10}x+\frac{5}{10}$ is the first line

$\displaystyle ax+by=5\Rightarrow\ y=-\frac{a}{b}x-\frac{5}{b}$ is the second line.

For the slopes to be equal, $\displaystyle -\frac{a}{b}=\frac{5}{10}\Rightarrow\ b=-2a$

This will cause the lines to be the same (touching at all points) if a=5,
hence "a" must not be 5.

There is an infinite number of solutions if $\displaystyle a=5$ and $\displaystyle b=-2a$

as the equations will be the equations of the same line.

Then for y being any value "t", x will be 1+2y=1+2t

3. Oh well; below is another lengthy response.

The special case occurs when the pair of coefficients of x and y from the first equation is proportional to the pair from the second equation. In this case, this means (1, -2) is proportional to (a, b), which in turn means that there is a number k such that k * 1 = a and k * (-2) = b. If the equations are proportional in this way, then there is only one equation (more precisely, left part) because an equation can be multiplied by an arbitrary number.

Now, the right-hand sides determine whether the system has zero or infinitely many solutions in this special case. If not only the pairs but the whole triples (1, -2, 1) and (a, b, 5) are proportional, then there is literally only one equation. If a = 5 and b = -10, then the second equation is the first multiplied by 5, so it does not add any information. Then x = 1 + 2y, and for every y there is a corresponding x, i.e., there are infinitely many solutions.

If the triples are not proportional, i.e., when we still have k * 1 = a and k * (-2) = b but k * 1 <> 5, then there are no solutions. Indeed, multiplying the first equation by k we get ax + by = k, while the second equation says ax + by = 5 -- a contradiction.

Finally, if (1, -2) and (a, b) are not proportional, then there is exactly one solution. It can be found by isolating x in the first equation and substituting it into the second one.

It's left to you to express the phrase "(1, -2) and (a, b) are proportional" as an equation. Note that when this equation is false, it guarantees that the denominator of the solution for x and y is not zero.

4. Yet a third answer saying basically the same thing!

To determine for what values of a and b
x - 2y = 1
ax + by = 5
has or has not an answer, try solving the equations.

Multiply the first equation by b to get bx- 2by= b. Multiply the second equation by 2 to get 2ax+ 2by= 10.

The point of that is that now y has the same coefficient, 2b, both positive and negative. Adding these two equations will eliminate b: (bx- 2by)+ (2ax+ 2by)= bx+ 2ax= (b+ 2a)x= 10+ b. Now divide both sides by b+ 2a to get $\displaystyle x= \frac{10+ b}{b+ 2a}$.

If $\displaystyle b+ 2a\ne 0$ we can do that and get "the" answer- a unique solution.

If $\displaystyle b+ 2a= 0$ the equation becomes 0x= 10+ b and then there are two possibilities:
i) 10+ b= 0 so the equation says 0x= 0 which is true no matter what x is. There are an infinite number of solutions.
ii) $\displaystyle 10+ b\ne 0$ so the equations says 0x= non-zero which is impossible. There is no solution.