how would you do this question?
ax − 7y = 0
2x + (a − 9)y = 0
Find the value(s) of a, where a is a real constant. Consider a set of simultaneous equations that
have a unique solution.when is there a unique solution?
There is a method of this that will work in all cases - I will show you this method. However, there IS another method that will work brilliantly with this example that is a lot less effort - and I'll follow up the first method with this.
For the first method, let's put these simultaneous equations into matrix form:
We know that a solution for this will ONLY exist if the determinant of the first matrix is non-zero (or if x and y are 0, but that's trivial). So, if the determinant IS zero, then there is NO unique solution, and every other time, there IS a unique solution. So, let's find when the determinant is zero and discount those solutions:
-(-7)(2)=a^2-9a+14=(a-7)(a-2)=0<br />\\ \therefore a=2,7)
So, there exists a unique solution when
On our second method, let's transpose both equations like so:
y=0\implies 2x=-(a-9)y\implies y=\frac{-2}{a-9}x)
Now, we know that if the gradients are equal, then there are either infinitely many or no solutions. So, let's find out when the gradients are equal, and exclude those values:
=-14<br />\\ a^2-9a+14=0<br />\\ (a-7)(a-2)=0<br />\therefore a=2,7)
Which reflects what we got earlier, so we once again say that
Note: the second method will always work as well, it's just that it's not as nice if the two lines don't intercept at (0, 0).
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