Show that if
$\displaystyle A = \[ \left( \begin{array}{cc}
a & b \\
c & d \end{array} \right)\] $
then Ax=0 has only the trivial solution if and only if ad-bc ≠ 0
Would it help to write Ax= 0 as
$\displaystyle \begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}ax+ by \\ cx+ dy\end{pmatrix}= \begin{pmatrix}0 \\ 0 \end{pmatrix}$
which is the same as
ax+ by= 0, cx+ dy= 0.
How would you solve those two equations?
It is the fact that the inverse exists.
I tried and I get stuck. So hold on I think I'm heading in the wrong direction here. To solve these equations means find the value of x and y that satisfies them. I've been trying to add the equations to each other to cancel off x and y, this is wrong isn't it?
Let's do this "long hand." Let's solve Ax = 0 and see what happens.
$\displaystyle \diplaystyle \left ( \begin{array}{cc} a & b \\ c & d \end{array} \right ) \left ( \begin{array}{c} x \\ y \end{array} \right ) = \left ( \begin{array}{c} ax + by \\ cx + dy \end{array} \right ) = 0$
Solving the top equation for y gives:
$\displaystyle \displaystyle y = -\frac{a}{b} \cdot x$
Putting this into the bottom equation:
$\displaystyle \displaystyle cx + d \left ( - \frac{a}{b} \cdot x \right ) = 0$
$\displaystyle \displaystyle \left ( c - \frac{ad}{b} \right ) x = 0$
$\displaystyle \displaystyle \frac{cb - ad}{b} \right ) x = 0$
So either ad - bc = 0 or x = 0 (and hence y = 0).
This is, of course, much easier to do using matrix properties, but I thought it might be informative to show you the "algebraic" version for 2 x 2s.
-Dan
Thanks I get it. How would this be done using the matrix properties? That's what I tried to use but go messed up.
EDIT: I got messed up because how can you possibly solve for a variable without an augmented matrix?
Well, you can use an augmented matrix- in this case it would be
$\displaystyle \begin{bmatrix}{a & b & 0 \\ c & d & 0\end{bmatrix}$
because you are solving
$\displaystyle \begin{bmatrix}{a & b & 0 \\ c & d & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$
To "row reduce that you would divide the first row by a, to get a "1" in the first row, first column, and then subtract that new first row, multiplied by c, from the second row:
$\displaystyle \begin{bmatrix}{1 & \frac{b}{a} & 0 \\ 0 & d- \frac{bc}{a} & 0\end{bmatrix}$
or
$\displaystyle \begin{bmatrix}1 & \frac{b}{a} & 0 \\ 0 & \frac{ad- bc}{a} & 0 \end{bmatrix}$
The next step would be to get a "1" in the second row, second column, by dividing the second row by $\displaystyle \frac{ad- bc}{a}$.
Now do you see why "ad- bc" being 0 or not is important?
To solve the equation
$\displaystyle \begin{bmatrix}{a & b & 0 \\ c & d & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$
without using an "augmented matrix" or "row reduction", you could find the inverse matrix and multiply both sides by it:
$\displaystyle A^{-1}(AX)= A^{-1}B$
so $\displaystyle X= A^{-1}B$
as long as A has an inverse and so we would have that single solution. Of course, if B is the 0 vector, then $\displaystyle X= A^{-1}0= 0$ so that single solution is the 0 or "trivial" solution. Such an equation will have a "non-trivial" solution, then, if and only if A does not have an inverse. How can you determine if a 2 by 2 matrix has an inverse or not?