# Thread: trivial solution for a homogeneous system

1. ## trivial solution for a homogeneous system

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

2. Originally Posted by Jskid
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

$\displaystyle \displaystyle{ad-bc\neq 0\Longleftrightarrow A^{-1}\,\,exists\,\,\Longleftrightarrow A^{-1}\left(Ax\right)=A^{-1}(0)$ , and etc.

Tonio

3. Originally Posted by tonio
$\displaystyle \displaystyle{ad-bc\neq 0\Longleftrightarrow A^{-1}\,\,exists\,\,\Longleftrightarrow A^{-1}\left(Ax\right)=A^{-1}(0)$ , and etc.

Tonio
This isn't apparent to me, could someone please elaborate?

4. Can you say what exactly isn't apparent to you? Is it the fact that the inverse exists? Or are you not sure where tonio's last equation leads you? Or something else?

5. 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?

6. Originally Posted by Ackbeet
Can you say what exactly isn't apparent to you? Is it the fact that the inverse exists? Or are you not sure where tonio's last equation leads you? Or something else?
It is the fact that the inverse exists.

Originally Posted by HallsofIvy
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?
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?

7. Originally Posted by Jskid
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
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

8. 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?

9. 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?

10. Got it. One last thing concerning me is when you said shouldn't that be $\displaystyle \left[\begin{array}{cc} a & b \\ c & d \\ \end{array} \right]?$

11. Originally Posted by Jskid
Got it. One last thing concerning me is when you said shouldn't that be $\displaystyle \left[\begin{array}{cc} a & b \\ c & d \\ \end{array} \right]?$
Yes, you are correct.

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### Let T:R²---R² be a given by T(x,y)=(ax by,cx dy). Then show that T is an isomorphism iff ad-bc=0

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