I have the following matrix to solve

$\begin{bmatrix}2 & 1\\-4 & -2\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$

I have a solution of $\begin{bmatrix}x\\y\end{bmatrix} = c\begin{bmatrix}1\\-2\end{bmatrix}$

How did they get that solution? the determinant of the matrix is $0$, so why will there be non trivial solutions? I am currently taking Differential equations and I understand how to solve DEs using matrix, but this is a simple algebra problem and I am not sure how they got that? why is there a $c$ in the solution?

  • $\begingroup$ oh yes, the 2 is also negative in the matrix $\endgroup$ – Edgar Torres Nov 28 '16 at 3:16
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    $\begingroup$ The fact that the determinant is $0$ tells you that an inverse transformation cannot exist. Here, you are just asked to find the kernel of the matrix, which exists for all linear transformations, isomorphic or not. Plug in the solution vector and you can easily verify that it is the kernel. So is every scalar multiple of that vector. Kernel's are usually most easily found using the reduced row echelon form of the matrix. $\endgroup$ – infinitylord Nov 28 '16 at 3:18

Starting with the system of 2 equations in 2 variables

$$\begin{bmatrix}2 & 1\\-4 & -2\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$$

Add 2 times equation 1 to equation 2

$$\begin{bmatrix}2 & 1\\0 & 0\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$$

Now let's consider these two equations individually. The second is just $0=0$. That gives us exactly no information. The first equation, $2x+y=0$, is a line. Let's parametrize it.

Let $x=c$. Then $y = -2c$. So writing this back in matrix form we have

$$\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}c \\ -2c\end{bmatrix} = c\begin{bmatrix}1 \\ -2\end{bmatrix}$$

  • $\begingroup$ Thanks!! I understand now, but can c be 0 ? I tried to to solve by using 2x+1=0 and -4x-2y=0. I multiplied the first equation by two, and then I added both equations which results all 0s. is that not related to matrices? $\endgroup$ – Edgar Torres Nov 28 '16 at 3:32
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    $\begingroup$ $c$ can be any number. $\begin{bmatrix}x \\ y\end{bmatrix} = c\begin{bmatrix}1 \\ -2\end{bmatrix}$ means that there are actually an infinite number of solutions -- one for every value of $c$ that you could plug in. As for what you tried, when you add the two new equations together, you can replace one of the other two equations with it, but you should always have two equations. So the fact that one of the equations is $0=0$ just says that the other one carries all the information contained in the system of equations. $\endgroup$ – user137731 Nov 28 '16 at 3:35
  • $\begingroup$ You're probably going to want to learn a bit about Gaussian elimination so that you understand how to solve these matrix equations for this section in your ODEs class. $\endgroup$ – user137731 Nov 28 '16 at 3:38
  • $\begingroup$ aaa I understand now, thanks!! I remember doing gaussian elimination in college algebra, I just didn't knew that I had to use that method, my book doesn't say that, thanks again!! $\endgroup$ – Edgar Torres Nov 28 '16 at 3:40

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