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Solutions of Ax = 0 in parametric vector form

Hi everyone! So here is the question:

Describe all solutions to Ax = 0 in parametric vector form, where A is row equivalent to the given matrix:

Attachment 28470

My thoughts so far:

I know that when two matrices are "row equivalent" it means that one of them can be changed to the other by doing row operations. So I assume here, that I can use the matrix in the picture above, insert it where A should be and do row operations on it. If I do that I get the row reduced augmented matrix:

Attachment 28471

Now, what I find strange here is that, first of all, there seems to be an x_{1}, x_{2}, x_{3}, and x_{4}, but I only have two rows? Somehow I feel like there should have been 4 rows in this matrix, because if I think about it, if this was an equation containing 4 unknowns, I would have to set up 4 simulations equations to solve it. Here I can only set up two which will be:

x_{1} + 0_{x2} + 9_{x3} -8_{x4} = 0

0_{x1} + 1_{x2} -4_{x3} + 5_{x5} = 0

I feel like this is impossible? I must be doing something wrong here, but I can't figure out what.

Thank you everyone for reading ^_^

Re: Solutions of Ax = 0 in parametric vector form

Hey Nora314.

What you have done is right: you basically have to introduce free parameters to account for the low number of rows relative to the number of columns.

Try introducing 2 parameters ( say u and v) to take into account the large number of columns. (Example x2 - 4v + 5u = 0).

Re: Solutions of Ax = 0 in parametric vector form

Thank you so much! Now I get that it does does not matter how many rows I have compared to columns, because I can always make up a free variable. What I did than is that I put x3 and x4 to be free variables. I wrote x1 and x2 in terms of x3 and x4 and than expressed it all as a parametric vector equation with x3 and x4 as weights.

I got the right answer than :)

Re: Solutions of Ax = 0 in parametric vector form

Or you can translate the matrix equation into two equations. Taking the unknown vector to be $\displaystyle \begin{bmatrix}a \\ b \\ c \\ d\end{bmatrix}$, this becomes

$\displaystyle \begin{bmatrix}1 & 3 & -3 & 7 \\ 0 & 1 & -4 & 5\end{bmatrix}\begin{bmatrix}a \\ b \\ c \\ c \end{bmatrix}= \begin{bmatrix}a+ 3b- 3c+ 7d \\ b- 4c+ d\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$ or just a+ 3b- 3c+ 7d= 0, b- 4c+ d= 0. Since there is no a in the second equation, we can solve it for b= 4c- d. Then the first equation becomes a+ 3(4c- d)- 3c+ 7d= a+ (12- 3)c+ (7- 3)d= a+ 9c+ 4d= 0 or a= -9c- 4d.

Then we can write $\displaystyle \begin{bmatrix}a \\ b \\ c \\ c\end{bmatrix}= \begin{bmatrix}-9c- 4d \\ 4c- d\\ c \\ d\end{bmatrix}= c\begin{bmatrix}-9 \\ 4 \\ 1 \\ 0\end{bmatrix}+ d\begin{bmatrix}-9 \\ -1 \\ 0 \\ 1\end{bmatrix}$

Re: Solutions of Ax = 0 in parametric vector form

Thank you so much for showing me another way to solve the problem, I found that method to be really intuitive.