**Give an example of a matrix A for which the system Ax= b can be solved for any b, but not necessarily uniquely.**

Is there a way to do this that does

*not*involve "brute force"?

Is there a way to do this that does

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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Maybe, but I'm not seeing it.Give an example of a matrix A for which the system Ax= b can be solved for any b, but not necessarily uniquely.

Is there a way to do this that doesnotinvolve "brute force"?

In any case, "brute force" isn't too bad here, since you can pick the matrix to be any size you like. So let A, x, and b be the following system:

```
[ a b ][x]
[ c d ][y] = [t]
```

This system will be dependent (and thus not uniquely solvable) if the second row is a multiple of the first, so you can pick anything you like for the first row, and then make the second a muliple of it.