Constructing a 3x3 Linear system question

[Mohamed Abdul]

Homework Statement

Construct a 3 × 3 example of a linear system that has 9 different coefficients on the left hand side but rows 2 and 3 become zero in elimination. If the right hand sude of your system is <b1,b2,b3> (Imagine this is a column vector) then how many solutions does your system have for (i) b1 = 1, b2 = 10, b3 = 100, and (ii) b1 = b2 = b3 = 0?

Homework Equations

Gaussian elimination method I used here:
http://mathworld.wolfram.com/GaussianElimination.html

The Attempt at a Solution

I've tried setting up a matrix in the form C|d, but I'm not sure what to put for my coefficients. Should I be putting actual numbers in that would make the system 0? I'm just really stuck on this problem.

 

[hilbert2]

For example, if the second row of the matrix has elements that are 2 times the elements of first row, and the third row is 3 times the first row, as in

$\begin{bmatrix}1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9\end{bmatrix}$

and the RHS column vector is like this too, then the second and third rows don't contain any new information after reading the first row, and can be converted to the identity 0 = 0.

 

[Mohamed Abdul]

For example, if the second row of the matrix has elements that are 2 times the elements of first row, and the third row is 3 times the first row, as in

$\begin{bmatrix}1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9\end{bmatrix}$

and the RHS column vector is like this too, then the second and third rows don't contain any new information after reading the first row, and can be converted to the identity 0 = 0.

So I can write the matrix as you have above with actual numbers since they are just coefficients, then? Additionally, when solving the second part of the problem, would I find an element of column b, say b2 - b1, and set it equal to zero, plugging in the points. But would I then need an element of the column with b1 b2 and b3 all present?

 

[hilbert2]

Yes, you can write it with either actual numbers or as arbitrary symbols like $a,2a,3a$.

Now if the full system is something like

$\begin{bmatrix}1 & 2 & 3 | 1 \\ 2 & 4 & 6 | 1 \\ 3 & 6 & 9 | 1\end{bmatrix}$

you can immediately subtract two times the first row from the second row and get an identically false equation $0 = -1$, which means that the system doesn't have a solution at all. For some other choices of RHS, the equation can have an infinite number of solutions.

 

[Mohamed Abdul]

Yes, you can write it with either actual numbers or as arbitrary symbols like $a,2a,3a$.

Now if the full system is something like

$\begin{bmatrix}1 & 2 & 3 | 1 \\ 2 & 4 & 6 | 1 \\ 3 & 6 & 9 | 1\end{bmatrix}$

you can immediately subtract two times the first row from the second row and get an identically false equation $0 = -1$, which means that the system doesn't have a solution at all. For some other choices of RHS, the equation can have an infinite number of solutions.

This is what I got for my reduced matrix then. So to answer part b, which one of the b columns would I plug values into see if it's consistent considering none of them have b1 b2 and b3 in them?