# Need help with a problem. Ax=b

• January 13th 2009, 03:22 PM
prometheos
Need help with a problem. Ax=b
We just started class and our first HW problem over a chapter we haven't gotten to yet has me stumped.

The instructor made this one up I believe, and the following is from the chalkboard;

[1 2 0 1 3 | 4 ] is the reduced row echelon form if [A.b] for the eqn.
[0 0 1 2 4 | -3] Ax=b. Find solution for x.
[0 0 0 0 0 | 0 ]

From what I can gather by reading ahead in the book is for an Ax=b situation the solution is given by x=A^-1 * b... or x equals the inverse of A times b. From what I have read, inverses can only be calculated from matrices that are square or nxn dimensions. Therefore, I am stumped.

Any help is greatly appreciated, even if you just show me a general method, so I can solve it on my own.
• January 13th 2009, 03:56 PM
Prove It
Quote:

Originally Posted by prometheos
We just started class and our first HW problem over a chapter we haven't gotten to yet has me stumped.

The instructor made this one up I believe, and the following is from the chalkboard;

[1 2 0 1 3 | 4 ] is the reduced row echelon form if [A.b] for the eqn.
[0 0 1 2 4 | -3] Ax=b. Find solution for x.
[0 0 0 0 0 | 0 ]

From what I can gather by reading ahead in the book is for an Ax=b situation the solution is given by x=A^-1 * b... or x equals the inverse of A times b. From what I have read, inverses can only be calculated from matrices that are square or nxn dimensions. Therefore, I am stumped.

Any help is greatly appreciated, even if you just show me a general method, so I can solve it on my own.

It's going to have infinitely many solutions.
• January 13th 2009, 04:03 PM
NonCommAlg
Quote:

Originally Posted by prometheos
We just started class and our first HW problem over a chapter we haven't gotten to yet has me stumped.

The instructor made this one up I believe, and the following is from the chalkboard;

[1 2 0 1 3 | 4 ] is the reduced row echelon form if [A.b] for the eqn.
[0 0 1 2 4 | -3] Ax=b. Find solution for x.
[0 0 0 0 0 | 0 ]

From what I can gather by reading ahead in the book is for an Ax=b situation the solution is given by x=A^-1 * b... or x equals the inverse of A times b. From what I have read, inverses can only be calculated from matrices that are square or nxn dimensions. Therefore, I am stumped.

Any help is greatly appreciated, even if you just show me a general method, so I can solve it on my own.

just write $Ax=b$ as a system of equations and solve it: $\begin{cases} x_1 + 2x_2 + x_4 + 3x_5=4 \\ x_3 + 2x_4 + 4x_5 =-3 \end{cases}.$ we have two equations and 5 variables.

the solutions are: $x_1=4-2x_2-x_4-x_5, \ x_3=-3-2x_4-4x_5.$ note that $x_2, x_4, x_5$ are free variables.
• January 13th 2009, 05:23 PM
prometheos
Ah, I think I see now what my problem was. The wording of the question led me to believe it wasn't a simple solution. Go go new math class language. Thank you.