Understanding Matrices Infinite: Case 3

[jwxie]

I don't understand the asepcts of matrices infinite.

Let say I have 3 x 3 matrix

A = [ (A1) 2 3; 2 3 4; 4 4 5 ]
b = [ (b1); 2; 3]

I want to find all three conditions:
case 1 det =/ 0 - > unique solution
case 2 det = 0, no solution
case 3 infinite solution

I am actually writing a program for these 3 cases. I am only allow to change A1 and b2 (they are variables). I know how to program them. But what I don't understand is the concept of matrices in terms of the 3 cases.

Can someone explains, how does one gets case 3?
i see how case 2 you can use det to determine something solution or no solution.
what about case 3?

 

[HallsofIvy]

I don't understand the asepcts of matrices infinite.

Let say I have 3 x 3 matrix

A = [ (A1) 2 3; 2 3 4; 4 4 5 ]
b = [ (b1); 2; 3]

I have no idea what this means. Do you mean you have the matrix equation Ax= b
\begin{bmatrix}A_1 & 2 & 3 \\ 2 & 3 & 4 \\ 4 & 4 & 5\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}b_1 \\ 2 \\ 3\end{bmatrix}?

I want to find all three conditions:
case 1 det =/ 0 - > unique solution
case 2 det = 0, no solution
case 3 infinite solution

I am actually writing a program for these 3 cases. I am only allow to change A1 and b2 (they are variables). I know how to program them. But what I don't understand is the concept of matrices in terms of the 3 cases.

Can someone explains, how does one gets case 3?
i see how case 2 you can use det to determine something solution or no solution.
what about case 3?

A matrix equation, Ax= b, has a unique solution if and only if its determinant is non-zero. That means you can row-reduce A without getting a row consisting only of "0"s. If the determinant is 0, then row-reducing A will give at least one row consisting of "0"s. If the corresponding rows of b, row-reduced along with A, are also 0 there are an infinite number of solutions. If any of them are not 0, there is no solution.

 

[jwxie]

Yes. Thanks for the response. Here is the Matrix

3 2 0 | 4
0 1 2 | 0
4 0 0 | 0

I was looking at the reduction form - so if I make the first row reduces by 1/3 I will get 1 2/3 0 and 4/3
Now, I have the first two rows fit in the diagonally form of 1 1 1. But the third room, how do I make the 3rd zero into 1 so I have 1 1 1 for each row?

Work out det(A) as a function of alpha. All values where det(A) is not 0 are unique solutions.

For the one value of alpha where det(A) = 0, I would use row reduction on the system to find the values of beta that produce an inconsistent (no solutions) or consistent (infinite solutions) system.

You have your cases listed incorrectly, which will make it hard to solve your problem. They should be:

case 1 det A /= 0 -> unique solution
case 2 det A = 0, no solution
case 3 det A = 0, infinite solutions

The determinant of A depends only on A, so you need b to distinguish between cases 2 and 3.

 

[HallsofIvy]

Yes. Thanks for the response. Here is the Matrix

3 2 0 | 4
0 1 2 | 0
4 0 0 | 0

I was looking at the reduction form - so if I make the first row reduces by 1/3 I will get 1 2/3 0 and 4/3
Now, I have the first two rows fit in the diagonally form of 1 1 1. But the third room, how do I make the 3rd zero into 1 so I have 1 1 1 for each row?

What happened to A1 and b1?

Any way, you can, as you say, multiply the first row by 1/3 to get a 1 in the first cell of the first row. But you should not worry about the other columns until you have completed the first column. You want
\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}
and already have a "0" in the second row so just subtract 4 times that (new) first row from the third:
\left[\begin{array}{ccc}1 & \frac{2}{3} & 0 \\ 0 & 1 & 2 \\ 0 & -\frac{8}{3} & 0\end{array}\right |\left|\begin{array}{c}\frac{4}{3} \\ 0 \\ -\frac{16}{3}\end{array}\right]

Now, you already have a "1" in the "pivot" position, the second cell in the second row. To get rid of that "-8/3" add 8/3 times the second row to the third:
\left[\begin{array}{ccc}1 & \frac{2}{3} & 0 \\ 0 & 1 & 2 \\ 0 & 0 & \frac{16}{3}\end{array}\right |\left|\begin{array}{c}\frac{4}{3} \\ 0 \\ -\frac{16}{3}\end{array}\right]

Notice that, in doing that the final pivot position, the third cell in the third row, is no longer 0. The final step to get it in "echelon" form would be to divide the third row by 16/3.
If you want it in diagonal form, it is easy to "back substitute".

By the way, in talking about matrices, the standard English term is "cell", not "room".