View Full Version : Mathematic exam in Ukraine
Nadin
Jan 7, 2008, 02:02 PM
:) Hello. I am Ukrainian, so I apologize to my English. Help me, please, to do my work.
{x-4y+2z=-5,
{4x+y-3z=-3,
{2x+3y+4z=1.
(1 -4 2) (x) (-5)
(4 1 -3) (y) (-3)
(2 3 4) (z) (1)
AX=B | A (-1)
X=A(-1) B
A=
(1 -4 2)
(4 1 -3)
(2 3 4)
(A|E) --> (E|A(-1))
What next?
Nadin
Jan 7, 2008, 04:23 PM
:confused: People, help me, please. I really don't understand it. If I wrote matrix correct?
Nadin
Jan 7, 2008, 04:49 PM
If you can't read correctly you can see this sum on the paper. Please, do this sum!!
Please read this ANNOUNCEMENT (https://www.askmehelpdesk.com/math-sciences/announcement-u-b-read-first-expectations-homework-help-board-b-u.html)
kp2171
Jan 7, 2008, 05:31 PM
I was wrong... she is attempting it
kp2171
Jan 7, 2008, 05:32 PM
Scanned through too fast at first and thought it was just the original question posted, not any of her work...
My mistake.
Nadin
Jan 7, 2008, 05:37 PM
:confused: OK. I apologize. I am really, really sorry. This math just confused me and in my country is 2:33 a.m. It is too late. So, do I deserve a forgiveness and can I wait for anybody help?
galactus
Jan 7, 2008, 05:39 PM
What is it you're trying to do exactly? Solve the system? Find the inverse?
\left[\begin{array}x\\y\\z\end{array}\right]=A^{-1}B
\left[\begin{array}x\\y\\z\end{array}\right]=\underbrace{\begin{bmatrix}1&-4&2\\4&1&-3\\2&3&4\end{bmatrix}^{-1}}_{\text{A^-1}}\cdot\overbrace{\begin{bmatrix}-5\\-3\\1\end{bmatrix}}^{\text{B}}
Find A^{-1}, the inverse of your given matrix, and multiply by B to find the solutions to the system.
You can also just use reduced row echelon to find the solutions. Pick your poison.
If you want to know how to find an inverse, it's tedious regardless of how you go about it.
One way is elementary row operations, which looks like what you're attempting?
\left[\begin{array}{ccc|ccc}1&-4&2&1&0&0\\4&1&-3&0&1&0\\2&3&4&0&0&1\end{array}\right]
Now, use Gaussian elimination to switch the matrices around. That is, do the row operations until the identity matrix is on the left of the vertical line. Then your inverse will be on the right. See what I mean?
Nadin
Jan 7, 2008, 05:55 PM
:rolleyes: I try to write it with matrix form and solve it with turned matrix. I wrote it with matrix. But you know, I just can't get how to solve matrix. I try to understand!
Nadin
Jan 7, 2008, 06:11 PM
:rolleyes: probably:
(1 -4 2 | 1 0 0) (1 -4 2 | 1 0 0)
(4 1 -3 | 0 1 0) (0 17 -11| -4 1 0)
(2 3 4 | 0 0 1) (0 11 0 | -2 0 1)
I think, I start to understand. Yes?
Nadin
Jan 7, 2008, 06:27 PM
:confused: No, I can't do that. It is too hard for me. I understand that I need to see:
(1 0 0 |? )
(0 1 0 |? )
(0 0 1 |? )
But how to make it?:p
Nadin
Jan 8, 2008, 07:10 AM
:rolleyes: So, can anyboby help me to understand matrix and solve this sum??
Nadin
Jan 8, 2008, 02:20 PM
Hello. I think I done it. I am not sure if I did it right, but I hope it's right.
galactus
Jan 8, 2008, 04:14 PM
Yes, Nadin, that is correct. Good job:) :D
Nadin
Jan 9, 2008, 08:24 AM
:) :cool: :D ;) OK. Thank you very much.