Message Boards Message Boards

0
|
2066 Views
|
1 Reply
|
1 Total Likes
View groups...
Share
Share this post:

How exactly would you do this on mathematica

Anonymous User
Anonymous User
Posted 11 years ago

For which values of a will the following systems of linear equations have no solutions, one solution , or an infinite number of solutions? x+2y-3z=4 2x-y+5z=2 4x+3y+a^2z=a+3

POSTED BY: Anonymous User

Several functions in Mathematica can do it. Example: use the Reduce function:

Reduce[{x + 2 y - 3 z == 4 2 x - y + 5 z == 2 4 x + 3 y + a^2 z == a + 3}, {x, y, z}]

enter image description here

From the result you should always get the linear system solvable for $a \in \mathbb{R}$

POSTED BY: Shenghui Yang
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract