Ask Your Question
0

Linear Algebra Conditions

asked 2017-02-15 12:30:02 +0200

this post is marked as community wiki

This post is a wiki. Anyone with karma >750 is welcome to improve it.

imgur.com/a/xIydC The answers is b) ab≠1, but I have no clue how to get to that answer... Can someone help me? :D

edit retag flag offensive close merge delete

Comments

This looks like homework. Could you please tell us how it is related to Sage, and what were your attempts in solving this ?

tmonteil gravatar imagetmonteil ( 2017-02-15 13:37:59 +0200 )edit

2 Answers

Sort by » oldest newest most voted
0

answered 2017-02-16 00:30:30 +0200

Daniel L gravatar image

The system of equations

x+by=-1

2ax+2y=5

is equivalent to the matrix-vector equation Ax=b, where A = matrix(2,1,(1,b,2a,2)), x = (x,y) and b = (-1,5). This equation has a unique solution exactly when det(A) is non-zero.

Now, det(A) = 2-2ab, which is nonzero precisely when ab != 1.

edit flag offensive delete link more

Comments

to echo tmoneil's comment, this post is more suited to mathstackexchange

Daniel L gravatar imageDaniel L ( 2017-02-16 00:33:42 +0200 )edit
0

answered 2017-02-23 15:32:07 +0200

Terry D. Destefano gravatar image

updated 2017-02-23 15:32:29 +0200

http://domyhomeworkfor.me/ told me this can easily be done with determinants. If a square matrix's determinant does not equal zero, then that square matrix will have an inverse hence having a unique solution. Since this is a 2x2 matrix, just compute the determinant with the condition that it cannot equal zero: (1)(2)-(2ab) =/= 0 2 =/= 2ab 1=/= ab

edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2017-02-15 12:30:02 +0200

Seen: 620 times

Last updated: Feb 23 '17