# Linear Algebra Conditions

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 close merge delete

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

( 2017-02-15 13:37:59 +0100 )edit

Sort by » oldest newest most voted

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

more

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.

more

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

( 2017-02-16 00:33:42 +0100 )edit