foxsboris.naumov1994 Posted February 15, 2017 Share Posted February 15, 2017 http://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? Link to comment Share on other sites More sharing options...
imatfaal Posted February 15, 2017 Share Posted February 15, 2017 please post question here - lots of members will not go to third party site to download Link to comment Share on other sites More sharing options...
zztop Posted February 15, 2017 Share Posted February 15, 2017 http://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? Hint: the determinant of the system must not be zero Link to comment Share on other sites More sharing options...
foxsboris.naumov1994 Posted February 23, 2017 Author Share Posted February 23, 2017 (edited) Well here is the answer SNIP gave 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 Edited June 5, 2022 by Phi for All ad links removed by moderator Link to comment Share on other sites More sharing options...
zztop Posted February 23, 2017 Share Posted February 23, 2017 On 2/23/2017 at 7:45 AM, foxsboris.naumov1994 said: Well here is the answer SNIP gave 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 Did you look at the post just above yours? Link to comment Share on other sites More sharing options...
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