ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Thu, 23 Feb 2017 15:32:07 +0100Linear Algebra Conditionshttps://ask.sagemath.org/question/36625/linear-algebra-conditions/ 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 Wed, 15 Feb 2017 12:30:02 +0100https://ask.sagemath.org/question/36625/linear-algebra-conditions/Comment by tmonteil for <p>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 </p>
https://ask.sagemath.org/question/36625/linear-algebra-conditions/?comment=36627#post-id-36627This looks like homework. Could you please tell us how it is related to Sage, and what were your attempts in solving this ?Wed, 15 Feb 2017 13:37:59 +0100https://ask.sagemath.org/question/36625/linear-algebra-conditions/?comment=36627#post-id-36627Answer by Daniel L for <p>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 </p>
https://ask.sagemath.org/question/36625/linear-algebra-conditions/?answer=36634#post-id-36634The 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.
Thu, 16 Feb 2017 00:30:30 +0100https://ask.sagemath.org/question/36625/linear-algebra-conditions/?answer=36634#post-id-36634Comment by Daniel L for <p>The system of equations</p>
<p>x+by=-1</p>
<p>2ax+2y=5</p>
<p>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.</p>
<p>Now, det(A) = 2-2ab, which is nonzero precisely when ab != 1.</p>
https://ask.sagemath.org/question/36625/linear-algebra-conditions/?comment=36635#post-id-36635to echo tmoneil's comment, this post is more suited to mathstackexchangeThu, 16 Feb 2017 00:33:42 +0100https://ask.sagemath.org/question/36625/linear-algebra-conditions/?comment=36635#post-id-36635Answer by Terry D. Destefano for <p>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 </p>
https://ask.sagemath.org/question/36625/linear-algebra-conditions/?answer=36711#post-id-36711[domyhomeworkfor.me](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=/= abThu, 23 Feb 2017 15:32:07 +0100https://ask.sagemath.org/question/36625/linear-algebra-conditions/?answer=36711#post-id-36711