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Linear Algebra Conditions

asked 2017-02-15 12:30:02 +0100

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This post is a wiki. Anyone with karma >750 is welcome to improve it. The answers is b) ab≠1, but I have no clue how to get to that answer... Can someone help me? :D

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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 +0100 )edit

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answered 2017-02-16 00:30:30 +0100

Daniel L gravatar image

The system of equations



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.

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

Daniel L gravatar imageDaniel L ( 2017-02-16 00:33:42 +0100 )edit

answered 2017-02-23 15:32:07 +0100

Terry D. Destefano gravatar image

updated 2017-02-23 15:32:29 +0100 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

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Asked: 2017-02-15 12:30:02 +0100

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Last updated: Feb 23 '17