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Review of determinants and their properties A If A is a square n n matrix, it represents a linear function Rn Rn (with the same Rn as input and as...
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Linear algebra, 7 problems total (you can ignore the pages where my

professors goes on and on about stuff). I need a detailed solution guide. Preferably in digital format, but I do accept handwritten material. thanks

Review of determinants and their properties If A is a square n × n matrix, it represents a linear function R n A -→ R n (with the same R n as input and as output). The determinant of A is deFned as the factor by which this function multiplies volumes. SpeciFcally, if B in is some n -dimensional region in R n , and B out is the corresponding region obtained by applying A B in A R n B out R n Then det A = ± volume( B out ) volume( B in ) . The ± sign out front has to do with orientation, and will be discussed momentarily. Also note that “volume” is interpreted in an n -dimensional sense; for example in R 1 “volume” means length, in R 2 “volume” means area, in R 3 “volume” means volume, in R 4 “volume” is a 4d volume, etc. .. The deFnition of det A above does not depend on the choice of region B in . The volume of any region is multiplied by the same amount. ±or computations it’s convenient to choose B in to be the unit cube (or square, or hypercube, depending on the dimension). The unit cube has all its edges parallel to the standard basis vectors, and its volume is 1 B in = unit cube : edges 1 0 . . . 0 , 0 1 . . . 0 ,..., 0 0 . . . 1 , volume=1 When B in is the unit cube, B out is a “parallelepiped” whose edges are parallel to the column vectors of A . Thus if A = ( v 1 v 2 ··· v n ) then B out = parallelepiped: edges v 1 , v 2 ,..., v n Then we Fnd that det( A ) = ± volume(parallelepiped with edges v 1 , v 2 ,..., v n ) volume(unit cube) = ± volume(parallelepiped with edges v 1 , v 2 ,..., v n ) . ±or example, suppose we have a matrix A = ± a b c d ² that encodes a function from R 2 to R 2 . This function maps the unit square to a parallelogram with edges determined by the 1
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column vectors v 1 = ± a c ² and v 2 = ± b d ² v 1 v 2 1 0 0 1 R 2 A = ( v 1 v 2 ) R 2 On the previous homework, you found the volume of this parallelogram, and thus computed det A = ± ( ad - bc ) . For a picture of a 3d parallelepiped, see <en.wikipedia.org/wiki/Parallelepiped> The signs It’s finally time to talk about the ± sign. The sign is determined by two properties. First, consider the identity matrix A = I . The parallelepiped with edges parallel to the columns of I is just the unit cube itself, which has volume 1; so the definition above says det I = ± 1. One actually requires det I = 1 , with a plus sign. Second, the determinant is required to change sign whenever two columns of A are exchanged; for example, det ( v 1 v 2 ··· v n ) = - det ( v 2 v 1 ··· v n ) . (The sign changes when any two columns are swapped, not just the first two.) The change of sign in the second property might look strange, but it actually makes the determinant a nice smooth function. 1 To illustrate the change of sign, consider the sequence of parallelograms below. v 1 v 2 v 1 v 2 v 2 v 1 v 1 v 2 v 1 v 2 det( v 1 v 2 ) > 0 det( v 1 v 2 )=0 det( v 1 v 2 ) < 0 1 If you’ve taken calculus, compare the functions f ( x ) = | x | and f ( x ) = x . The first has a cusp at x = 0, where its derivative is discontinuous. The second has a continuous derivative – it’s nice and smooth. We want the determinant to be like the second function. 2
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