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 Review of Some Linear Algebra In this discussion, we expect some familiarity with matrices. For a review of the basics click here. We will rely heavily on calculators and computers to work out the problems. Consider some examples. 
 Example Solve the system of equations         4x + y +
3z  =  2 Solution We write this system as the matrix equation Ax = b where         
 To solve this matrix equation we take the inverse of both sides (which is possible since the determinant of A is -13 which is not equal to zero. We have x = A-1b Using a calculator we find that         
 Multiplying by b gives         
 What we mean by "x" is the vector <x,y,z>. The solution is x = 1 y = 4 z = -2 
 Example Find the solution of         3x + 2y - z  =  5 Solution A quick check shows that we cannot solve this problem in the same way, since the determinant of A is 0. Instead, we rref the augmented matrix         
 to get         
 Putting this back into equation form, we get x - z = -1 and y + z = 4 We write this as x = -1 + z y = 4 - z z = z Letting z = t be the parameter we get parametric equations for the solution set x = -1 + t y = 4 - t z = t 
 Recall that vectors v1, ..., vn are called linearly independent if c1v1 + ... + c2v2 = 0 implies that all of the constants ci are zero. A theorem from linear algebra tell us that if we have n vectors in Rn then they will be linearly independent if and only if the determinant of the matrix whose columns are these vectors has nonzero determinant. 
 Example Show that the vectors u = <1,4,-2> v = <0,3,5> and w = <1,2,3> are linearly independent. Solution We find the determinant         
 Since the determinant is nonzero, the vectors are linearly independent. 
 For systems of differential equations, eigenvalues and eigenvectors play a crucial role. We recall their definitions below 
 
 Example Find the eigenvalues and eigenvectors for         
 Solution If Av = lv then A - lI = 0 Taking determinants of both sides, we get (6 - l)(-1 - l) + 12 = 0 l2 - 5l + 6 = 0 (l - 2)(l - 3) = 0 The eigenvalues are l = 2 and l = 3 To find the eigenvectors, we plug the eigenvalues into the equation A - lI = 0 and find the null space of the left hand side. For the eigenvalue l = 2, we have           
 The first row gives y = -x so that an eigenvector corresponding to the eigenvalue l = 2 is         
 For the eigenvalue l = 3, we have           
 The first row gives 3y = -4x so that an eigenvector corresponding to the eigenvalue l = 3 is         
 Typically, we want to normalize the eigenvectors, that is find unit eigenvectors. We get           
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