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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