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
x - 2y - 5z  =  3
5x + 2z  =  1

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
2x + y - z  =   2
5x + 4y - z  =  11

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

 Definition:   Eigenvalues and Eigenvectors Let A be an n x n matrix.  Then l is an eigenvalue for A with eigenvector v if            Av  =  lv

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 