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.  



Solve the system of equations

        4x + y + 3z  =  2
        x - 2y - 5z  =  3
        5x + 2z  =  1


We write this system as the matrix equation 

        Ax  =  b



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



Find the solution of 

        3x + 2y - z  =  5
        2x + y - z  =   2
        5x + 4y - z  =  11


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.



Show that the vectors 

        u  =  <1,4,-2>    v  =  <0,3,5>    and    w  =  <1,2,3> 

are linearly independent.


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



Find the eigenvalues and eigenvectors for 




        Av  =  lv


        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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