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# 5.1: Review of 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

\begin{align} &4x &+y &&+3z &=2 \\ &x &-2y &&-5z &=3 \\ &5x & &&+2z &=1 \end{align}

Solution

We write this system as the matrix equation

$Ax = b$

where

$A = \begin{pmatrix} 4 &1 &3 \\ 1 &-2 &-5 \\ 5 &0 &2 \end{pmatrix} \;\;\; b=\begin{pmatrix} 2 \\ 3 \\1 \end{pmatrix}$

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

$x=A^{-1}b$

Using a calculator we find that

$A^{-1} = \dfrac{1}{13} \begin{pmatrix} 4 & 27 &-10 \\ 2 &7 &-5 \\ -1 & -23 & 9 \end{pmatrix}$

Multiplying by $$b$$ gives

$x=\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}$

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

\begin{align} &3x &+2y &&-z &=5 \\ &2x &+y &&-z &=2 \\ &5x &+4y &&-z &=11 \end{align}

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

$\left(\begin{array}{ccc|c} 3 &2 &-1 &5 \\ 2 &1 &-1 &2 \\ 5 &4 &-1 & 11 \end{array} \right)$

to get

$\left(\begin{array}{ccc|c} 1 &0 &-1 &-1 \\ 0 &1 &1 &4 \\ 0 &0 &0 &0 \end{array}\right)$

Putting this back into equation form, we get

$x-z=-1 \;\;\; \text{and} \;\;\; y+z =4$

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 $$v_1,...,v_n$$ are called linearly independent if

$c_1v_1+...+c_nv_n=0$

implies that all of the constants $$c_i$$ are zero.  A theorem from linear algebra tell us that if we have $$n$$ vectors in $$\mathbb{R}^n$$ 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

$u = <1,4,-2> \;\;\; v = <0,3,5> \;\;\; \text{and} \;\;\; w = <1,2,3>$

are linearly independent.

Solution

We find the determinant

$det\begin{pmatrix} 1 &0 &1 \\ 4 &3 &2 \\ -2 &5 &3 \end{pmatrix} =25$

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 $$n \times n$$ matrix.  Then $$\lambda$$ is an eigenvalue for $$A$$ with eigenvector $$v$$ if

$Av = \lambda v$

Example

Find the eigenvalues and eigenvectors for

$A=\begin{pmatrix} 6 &4 \\ -3 & -1 \end{pmatrix}$

Solution

If

$Av = \lambda v$

then

$A -\lambda = 0$

Taking determinants of both sides, we get

\begin{align}(6 - \lambda)(-1 - \lambda) + 12 &= 0 \\ \lambda ^2 - 5\lambda + 6 &= 0 \\ (\lambda - 2)(\lambda - 3) &= 0 \end{align}

The eigenvalues are

$\lambda=2 \;\;\; \text{and} \;\;\; \lambda=3$

To find the eigenvectors, we plug the eigenvalues into the equation

$A -\lambda = 0$

and find the null space of the left hand side.  For the eigenvalue $$\lambda = 2$$, we have

$A-\lambda I = \begin{pmatrix} 4 &4 \\ -3 &-3 \end{pmatrix}$

The first row gives

$y = -x$

so that an eigenvector corresponding to the eigenvalue $$\lambda = 2$$ is

$v_2 = \begin{pmatrix}1\\-1 \end{pmatrix}$

For the eigenvalue $$\lambda = 3$$, we have

$A-\lambda I = \begin{pmatrix} 3&4 \\ -3 &-4 \end{pmatrix}$

The first row gives

$3y = -4x$

so that an eigenvector corresponding to the eigenvalue $$\lambda = 3$$ is

$v_3 = \begin{pmatrix}3\\-4\end{pmatrix}$

Typically, we want to normalize the eigenvectors, that is find unit eigenvectors.  We get

$u_2 =\begin{pmatrix}\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{pmatrix} \;\;\; \text{and} \;\;\; u_3 =\begin{pmatrix}\frac{3}{5}\\-\frac{-4}{5}\end{pmatrix}$