Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

5.1: Review of Linear Algebra

In this discussion, we expect some familiarity with matrices. We will rely heavily on calculators and computers to work out the problems. Consider some examples.

Example \(\PageIndex{1}\)

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 \(\PageIndex{2}\)

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 \(\PageIndex{3}\)

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 \(\PageIndex{4}\)

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

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

Contributors