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Mathematics LibreTexts

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

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

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