MAT223: 7_Matrix

Matrix §

Coefficient matrix §

• Every system (or single) of linear equations can be rewritten as a matrix equation of the form

  $A\vec{x}=\vec{b}$, where $A$ is a coefficient matrix, $\vec{x}$ is a column vector of variables, and $\vec{b}$ is a column vector of constants

  $$A:\{\begin{array}{c}x+2y-2z=-15\\ 2x+y-5zz=-21\\ x-4y+z=18\end{array}\text{can be writen as}\left[\begin{array}{ccc}1& 2& -2\\ 2& 1& -5\\ 1& -4& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ x\end{array}\right]=\left[\begin{array}{c}-15\\ -21\\ 18\end{array}\right]$$

The column Picture §

• Using the column interpretation of matrix to rewrite system $A$:

  $$\left[\begin{array}{ccc}1& 2& -2\\ 2& 1& -5\\ 1& -4& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ x\end{array}\right]=x\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]+y\left[\begin{array}{c}2\\ 1\\ -4\end{array}\right]+z\left[\begin{array}{c}-2\\ -5\\ 1\end{array}\right]=\left[\begin{array}{c}-15\\ -21\\ 18\end{array}\right]$$

• This interpretation is mostly used to find the solutions to the system $A$

  • What linear combinations of the columns of $A$ would give the constant vector

The Row Picture §

• Let ${\vec{r}}_1,{\vec{r}}_3,{\vec{r}}_3$ be the rows of the coefficient matrix for $A$:

  $$\left[\begin{array}{ccc}1& 2& -2\\ 2& 1& -5\\ 1& -4& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ x\end{array}\right]=\left[\begin{array}{c}{\vec{r}}_1\\ {\vec{r}}_2\\ {\vec{r}}_3\end{array}\right]\vec{u}=\left[\begin{array}{c}{\vec{r}}_1\cdot \vec{u}\\ {\vec{r}}_2\cdot \vec{u}\\ {\vec{r}}_3\cdot \vec{u}\end{array}\right]=\left[\begin{array}{c}-15\\ -21\\ 18\end{array}\right]$$

• This interpret the solutions to the system $A$ as vectors whose dot product with ${\vec{r}}_1$ is $-15$, ${\vec{r}}_2$ is $-21$, ${\vec{r}}_3$ is $18$. This perspective is useful with the additional geometric definition of the dot product.

  • Especially when the right side of the equations is all zeros

Homogeneous Systems §

$$B:\begin{array}{cc}x+2y-2z& =0\\ 2x+y-5zz& =0\\ x-4y+z& =0\end{array}⟺\left[\begin{array}{ccc}1& 2& -2\\ 2& 1& -5\\ 1& -4& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ x\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$$

• This relates to whether the column of vectors of the coefficient matrix are linearly independent
• Let ${\vec{r}}_1,{\vec{r}}_3,{\vec{r}}_3$ be the rows of the coefficient matrix for $B$, what vectors are orthogonal to them

Find all vectors orthogonal to $\vec{a}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ and $\vec{b}=\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]$ §

To find all vectors orthogonal to $\vec{a}$ and $\vec{b}$ we need to find vectors $\vec{x}$ satisfying $\vec{a}\cdot \vec{x}=0$ , $\vec{a}\cdot \vec{x}=0$. This is equivalent to solving the matrix equation

$$\left[\begin{array}{c}\vec{a}\\ \vec{b}\end{array}\right]\vec{u}=\left[\begin{array}{c}\vec{a}\cdot \vec{x}\\ \vec{b}\cdot \vec{x}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ x\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$

By row reducing $A$, we get

$$\text{rref}(A)=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\end{array}\right]$$

And so the complete solution expressed in vector form is

$$\vec{x}=t\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$$

Let $\mathcal{Q}$ be the hyperplane specified in vector form by $\vec{x}=t\left[\begin{array}{c}1\\ 1\\ -1\\ 1\end{array}\right]+s\left[\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right]+r\left[\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right]+\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$, find a normal vector for $\mathcal{Q}$ and write $\mathcal{Q}$ in normal form §

Like the above example, since normal vectors for $\mathcal{Q}$ need to be orthogonal to ${\vec{d}}_1=\left[\begin{array}{c}1\\ 1\\ -1\\ 1\end{array}\right]$, ${\vec{d}}_2=\left[\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right]$, ${\vec{d}}_3=\left[\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right]$, find the normal vectors by solving

$$\left[\begin{array}{cccc}1& 1& -1& 1\\ 0& 1& 0& 1\\ 2& 0& 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ x\\ w\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$$

By row reducing $A$, we get

$$\text{rref}(A)=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 1\\ 0& 0& 1& 0\end{array}\right]$$

And so we get that the complete solution expressed in vector form is

$$\vec{x}=t\left[\begin{array}{c}0\\ -1\\ 0\\ 1\end{array}\right]$$