- #1

- 93

- 0

Why do people sometimes use systems of linear equations to describe something and other times matrices?

Is it all just a way of writing things down faster or are there things you could do to matrices that you couldn't do to linear equations?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter japplepie
- Start date

- #1

- 93

- 0

Why do people sometimes use systems of linear equations to describe something and other times matrices?

Is it all just a way of writing things down faster or are there things you could do to matrices that you couldn't do to linear equations?

- #2

Mark44

Mentor

- 35,128

- 6,873

Mostly matrices are a shorthand way of writing a system of linear equations, but there is one other advantage for certain systems : the ability to use a matrix inverse to solve the system.

Why do people sometimes use systems of linear equations to describe something and other times matrices?

Is it all just a way of writing things down faster or are there things you could do to matrices that you couldn't do to linear equations?

For example, suppose we have this system:

2x + y = 5

x + 3y = 5

This system can be written in matrix form as:

##\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 5 \\5 \end{bmatrix}##

Symbolically, the system is A

Because I cooked this example up, I know that A has an inverse; namely ##A^{-1} = \frac 1 5 \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}##

If I apply this inverse to both sides of A

##= \begin{bmatrix} 2 \\1 \end{bmatrix}##

From this I see that x = 2 and y = 1. You can check that this is a solution by substituting these values in the system of equations.

- #3

- 93

- 0

I see, thank you very much!Mostly matrices are a shorthand way of writing a system of linear equations, but there is one other advantage for certain systems : the ability to use a matrix inverse to solve the system.

For example, suppose we have this system:

2x + y = 5

x + 3y = 5

This system can be written in matrix form as:

##\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 5 \\5 \end{bmatrix}##

Symbolically, the system is Ax=b, where A is the matrix of coefficients on the left, andbis the column vector whose entries are 5 and 5. (xis the column vector of variables x and y.)

Because I cooked this example up, I know that A has an inverse; namely ##A^{-1} = \frac 1 5 \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}##

If I apply this inverse to both sides of Ax=b, I get ##A^{-1}Ax = A^{-1}b = \frac 1 5 \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 5 \\5 \end{bmatrix}##

##= \begin{bmatrix} 2 \\1 \end{bmatrix}##

From this I see that x = 2 and y = 1. You can check that this is a solution by substituting these values in the system of equations.

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 964

- #5

FactChecker

Science Advisor

Gold Member

- 6,369

- 2,514

Share: