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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?

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- Thread starter japplepie
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- #1

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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

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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

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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.

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HallsofIvy

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