# Why are linear equations usually written down as matrices?

I've been taught that for any system of linear equations, it has a corresponding matrix.

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?

Mark44
Mentor
I've been taught that for any system of linear equations, it has a corresponding matrix.

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?
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, and b is the column vector whose entries are 5 and 5. (x is 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.

• japplepie and suremarc
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, and b is the column vector whose entries are 5 and 5. (x is 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.
I see, thank you very much!

HallsofIvy