A system of linear equations is a collection of linear equations involving the same set of variables:
\[\begin{alignat}{7} 3x &\; + \;& 2y &\; - \;& z &\; = \;& 1 & \\ 2x &\; - \;& 2y &\; + \;& 4z &\; = \;& -2 & \\ -x &\; + \;& \tfrac{1}{2} y &\; - \;& z &\; = \;& 0 & \end{alignat}\]
More generally, we can write
\[\begin{alignat}{7} a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& b_1 \\ a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& b_2 \\ \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && &&& \;\vdots \\ a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& b_m \\ \end{alignat}\]
where we all parameters \(a_{ij}\) and \(b_i\) are known and we would like to find \(x_j\) that satisfy all these equations. If we have the same number \(n\) of unknown variables \(x_j\) as number of equations \(m\), and all these equations are independent, then there is a unique solution.
This is a fundamental problem in the domain of linear algebra, and we can use its power to find the solution. Accordingly we can write the equivalent problem with matrices and vectors:
\[\mathbf{A}= \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix},\quad \mathbf{x}=\begin{bmatrix}x_1\\x_2\\ \vdots \\x_n\end{bmatrix},\quad \mathbf{b}=\begin{bmatrix}b_1\\b_2\\ \vdots \\b_m\end{bmatrix}\]
such that
\[\mathbf{A}\mathbf{x}=\mathbf{b}\]
The initial example system would then look like this:
\[\begin{bmatrix}3 & 2 & -1 \\2 & -2 & 4 \\-1 & \tfrac{1}{2} & -1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} \;=\; \begin{bmatrix}1\\-2\\0\end{bmatrix}\]
Which we can solve explicitly with the LU-decomposition, or simply by using the Solve method:
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The resulting \(\mathbf{x}\) is \([1,\;-2,\;-2]\), hence the solution \(x=1,\;y=-2,\;z=-2\).
In F# the syntax is a bit lighter:
let A = matrix [[ 3.0; 2.0; -1.0 ]
[ 2.0; -2.0; 4.0 ]
[ -1.0; 0.5; -1.0 ]]
let b = vector [ 1.0; -2.0; 0.0 ]
let x = A.Solve(b) // 1;-2;-2
In practice, a linear equation system to be solved is often not in the standard form required to use the linear algebra approach. For example, let's have a look at the following system:
\[\begin{bmatrix}1 & 2 & 3 & 4\\2 & 3 & 4 & 5\\3 & 4 & 5 & 6\\4 & 5 & 6 & 7\end{bmatrix} \begin{bmatrix}0\\0\\V\\T\end{bmatrix} \;=\; \begin{bmatrix}F\\M\\20\\0\end{bmatrix}\]
The first two values of the solution vector \([0,\;0,\;V,\;T]\) are constant zero, so we can simplify the system to:
\[\begin{bmatrix}3 & 4\\4 & 5\\5 & 6\\6 & 7\end{bmatrix} \begin{bmatrix}V\\T\end{bmatrix} \;=\; \begin{bmatrix}F\\M\\20\\0\end{bmatrix}\]
Then we need to subtract the two unknowns from the right side back from the left (so that they become zero on the right side), by introducing a new column each. First we subtract \([F,\;0,\;0,\;0]^T\) from both sides:
\[\begin{bmatrix}3 & 4 & -1\\4 & 5 & 0\\5 & 6 & 0\\6 & 7 & 0\end{bmatrix} \begin{bmatrix}V\\T\\F\end{bmatrix} \;=\; \begin{bmatrix}0\\M\\20\\0\end{bmatrix}\]
Then we subtract \([0,\;M,\;0,\;0]^T\) from both sides the same way:
\[\begin{bmatrix}3 & 4 & -1 & 0\\4 & 5 & 0 & -1\\5 & 6 & 0 & 0\\6 & 7 & 0 & 0\end{bmatrix} \begin{bmatrix}V\\T\\F\\M\end{bmatrix} \;=\; \begin{bmatrix}0\\0\\20\\0\end{bmatrix}\]
Which is in standard from, so we can solve normally:
let A' = matrix [[ 3.0; 4.0; -1.0; 0.0 ]
[ 4.0; 5.0; 0.0; -1.0 ]
[ 5.0; 6.0; 0.0; 0.0; ]
[ 6.0; 7.0; 0.0; 0.0 ]]
let b' = vector [ 0.0; 0.0; 20.0; 0.0 ]
let x' = A'.Solve(b') // -140; 120; 60; 40