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<h1><a name="Linear-Equation-Systems" class="anchor" href="#Linear-Equation-Systems">Linear Equation Systems</a></h1>
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<p>A system of linear equations is a collection of linear equations involving the same set of variables:</p>
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<p><span class="math">\[\begin{alignat}{7}
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3x &\; + \;& 2y &\; - \;& z &\; = \;& 1 & \\
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2x &\; - \;& 2y &\; + \;& 4z &\; = \;& -2 & \\
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-x &\; + \;& \tfrac{1}{2} y &\; - \;& z &\; = \;& 0 &
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\end{alignat}\]</span></p>
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<p>More generally, we can write</p>
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<p><span class="math">\[\begin{alignat}{7}
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a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& b_1 \\
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a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& b_2 \\
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\vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && &&& \;\vdots \\
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a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& b_m \\
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\end{alignat}\]</span></p>
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<p>where we all parameters <span class="math">\(a_{ij}\)</span> and <span class="math">\(b_i\)</span> are known and we would like to find <span class="math">\(x_j\)</span> that satisfy
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all these equations. If we have the same number <span class="math">\(n\)</span> of unknown variables <span class="math">\(x_j\)</span> as number of
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equations <span class="math">\(m\)</span>, and all these equations are independent, then there is a unique solution.</p>
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<p>This is a fundamental problem in the domain of linear algebra, and we can use its power to find the solution.
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Accordingly we can write the equivalent problem with matrices and vectors:</p>
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<p><span class="math">\[\mathbf{A}=
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\begin{bmatrix}
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a_{11} & a_{12} & \cdots & a_{1n} \\
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a_{21} & a_{22} & \cdots & a_{2n} \\
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\vdots & \vdots & \ddots & \vdots \\
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a_{m1} & a_{m2} & \cdots & a_{mn}
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\end{bmatrix},\quad
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\mathbf{x}=\begin{bmatrix}x_1\\x_2\\ \vdots \\x_n\end{bmatrix},\quad
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\mathbf{b}=\begin{bmatrix}b_1\\b_2\\ \vdots \\b_m\end{bmatrix}\]</span></p>
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<p>such that</p>
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<p><span class="math">\[\mathbf{A}\mathbf{x}=\mathbf{b}\]</span></p>
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<p>The initial example system would then look like this:</p>
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<p><span class="math">\[\begin{bmatrix}3 & 2 & -1 \\2 & -2 & 4 \\-1 & \tfrac{1}{2} & -1\end{bmatrix}
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\begin{bmatrix}x\\y\\z\end{bmatrix}
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\;=\;
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\begin{bmatrix}1\\-2\\0\end{bmatrix}\]</span></p>
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<p>Which we can solve explicitly with the LU-decomposition, or simply by using the Solve method:</p>
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<table class="pre"><tr><td class="lines"><pre class="fssnip"><span class="l">1: </span>
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<span class="l">2: </span>
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<span class="l">3: </span>
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<span class="l">4: </span>
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<span class="l">5: </span>
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<span class="l">6: </span>
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<span class="l">7: </span>
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</pre></td>
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<td class="snippet"><pre class="fssnip highlighted"><code lang="csharp"><span class="k">var</span> A <span class="o">=</span> Matrix<<span class="k">double</span>>.Build.DenseOfArray(<span class="k">new</span> <span class="k">double</span>[,] {
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{ <span class="n">3</span>, <span class="n">2</span>, <span class="n">-1</span> },
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{ <span class="n">2</span>, <span class="n">-2</span>, <span class="n">4</span> },
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{ <span class="n">-1</span>, <span class="n">0.5</span>, <span class="n">-1</span> }
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});
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<span class="k">var</span> b <span class="o">=</span> Vector<<span class="k">double</span>>.Build.Dense(<span class="k">new</span> <span class="k">double</span>[] { <span class="n">1</span>, <span class="n">-2</span>, <span class="n">0</span> });
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<span class="k">var</span> x <span class="o">=</span> A.Solve(b);
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</code></pre></td></tr></table>
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<p>The resulting <span class="math">\(\mathbf{x}\)</span> is <span class="math">\([1,\;-2,\;-2]\)</span>, hence the solution <span class="math">\(x=1,\;y=-2,\;z=-2\)</span>.</p>
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<p>In F# the syntax is a bit lighter:</p>
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<table class="pre"><tr><td class="lines"><pre class="fssnip"><span class="l">1: </span>
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<span class="l">2: </span>
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<span class="l">3: </span>
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<span class="l">4: </span>
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<span class="l">5: </span>
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</pre></td>
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<td class="snippet"><pre class="fssnip highlighted"><code lang="fsharp"><span class="k">let</span> <span class="i">A</span> <span class="o">=</span> <span class="i">matrix</span> [[ <span class="n">3.0</span>; <span class="n">2.0</span>; <span class="o">-</span><span class="n">1.0</span> ]
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[ <span class="n">2.0</span>; <span class="o">-</span><span class="n">2.0</span>; <span class="n">4.0</span> ]
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[ <span class="o">-</span><span class="n">1.0</span>; <span class="n">0.5</span>; <span class="o">-</span><span class="n">1.0</span> ]]
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<span class="k">let</span> <span class="i">b</span> <span class="o">=</span> <span class="i">vector</span> [ <span class="n">1.0</span>; <span class="o">-</span><span class="n">2.0</span>; <span class="n">0.0</span> ]
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<span class="k">let</span> <span class="i">x</span> <span class="o">=</span> <span class="i">A</span><span class="o">.</span><span class="i">Solve</span>(<span class="i">b</span>) <span class="c">// 1;-2;-2</span>
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</code></pre></td>
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</tr>
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</table>
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<h2><a name="Normalizing-Equation-Systems" class="anchor" href="#Normalizing-Equation-Systems">Normalizing Equation Systems</a></h2>
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<p>In practice, a linear equation system to be solved is often not in the standard form required
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to use the linear algebra approach. For example, let's have a look at the following system:</p>
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<p><span class="math">\[\begin{bmatrix}1 & 2 & 3 & 4\\2 & 3 & 4 & 5\\3 & 4 & 5 & 6\\4 & 5 & 6 & 7\end{bmatrix}
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\begin{bmatrix}0\\0\\V\\T\end{bmatrix}
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\;=\;
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\begin{bmatrix}F\\M\\20\\0\end{bmatrix}\]</span></p>
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<p>The first two values of the solution vector <span class="math">\([0,\;0,\;V,\;T]\)</span> are constant zero, so we can simplify
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the system to:</p>
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<p><span class="math">\[\begin{bmatrix}3 & 4\\4 & 5\\5 & 6\\6 & 7\end{bmatrix}
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\begin{bmatrix}V\\T\end{bmatrix}
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\;=\;
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\begin{bmatrix}F\\M\\20\\0\end{bmatrix}\]</span></p>
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<p>Then we need to subtract the two unknowns from the right side back from the left (so that they
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become zero on the right side), by introducing a new column each. First we subtract
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<span class="math">\([F,\;0,\;0,\;0]^T\)</span> from both sides:</p>
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<p><span class="math">\[\begin{bmatrix}3 & 4 & -1\\4 & 5 & 0\\5 & 6 & 0\\6 & 7 & 0\end{bmatrix}
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\begin{bmatrix}V\\T\\F\end{bmatrix}
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\;=\;
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\begin{bmatrix}0\\M\\20\\0\end{bmatrix}\]</span></p>
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<p>Then we subtract <span class="math">\([0,\;M,\;0,\;0]^T\)</span> from both sides the same way:</p>
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<p><span class="math">\[\begin{bmatrix}3 & 4 & -1 & 0\\4 & 5 & 0 & -1\\5 & 6 & 0 & 0\\6 & 7 & 0 & 0\end{bmatrix}
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\begin{bmatrix}V\\T\\F\\M\end{bmatrix}
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\;=\;
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\begin{bmatrix}0\\0\\20\\0\end{bmatrix}\]</span></p>
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<p>Which is in standard from, so we can solve normally:</p>
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<table class="pre"><tr><td class="lines"><pre class="fssnip"><span class="l">1: </span>
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<span class="l">2: </span>
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<span class="l">3: </span>
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<span class="l">4: </span>
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<span class="l">5: </span>
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<span class="l">6: </span>
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</pre></td>
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<td class="snippet"><pre class="fssnip highlighted"><code lang="fsharp"><span class="k">let</span> <span class="i">A'</span> <span class="o">=</span> <span class="i">matrix</span> [[ <span class="n">3.0</span>; <span class="n">4.0</span>; <span class="o">-</span><span class="n">1.0</span>; <span class="n">0.0</span> ]
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[ <span class="n">4.0</span>; <span class="n">5.0</span>; <span class="n">0.0</span>; <span class="o">-</span><span class="n">1.0</span> ]
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[ <span class="n">5.0</span>; <span class="n">6.0</span>; <span class="n">0.0</span>; <span class="n">0.0</span>; ]
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[ <span class="n">6.0</span>; <span class="n">7.0</span>; <span class="n">0.0</span>; <span class="n">0.0</span> ]]
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<span class="k">let</span> <span class="i">b'</span> <span class="o">=</span> <span class="i">vector</span> [ <span class="n">0.0</span>; <span class="n">0.0</span>; <span class="n">20.0</span>; <span class="n">0.0</span> ]
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<span class="k">let</span> <span class="i">x'</span> <span class="o">=</span> <span class="i">A'</span><span class="o">.</span><span class="i">Solve</span>(<span class="i">b'</span>) <span class="c">// -140; 120; 60; 40</span>
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</code></pre></td>
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