11.10: Solving Linear Equations Using Mathcad
- Page ID
- 135018
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
Numeric Methods: A system of equations is solved numerically using a Given/Find solve block. Mathcad requires seed values for each of the variables in the numeric method.
Seed values: x :=1 y :=1 z :=1
Given: \(5 \cdot x + 2 \cdot y + z = 36\) \(x + 7 \cdot y + 3 \cdot z = 63\) \(2 \cdot x + 3 \cdot y + 8 \cdot z = 81\)
Find (x, y, z) = \( \begin{pmatrix}
3.6\\
5.4\\
7.2
\end{pmatrix}\)
Other Given/Find solve blocks can be used.
Given \(\begin{pmatrix}
5 \cdot x + 2 \cdot y + z = 36\\
x + 7 \cdot y + 3 \cdot z = 63\\
2 \cdot x + 3 \cdot y + 8 \cdot z = 81
\end{pmatrix}
= \begin{pmatrix}
36\\
63\\
81
\end{pmatrix}\) Find(x, y, z) = \( \begin{pmatrix}
3.6\\
5.4\\
7.2
\end{pmatrix}\)
Given \(\begin{pmatrix}
5 & 2 & 1\\
1 & 7 & 3\\
2 & 3 & 8
\end{pmatrix} \cdot \begin{pmatrix}
x\\
y\\
z
\end{pmatrix} = \begin{pmatrix}
36\\
63\\
81
\end{pmatrix}\) Find(x, y, z) = \( \begin{pmatrix}
3.6\\
5.4\\
7.2
\end{pmatrix}\)
Matrix methods: The equations can also be solved using matrix algebra as shown below. In matrix form, the equations are written as MX = C. The solution vector is found by matrix mutiplication of by the inverse of M.
M:= \(\begin{pmatrix}
5 & 2 & 1\\
1 & 7 & 3\\
2 & 3 & 8
\end{pmatrix}\) C:= \(\begin{pmatrix}
36\\
63\\
81
\end{pmatrix}\) X := M-1 \(\cdot\) C X = \(\begin{pmatrix}
3.6\\
5.4\\
7.2
\end{pmatrix}\)
Confirm that a solution has been found:
M \( \cdot\) X = \( \begin{pmatrix}
36\\
63\\
81
\end{pmatrix}\)
Alternative matrix solution using the lsolve command.
X := lsolve(M,C) X = \( \begin{pmatrix}
3.6\\
5.4\\
7.2
\end{pmatrix}\) M \( \cdot\) X = \( \begin{pmatrix}
36\\
63\\
81
\end{pmatrix}\)
Live symbolic method: To use the live symbolic method within this Mathcad document recursive definitions are required clear previous values of x, y and z. This would not be necessary if x, y and z had not been previous defined.
x := x y := y z := z
\(\begin{pmatrix}
5 \cdot x + 2 \cdot y + z = 36\\
x + 7 \cdot y + 3 \cdot z = 63\\
2 \cdot x + 3 \cdot y + 8 \cdot z = 81
\end{pmatrix}
solve, \begin{pmatrix}
x\\
y\\
z
\end{pmatrix} \rightarrow \begin{pmatrix}
\frac{18}{5} & \frac{27}{5} & \frac{36}{5}
\end{pmatrix} = \begin{pmatrix}
3.6 & 5.4 & 7.2
\end{pmatrix}\)