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11.10: Solving Linear Equations Using Mathcad

  • Page ID
    135018
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    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

    \(\begin{pmatrix}
    5 \cdot x + 2 \cdot y + z = 36\\
    x + 7 \cdot y + 3 \cdot z = 63\\
    2 \cdot x + 3 \cdot y + 8 \cdot z = 81
    \end{pmatrix}
    solve, \begin{pmatrix}
    x\\
    y\\
    z
    \end{pmatrix} \rightarrow \begin{pmatrix}
    \frac{18}{5} & \frac{27}{5} & \frac{36}{5}
    \end{pmatrix} = \begin{pmatrix}
    3.6 & 5.4 & 7.2
    \end{pmatrix}\)


    This page titled 11.10: Solving Linear Equations Using Mathcad is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform.