Putting It Together: Solving Systems With Matrices
Our furniture-making company has two workshops (Atlanta and Boston) that specialize in creating chairs and tables. Suppose we have received an order of 620 chairs and 360 tables. If the production rates for each workshop are given in the table below, how many days should each workshop commit to fulfilling the order keeping in mind that we are striving for efficiency?Atlanta | Boston | |
Chairs | 20 per day | 30 per day |
Tables | 15 per day | 12 per day |
[latex]20A+30B=C[/latex]
[latex]15A+12B=T[/latex]
The system may then be expressed as an equation of matrices,[latex]\begin{bmatrix} {20} & {30} \\[0.3em] {15} & {12} \\[0.3em] \end{bmatrix} \begin{bmatrix} {A}\\[0.3em] {B}\\[0.3em] \end{bmatrix}=\begin{bmatrix} {C}\\[0.3em] {T}\\[0.3em] \end{bmatrix}[/latex]
Now to solve the system for the unknowns [latex]A[/latex] and [latex]B[/latex], we need the inverse of the coefficient matrix. Because the coefficient matrix is 2 by 2, we may use the special shortcut formula for the inverse.[latex]\begin{bmatrix} {20} & {30} \\[0.3em] {15} & {12} \\[0.3em] \end{bmatrix}^{-1}={\large\frac{1}{(20)(12)-(30)(15)}} \begin{bmatrix} {12} & {-30} \\[0.3em] {-15} & {20} \\[0.3em] \end{bmatrix}[/latex]
[latex]={\large\frac{1}{-210}} \begin{bmatrix} {12} & {-30} \\[0.3em] {-15} & {20} \\[0.3em] \end{bmatrix}[/latex]
[latex]=\begin{bmatrix} -\frac{2}{35} & \frac{1}{7} \\[0.3em] \frac{1}{14} & -\frac{2}{21} \\[0.3em] \end{bmatrix}[/latex]
This means that the required production levels ([latex]A[/latex] and [latex]B[/latex] ) can be found by the matrix multiplication,[latex]\begin{bmatrix} {A} \\[0.3em] {B} \\[0.3em] \end{bmatrix}= \begin{bmatrix} {20} & {30} \\[0.3em] {15} & {12} \\[0.3em] \end{bmatrix}^{-1} \begin{bmatrix} {C} \\[0.3em] {T} \\[0.3em] \end{bmatrix}=\begin{bmatrix} -\frac{2}{35} & \frac{1}{7} \\[0.3em] \frac{1}{14} & -\frac{2}{21} \\[0.3em] \end{bmatrix} \begin{bmatrix} {C} \\[0.3em] {T} \\[0.3em] \end{bmatrix}[/latex]
For example, in the case that [latex]C=620[/latex] and [latex]T=360[/latex], we find:[latex]\begin{bmatrix} {A} \\[0.3em] {B} \\[0.3em] \end{bmatrix}=\begin{bmatrix} -\frac{2}{35} & \frac{1}{7} \\[0.3em] \frac{1}{14} & -\frac{2}{21} \\[0.3em] \end{bmatrix} \begin{bmatrix} {620} \\[0.3em] {360} \\[0.3em] \end{bmatrix}[/latex]
[latex]=\begin{bmatrix} -\frac{2}{35}(620) & \frac{1}{7}(360) \\[0.3em] \frac{1}{14}(620) & -\frac{2}{21}(360) \\[0.3em] \end{bmatrix}[/latex]
[latex]=\begin{bmatrix} {16} \\[0.3em] {10} \\[0.3em] \end{bmatrix}[/latex]
Therefore, [latex]A=16[/latex] and [latex]B=10[/latex], which means that the Atlanta workshop should work on producing chairs and tables for 16 days and the Boston workshop should take 10 days. While this may seem like a lot of work just to solve a simple system of equations, the general equation in terms of [latex]C[/latex] and [latex]T[/latex] may now be used for the next order and the order after that! No matter what the particular numbers of chairs and tables required, the same matrix equation can determine how many hours each workshop should run to efficiently produce the furniture.Licenses & Attributions
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- Putting It Together: Solve Systems with Matrices. Authored by: Lumen Learning. License: CC BY: Attribution.
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- Man in Woodshop. Authored by: Senior Airman Matthew Lotz. License: Public Domain: No Known Copyright.