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Study Guides > College Algebra CoRequisite Course

Putting It Together: Solving Systems With Matrices

A man slides a long block of wood over a table equipped with tools to cut the wood.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
Maximum production rates for chairs and tables at each of the two workshops, Atlanta and Boston.   To make the problem more general, let’s set it up for an arbitrary order of  [latex]C[/latex] chairs and [latex]T[/latex] tables. As discussed at the beginning of this module, this scenario requires us to write a system of linear equations:

[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.

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