Use matrix multiplication to answer an application question

Items sold (Q)

\(\displaystyle \begin{bmatrix}12&8&20\\9&15&10\end{bmatrix}\)

Price of one item (P)

\(\displaystyle \begin{bmatrix}\$5\\\$4\\\$3\end{bmatrix}\)

A school canteen sells pies, rolls and juices. The matrices show how many of each item were sold on Monday and Tuesday, and the price of one of each item. What were Tuesday's takings?

A fruit stall sold 30 apples, 20 pears and 10 mangoes on Saturday, and 15 apples, 25 pears and 40 mangoes on Sunday. Apples cost $1 each and pears cost $2 each. Move the mango price up 50 cents at a time and watch how much each day's takings rises

For every 50 cents added to the mango price, how did the two totals move?

Use matrix multiplication in an application

In an application product, the shared category is the columns of the first matrix and rows of the second. Each product entry is a row-by-column total, and its labels come from the two outer categories

1. 1Put the shared category inside The shared category must be the columns of the first matrix and the rows of the second matrix \(\displaystyle A_{r\times n}B_{n\times c}\Rightarrow AB\text{ has order }r\times c\)
2. 2Use a row and a column Each product entry pairs one row of the first matrix with one column of the second matrix \(\displaystyle (AB)_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+a_{i3}b_{3j}\)
3. 3Add every paired product The products for all shared categories are added to form one entry \(\displaystyle 3\times4+2\times6+4\times3=36\)
4. 4Read the outer labels The product rows keep the row labels of the first matrix and the product columns keep the column labels of the second matrix \(\displaystyle AB:\quad\text{rows from }A,\quad\text{columns from }B\)

Construct a complete labelled application model

A community print stall sells posters, booklets and cards. On Wednesday it sells 18 posters, 12 booklets and 9 cards; on Saturday it sells 14 posters, 20 booklets and 6 cards. Each poster sells for $6 and uses $2 of materials, each booklet sells for $4 and uses $1 of materials, and each card sells for $2 and uses $1 of materials. Ignore other costs

1. 1Choose and construct the factors Build a quantity matrix with days as rows and product types as columns. Build a per-item matrix with product types as rows and sales/materials as columns. State every label, unit and dimension. Explain why the product order matters, then calculate every product entry and each day’s sales minus materials
2. 2Select technology and verify Choose a matrix calculator or spreadsheet and record the actual inputs and operation used. Independently check one entry by a dot product and check the two-day totals by aggregating quantities before multiplication. Investigate a $0.50 increase in the card selling price with counts and materials unchanged. State why sales minus materials is not necessarily net profit
3. 3Save a complete independent record Choose the representation and technology yourself. Save all labelled matrices, full products, formulas or commands, checks and written reasons before opening the model response. Full independent construction, actual external-tool operation and reporting are not automatically marked; review them yourself or with a teacher

Model response: Construct a complete labelled application model

Compare every entry, label and verification step, not only the final scalar answer

1. 1Factors and dimensions Use rows Wednesday, Saturday and shared order posters, booklets, cards. Q is 2 by 3; P is 3 by 2 with columns sales dollars per item, material dollars per item. QP is 2 by 2 with day rows and sales/materials dollar totals \(\displaystyle Q=\begin{bmatrix}18&12&9\\14&20&6\end{bmatrix},\quad P=\begin{bmatrix}6&2\\4&1\\2&1\end{bmatrix}\)
2. 2Whole product and checks The product rows are [174,57] and [176,54]. Wednesday sales are 18×6+12×4+9×2=174. Saturday materials are 14×2+20×1+6×1=54. The daily contributions after materials are $117 and $122. Aggregating counts gives [32,32,15]; multiplying by P gives [350,111], so the total contribution is $239. Multiplying P by Q is dimensionally defined here, but gives a 3 by 3 array and does not represent the required day-by-output totals: its inner labels have unrelated meanings \(\displaystyle QP=\begin{bmatrix}174&57\\176&54\end{bmatrix},\qquad QP\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}117\\122\end{bmatrix}\)
3. 3Reproducible tool record For a spreadsheet, enter Q in B2:D3 and P in F2:G4, keeping labels outside those numeric ranges. Use =MMULT(B2:D3,F2:G4) for the full product. Independently check Wednesday sales with =B2*F2+C2*F3+D2*F4. Record the actual tool’s input ranges, orientation and output rather than assuming that displayed labels participate in the calculation. A matrix calculator can store the same Q and P, then calculate Q*P.
4. 4Sensitivity and limits A $0.50 card-price rise adds 9×0.50=$4.50 on Wednesday and 6×0.50=$3.00 on Saturday. New sales are $178.50 and $179.00, total $357.50; materials stay $111. The total contribution becomes $246.50. This assumes quantities sold do not change. Rent, labour, tax and other costs were excluded, so calling this net profit would overstate the result

Given information

\(\displaystyle q=(18,12,9),\quad p=(6,4,2)^T\)

Find Wednesday sales in dollars from these independently stated counts and prices

Given information

\(\displaystyle q=(14,20,6),\quad p=(6,4,2)^T,\quad c=(2,1,1)^T\)

Find Saturday sales minus material costs, in dollars

Given information

\(\displaystyle q_{\rm total}=(32,32,15),\quad p-c=(4,3,1)^T\)

Find the two-day contribution after materials, in dollars

Given information

\(\displaystyle n_{\rm cards}=9+6,\quad \Delta p=0.50\)

How much do two-day sales increase, in dollars, if only the card price rises by $0.50?

In the print-stall model, P has product rows and sales/materials columns, while Q has day rows and product columns. Why is checking only that PQ is dimensionally defined insufficient?

Reorder a model without changing its meaning

The print stall sells posters, booklets and cards. Wednesday counts are 18, 12 and 9; Saturday counts are 14, 20 and 6. Their respective sales prices are $6, $4 and $2 per item, and material costs are $2, $1 and $1 per item. A new file stores product types in the order cards, posters, booklets instead of posters, booklets, cards. Keep the two day rows and sales/materials output columns unchanged

1. 1Construct and apply the change Construct a 3 by 3 permutation matrix C such that QC reorders the quantity columns correctly. Derive the corresponding reordered price matrix rather than guessing which side to multiply. Show all entries of both new factors and their full product. Diagnose multiplying reordered quantities by the unreordered prices
2. 2Verify and undo Calculate C squared, C cubed and its transpose. Use these to prove how to undo the change and to explain why applying this C twice does not restore the original order. Verify the product identity symbolically and numerically. Record the actual matrix-tool inputs or spreadsheet formulas you use
3. 3Save a complete independent record Choose the representation and technology yourself. Save all labelled matrices, full products, formulas or commands, checks and written reasons before opening the model response. Full independent construction, actual external-tool operation and reporting are not automatically marked; review them yourself or with a teacher

Model response: Reorder a model without changing its meaning

Compare every entry, label and verification step, not only the final scalar answer

1. 1Constructed reorder The first output column of QC selects old column three, the second selects old column one, and the third selects old column two. Therefore the new rows are [9,18,12] and [6,14,20]. C-transpose P reorders price rows to [2,1], [6,2], [4,1] \(\displaystyle C=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix},\quad QC=\begin{bmatrix}9&18&12\\6&14&20\end{bmatrix},\quad C^TP=\begin{bmatrix}2&1\\6&2\\4&1\end{bmatrix}\)
2. 2Invariant product and counterexample The correctly reordered factors still give the displayed QP. Reordering only Q gives the displayed (QC)P, which is a defined calculation with wrong pairings. This is an explicit counterexample to the claim that compatible dimensions alone preserve an application \(\displaystyle \begin{aligned}(QC)(C^TP)&=Q(CC^T)P=QP=\begin{bmatrix}174&57\\176&54\end{bmatrix}\\(QC)P&=\begin{bmatrix}150&48\\132&46\end{bmatrix}\end{aligned}\)
3. 3Inverse and powers C squared equals C transpose, not C and not the identity. C cubed equals the identity. Every row and every column contains exactly one 1, so C is a permutation matrix; checking row totals alone would be insufficient. Undo the quantity-column change by multiplying on the right by C transpose. In a matrix tool record C*C, C*C*C and transpose(C), then check both products C*C-transpose and C-transpose*C equal the identity \(\displaystyle C^2=C^T=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix},\quad C^3=I_3,\quad C^{-1}=C^T\)

Given information

\(\displaystyle Q=\begin{bmatrix}18&12&9\\14&20&6\end{bmatrix},\quad C=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}\)

Find entry (1,3) of QC

Given information

\(\displaystyle QC=\begin{bmatrix}9&18&12\\6&14&20\end{bmatrix},\quad P=\begin{bmatrix}6&2\\4&1\\2&1\end{bmatrix}\)

The columns of P are sales price and material cost, in that order. Find the first sales entry of the deliberately mismatched product (QC)P.

Given information

\(\displaystyle C=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}\)

Find the smallest positive integer k for which C to the power k is the identity

Q is a 2 by 3 matrix and C is a 3 by 3 permutation matrix. Which operation undoes Q changing to QC?

Recover a matrix and test whether it is identified

A separate stall sells two products. Batch A contains 3 of the first and 2 of the second; batch B contains 1 of the first and 4 of the second. Total sales/material costs are $34/$16 for batch A and $38/$22 for batch B. Assume constant per-item prices and costs, exact totals, no discounts and no fixed costs

1. 1Construct, solve and check Construct a quantity matrix A and totals matrix B with explicitly labelled rows and columns. Let X hold each product’s sales price and material cost. Form the correct matrix equation, test invertibility, choose a tool and recover every entry of X. Verify the full original equation and two scalar equations independently. Compare the opposite multiplication order and explain why it is not the required solution
2. 2Investigate missing or inconsistent information Replace batch B by twice batch A: quantities [6,4] and totals sales/materials [68,32]. Determine whether the unit values are uniquely identified and give two different non-negative matrices that fit. Then change only the second sales total from $68 to $69 and decide what happens. Explain why a missing inverse alone does not distinguish these cases
3. 3Preserve the side of an inverse As a separate algebraic check, use the same A with an unknown matrix Y satisfying YA=H, where H has rows [7,8] and [11,14]. Solve every entry and verify by multiplication. Explain why the inverse goes on the other side than in AX=B.
4. 4Save a complete independent record Choose the representation and technology yourself. Save all labelled matrices, full products, formulas or commands, checks and written reasons before opening the model response. Full independent construction, actual external-tool operation and reporting are not automatically marked; review them yourself or with a teacher

Model response: Recover a matrix and test whether it is identified

Compare every entry, label and verification step, not only the final scalar answer

1. 1Model and exact inverse Rows of A are batches A/B; columns are first/second product. Rows of X are products and columns are sales/materials dollars per item. Thus AX=B. The determinant of A is 3×4−2×1=10, so its inverse exists. The solution is X=A-inverse B, not B A-inverse \(\displaystyle A=\begin{bmatrix}3&2\\1&4\end{bmatrix},\ B=\begin{bmatrix}34&16\\38&22\end{bmatrix},\ A^{-1}=\frac1{10}\begin{bmatrix}4&-2\\-1&3\end{bmatrix},\ X=\begin{bmatrix}6&2\\8&5\end{bmatrix}\)
2. 2Independent verification and technology Substitution gives the displayed AX. In particular 3×6+2×8=34 and 1×2+4×5=22. The displayed B A-inverse instead fails AX=B. Enter A in B2:C3 and B in E2:F3: record =MDETERM(B2:C3), =MINVERSE(B2:C3), and =MMULT(MINVERSE(B2:C3),E2:F3), with labels outside the numeric ranges. Check the original product using the recovered matrix. If decimal output appears, retain enough precision for a residual check and recover exact entries here; a displayed zero is not by itself an exact proof \(\displaystyle AX=\begin{bmatrix}34&16\\38&22\end{bmatrix},\qquad BA^{-1}=\begin{bmatrix}12&-2\\13&-1\end{bmatrix}\)
3. 3Singular does not mean one fixed outcome With quantity rows [3,2] and [6,4], the determinant is zero and the second equation repeats twice the first. For the displayed singular quantity matrix A_s, both displayed candidates X_1 and X_2 give the same displayed totals, so no unique unit values are identified. If the second sales total becomes 69, the sales equations contradict doubling, so there is no exact solution under the stated assumptions. More independent data or a revised model are needed; do not invent an inverse \(\displaystyle \begin{aligned}X_1&=\begin{bmatrix}6&2\\8&5\end{bmatrix}, &X_2&=\begin{bmatrix}8&4\\5&2\end{bmatrix}\\A_s&=\begin{bmatrix}3&2\\6&4\end{bmatrix}, &A_sX_1=A_sX_2&=\begin{bmatrix}34&16\\68&32\end{bmatrix}\end{aligned}\)
4. 4Right-side equation From YA=H, multiply on the right: Y=H A-inverse as displayed. Directly multiplying YA recovers the displayed H. The displayed A-inverse H does not solve the stated right-side equation. Matrix factors cannot simply be commuted \(\displaystyle \begin{aligned}YA=H&\Longrightarrow YAA^{-1}=HA^{-1}\Longrightarrow Y=HA^{-1}\\Y=HA^{-1}&=\begin{bmatrix}2&1\\3&2\end{bmatrix}\\YA=H&=\begin{bmatrix}7&8\\11&14\end{bmatrix}\\A^{-1}H&=\begin{bmatrix}0.6&0.4\\2.6&3.4\end{bmatrix}\end{aligned}\)
5. 5What this foundation does not establish This investigation tests representations and operations. It does not yet construct a transition or population model, establish a long-run equilibrium, or evaluate culling/restocking policies. Those require the subsequent Chapter 11 assumptions and state conventions. It also does not replace General’s exclusions on whole-matrix network algorithms

Given information

\(\displaystyle A=\begin{bmatrix}3&2\\1&4\end{bmatrix}\)

Find the determinant of A.

Given information

\(\displaystyle A=\begin{bmatrix}3&2\\1&4\end{bmatrix},\quad B=\begin{bmatrix}34&16\\38&22\end{bmatrix}\)

In AX=B, rows of X are the first and second products, and its columns are sales price and material cost. Recover the second product’s material cost in dollars

Given information

\(\displaystyle A=\begin{bmatrix}3&2\\1&4\end{bmatrix},\quad H=\begin{bmatrix}7&8\\11&14\end{bmatrix}\)

Find entry (2,1) of Y in YA=H.

The quantity rows are (3,2) and (6,4). The first totals row is (34,16); the second is either (68,32) or (69,32), with sales first and materials second. Why do these cases have different outcomes?