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Difficulty: 1 – 5

53070512

Two electronics stores record sales of two tablet models, \(T_1\) and \(T_2\), during two consecutive weeks. Rows represent weeks 1 and 2, and columns represent \(T_1\) and \(T_2\): \(A=\begin{pmatrix}24&18\\30&12\end{pmatrix}\), \(B=\begin{pmatrix}16&22\\25&20\end{pmatrix}\). a) Find the matrix \(S\) of combined sales for both stores. b) Find \(D=A-B\), and interpret a negative entry in context. c) State the dimension requirement for adding or subtracting two matrices. d) What would a \(2\times2\) zero matrix represent as a store’s sales matrix?

Hints

- Add or subtract entries in matching positions. - Use the row and column labels to interpret each entry. - Every entry must have a corresponding entry in the other matrix.

Solution

1. Add corresponding entries: \(S=A+B=\begin{pmatrix}40&40\\55&32\end{pmatrix}\). 2. Subtract corresponding entries: \(D=A-B=\begin{pmatrix}8&-4\\5&-8\end{pmatrix}\). A negative entry means store 2 sold more units than store 1 for that model during that week. 3. Matrices can be added or subtracted only when they have the same number of rows and the same number of columns. 4. A zero sales matrix means the store sold no units of either model in either week.

Answer

a) \(S=\begin{pmatrix}40&40\\55&32\end{pmatrix}\). b) \(D=\begin{pmatrix}8&-4\\5&-8\end{pmatrix}\). A negative entry means store 2 had greater sales in that position. c) The matrices must have the same dimensions. d) No tablets were sold in either week.

53070812

A farm offers three produce-box types—Basic, Family, and Gourmet—through weekly and biweekly subscriptions. Rows represent the two delivery schedules, and columns represent the box types in the order listed. The initial subscriber matrix is \(A=\begin{pmatrix}45&60&30\\25&40&20\end{pmatrix}\). a) How many subscribers ordered a Gourmet box in total? b) A promotion increases every category by \(20\%\). Find the new matrix \(A_{\mathrm{new}}\). c) Additional summer subscribers are represented by \(N=\begin{pmatrix}10&15&5\\5&10&8\end{pmatrix}\). Find the summer subscriber matrix \(A_{\mathrm{total}}\).

Hints

- Use the column labels to identify the Gourmet entries. - A \(20\%\) increase corresponds to a scalar factor of \(1.2\). - Add the matrices entrywise after applying the increase.

Solution

1. The Gourmet total is the sum of the third column: \(30+20=50\). 2. A \(20\%\) increase multiplies every entry by \(1.2\): \(A_{\mathrm{new}}=1.2A=\begin{pmatrix}54&72&36\\30&48&24\end{pmatrix}\). 3. Add the new subscribers entrywise: \(A_{\mathrm{total}}=A_{\mathrm{new}}+N=\begin{pmatrix}64&87&41\\35&58&32\end{pmatrix}\).

Answer

a) \(50\) subscribers. b) \(A_{\mathrm{new}}=\begin{pmatrix}54&72&36\\30&48&24\end{pmatrix}\). c) \(A_{\mathrm{total}}=\begin{pmatrix}64&87&41\\35&58&32\end{pmatrix}\).

53070912

A \(2\times5\) matrix \(A=(a_{ij})\) is defined by \(a_{ij}=2^i j\). Find all entries and write the matrix.

Hints

- The first index gives the row and the second gives the column. - Substitute each allowed value of \(i\) and \(j\) into the rule. - Check that the result has two rows and five columns.

Solution

1. For row \(i=1\), \(a_{1j}=2j\), giving \(2, 4, 6, 8, 10\). 2. For row \(i=2\), \(a_{2j}=4j\), giving \(4, 8, 12, 16, 20\). 3. Therefore, \(A=\begin{pmatrix}2&4&6&8&10\\4&8&12&16&20\end{pmatrix}\).

Answer

\(A=\begin{pmatrix}2&4&6&8&10\\4&8&12&16&20\end{pmatrix}\)

53071012

Construct the \(3\times3\) matrix \(M=(m_{ij})\) defined by \(m_{ij}=0.5\) when \(i=j\), and \(m_{ij}=0.25\) when \(i\ne j\).

Hints

- Entries with equal row and column indices are on the main diagonal. - Fill all remaining positions with the off-diagonal value. - Check that the matrix has three rows and three columns.

Solution

1. The entries with \(i=j\) lie on the main diagonal, so \(m_{11}=m_{22}=m_{33}=0.5\). 2. Every off-diagonal entry has value \(0.25\). 3. Therefore, \(M=\begin{pmatrix}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{pmatrix}\).

Answer

\(M=\begin{pmatrix}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{pmatrix}\)

53071412

A factory uses three raw materials, \(G_1, G_2, G_3\), to make four intermediate products, \(Z_1, Z_2, Z_3, Z_4\). In the \(3\times4\) matrix \(A\), the entry \(a_{ij}\) is the number of units of raw material \(G_i\) needed for one unit of product \(Z_j\). Construct \(A\) from these entries: \(a_{23}=8\), \(a_{11}=4\), \(a_{34}=7\), \(a_{12}=3\), \(a_{31}=2\), \(a_{22}=5\), \(a_{14}=6\), \(a_{33}=1\), \(a_{24}=2\). All other entries are \(0\).

Hints

- The first index gives the raw-material row, and the second gives the product column. - Draw an empty \(3\times4\) grid before inserting the values. - Fill every unspecified position with zero.

Solution

1. The matrix has three rows for the raw materials and four columns for the intermediate products. 2. Place each given entry in row \(i\), column \(j\), and fill the unspecified entries with zeros: \(A=\begin{pmatrix}4&3&0&6\\0&5&8&2\\2&0&1&7\end{pmatrix}\).

Answer

\(A=\begin{pmatrix}4&3&0&6\\0&5&8&2\\2&0&1&7\end{pmatrix}\)

53071612

Transportation costs, in dollars per ton, between four warehouses \(L_1, L_2, L_3, L_4\) are represented by \(K=\begin{pmatrix}0&10&15&20\\10&0&8&12\\15&8&0&5\\20&12&5&0\end{pmatrix}\). 1) A truck follows the route \(L_1\to L_2\to L_4\to L_3\to L_1\). Find the total transportation cost per ton for the trip. 2) Find each row sum and interpret it. Which warehouse is most central with respect to transportation cost? 3) A toll increase adds \(\$2.50\) per ton to every trip between different warehouses. Find the new matrix \(K_{\mathrm{new}}\). Does it remain symmetric?

Hints

- Read one matrix entry for each leg of the route. - A smaller row sum means lower combined costs to all other warehouses. - Change only off-diagonal entries for the toll increase.

Solution

1. Add the entries for the route: \(k_{12}+k_{24}+k_{43}+k_{31}=10+12+5+15=42\). The trip costs \(\$42\) per ton. 2. The row sums are \(45, 30, 28, 37\). Each row sum is the total cost per ton of shipping from that warehouse to each of the other warehouses once. Warehouse \(L_3\) is most central because its row sum, \(28\), is smallest. 3. Add \(2.50\) to every off-diagonal entry and leave the diagonal entries at \(0\): \(K_{\mathrm{new}}=\begin{pmatrix}0&12.50&17.50&22.50\\12.50&0&10.50&14.50\\17.50&10.50&0&7.50\\22.50&14.50&7.50&0\end{pmatrix}\). The matrix remains symmetric because the same amount was added to each symmetric pair.

Answer

1) \(\$42\) per ton. 2) Row sums: \(L_1: 45\), \(L_2: 30\), \(L_3: 28\), \(L_4: 37\). Warehouse \(L_3\) is most central. 3) \(K_{\mathrm{new}}=\begin{pmatrix}0&12.50&17.50&22.50\\12.50&0&10.50&14.50\\17.50&10.50&0&7.50\\22.50&14.50&7.50&0\end{pmatrix}\). It remains symmetric.

53071712

Find \(u\), \(v\), and \(w\) so that \(K=\begin{pmatrix}10&2u&v-3\\6&5&8\\4&w&7\end{pmatrix}\) is symmetric about its main diagonal.

Hints

- Match each entry above the main diagonal with the reflected entry below it. - Set the corresponding expressions equal. - The diagonal entries do not impose conditions on \(u\), \(v\), or \(w\).

Solution

1. A symmetric matrix satisfies \(k_{ij}=k_{ji}\). 2. Compare entries across the main diagonal: \(2u=6\), so \(u=3\). \(v-3=4\), so \(v=7\). \(w=8\). 3. The resulting matrix is \(\begin{pmatrix}10&6&4\\6&5&8\\4&8&7\end{pmatrix}\).

Answer

\(u=3\), \(v=7\), and \(w=8\).

53072112

A company uses three raw materials, \(R_1, R_2, R_3\), to make two intermediate products, \(Z_1, Z_2\). Those intermediate products are then used to make two final products, \(E_1, E_2\). The raw-material requirements per unit of intermediate product are \(A=\begin{pmatrix}3&1\\2&4\\0&5\end{pmatrix}\), and the intermediate-product requirements per unit of final product are \(B=\begin{pmatrix}2&6\\3&1\end{pmatrix}\). Find \(C=AB\), which gives the direct raw-material requirements per unit of each final product.

Hints

- Check the inner dimensions before multiplying. - Use row-by-column products. - Interpret row \(i\), column \(j\) as the amount of raw material \(R_i\) needed for final product \(E_j\).

Solution

1. The product has dimensions \((3\times2)(2\times2)=3\times2\). 2. Use row-by-column multiplication: \(c_{11}=3\cdot2+1\cdot3=9\), \(c_{12}=3\cdot6+1\cdot1=19\), \(c_{21}=2\cdot2+4\cdot3=16\), \(c_{22}=2\cdot6+4\cdot1=16\), \(c_{31}=0\cdot2+5\cdot3=15\), and \(c_{32}=0\cdot6+5\cdot1=5\). 3. Therefore, \(C=AB=\begin{pmatrix}9&19\\16&16\\15&5\end{pmatrix}\).

Answer

\(C=\begin{pmatrix}9&19\\16&16\\15&5\end{pmatrix}\)

53072312

A bakery makes three kinds of rolls. The ingredient amounts per roll are shown below: <table> <tr> <td></td> <th colspan="3">Roll type</th> </tr> <tr> <td></td> <td>Whole-grain</td> <td>Multigrain</td> <td>Pumpkin seed</td> </tr> <tr> <td>Flour (g)</td> <td>150</td> <td>120</td> <td>100</td> </tr> <tr> <td>Water (mL)</td> <td>90</td> <td>80</td> <td>70</td> </tr> <tr> <td>Seeds (g)</td> <td>10</td> <td>30</td> <td>25</td> </tr> </table> An order calls for \(60\) whole-grain rolls, \(50\) multigrain rolls, and \(40\) pumpkin seed rolls. Use matrix-vector multiplication to find the total amounts of flour, water, and seeds needed.

Hints

- Use ingredient types as rows and roll types as columns. - Multiply the production matrix by the order column vector. - Interpret each component of the result using the row labels.

Solution

1. The production matrix and order vector are \(A=\begin{pmatrix}150&120&100\\90&80&70\\10&30&25\end{pmatrix}\) and \(\mathbf{v}=\begin{pmatrix}60\\50\\40\end{pmatrix}\). 2. Compute \(\mathbf{r}=A\mathbf{v}\): Flour: \(150\cdot60+120\cdot50+100\cdot40=19{,}000\). Water: \(90\cdot60+80\cdot50+70\cdot40=12{,}200\). Seeds: \(10\cdot60+30\cdot50+25\cdot40=3100\). 3. Therefore, \(\mathbf{r}=\begin{pmatrix}19{,}000\\12{,}200\\3100\end{pmatrix}\), where the entries are flour in grams, water in milliliters, and seeds in grams.

Answer

\(\mathbf{r}=\begin{pmatrix}19{,}000\\12{,}200\\3100\end{pmatrix}\): \(19{,}000\,\text{g}\) of flour, \(12{,}200\,\text{mL}\) of water, and \(3100\,\text{g}\) of seeds.

53072812

Let \(A=\begin{pmatrix}1&2\\0&3\end{pmatrix}\) and \(B=\begin{pmatrix}4&-1\\2&5\end{pmatrix}\). a) Find \(AB\) and \(BA\). b) Determine whether multiplication is commutative for these matrices. c) Find \(A^2\).

Hints

- Use row-by-column multiplication. - Keep the matrix order unchanged in each product. - Two matrices are equal only when every corresponding entry is equal.

Solution

1. Use row-by-column multiplication: \(AB=\begin{pmatrix}8&9\\6&15\end{pmatrix}\). 2. Reversing the order gives \(BA=\begin{pmatrix}4&5\\2&19\end{pmatrix}\). 3. Since \(AB\ne BA\), matrix multiplication is not commutative for this pair. 4. Multiplying \(A\) by itself gives \(A^2=\begin{pmatrix}1&8\\0&9\end{pmatrix}\).

Answer

a) \(AB=\begin{pmatrix}8&9\\6&15\end{pmatrix}\) and \(BA=\begin{pmatrix}4&5\\2&19\end{pmatrix}\). b) No, because \(AB\ne BA\). c) \(A^2=\begin{pmatrix}1&8\\0&9\end{pmatrix}\).

53073312

A catering company offers three menus: Veggie, Classic, and Premium. Ingredient amounts per serving are shown in pounds: <table> <tr><th>Menu</th><th>Rice (lb)</th><th>Vegetables (lb)</th><th>Protein (lb)</th></tr> <tr><td>Veggie</td><td>\(0.2\)</td><td>\(0.4\)</td><td>\(0\)</td></tr> <tr><td>Classic</td><td>\(0.15\)</td><td>\(0.2\)</td><td>\(0.15\)</td></tr> <tr><td>Premium</td><td>\(0.1\)</td><td>\(0.3\)</td><td>\(0.25\)</td></tr> </table> Ingredient prices per pound are represented by \(\mathbf{p}=\begin{pmatrix}2.00\\3.50\\14.00\end{pmatrix}\). Construct the ingredient matrix \(M\), and use matrix-vector multiplication to find the ingredient cost of one serving of each menu.

Hints

- Use the same ingredient order in the matrix columns and price vector. - Each row-dot-vector product gives one menu’s ingredient cost. - Label the result entries in the same order as the menu rows.

Solution

1. Use menus as rows and ingredients as columns: \(M=\begin{pmatrix}0.2&0.4&0\\0.15&0.2&0.15\\0.1&0.3&0.25\end{pmatrix}\). 2. Multiply by the price vector: \(M\mathbf{p}=\begin{pmatrix}0.2\cdot2.00+0.4\cdot3.50+0\cdot14.00\\0.15\cdot2.00+0.2\cdot3.50+0.15\cdot14.00\\0.1\cdot2.00+0.3\cdot3.50+0.25\cdot14.00\end{pmatrix}\) \(=\begin{pmatrix}1.80\\3.10\\4.75\end{pmatrix}\). Thus, the entries are the ingredient costs for Veggie, Classic, and Premium, respectively.

Answer

\(M=\begin{pmatrix}0.2&0.4&0\\0.15&0.2&0.15\\0.1&0.3&0.25\end{pmatrix}\), and \(M\mathbf{p}=\begin{pmatrix}1.80\\3.10\\4.75\end{pmatrix}\). The costs are \(\$1.80\), \(\$3.10\), and \(\$4.75\) per serving.

53074112

Use matrix properties to evaluate the vector expression efficiently: \(\begin{pmatrix}0.6&0.1\\0.4&0.9\end{pmatrix}\begin{pmatrix}14\\37\end{pmatrix}+\begin{pmatrix}0.6&0.1\\0.4&0.9\end{pmatrix}\begin{pmatrix}86\\63\end{pmatrix}\).

Hints

- Look for a common matrix factor in the two terms. - Use the distributive property to factor out the matrix. - Add the vectors before performing the matrix multiplication.

Solution

1. Use the distributive property \(M\mathbf{v}+M\mathbf{w}=M(\mathbf{v}+\mathbf{w})\). 2. Add the vectors: \(\begin{pmatrix}14\\37\end{pmatrix}+\begin{pmatrix}86\\63\end{pmatrix}=\begin{pmatrix}100\\100\end{pmatrix}\). 3. Multiply: \(\begin{pmatrix}0.6&0.1\\0.4&0.9\end{pmatrix}\begin{pmatrix}100\\100\end{pmatrix}=\begin{pmatrix}70\\130\end{pmatrix}\).

Answer

\(\begin{pmatrix}70\\130\end{pmatrix}\)

53074212

Evaluate the expression efficiently: \(\begin{pmatrix}0.8&0.4\\0.2&0.6\end{pmatrix}\begin{pmatrix}12\\18\end{pmatrix}-\begin{pmatrix}0.3&0.4\\0.2&0.1\end{pmatrix}\begin{pmatrix}12\\18\end{pmatrix}\).

Hints

- Both terms have the same vector factor. - Subtract the matrices first by using the distributive property. - A diagonal matrix makes the final multiplication especially simple.

Solution

1. Use the distributive property in the form \(A\mathbf{x}-B\mathbf{x}=(A-B)\mathbf{x}\). 2. Subtract the matrices: \(\begin{pmatrix}0.8&0.4\\0.2&0.6\end{pmatrix}-\begin{pmatrix}0.3&0.4\\0.2&0.1\end{pmatrix}=\begin{pmatrix}0.5&0\\0&0.5\end{pmatrix}\). 3. Multiply by the vector: \(\begin{pmatrix}0.5&0\\0&0.5\end{pmatrix}\begin{pmatrix}12\\18\end{pmatrix}=\begin{pmatrix}6\\9\end{pmatrix}\).

Answer

\(\begin{pmatrix}6\\9\end{pmatrix}\)

53074312

Let \(A=\begin{pmatrix}3&-2\\4&1\end{pmatrix}\). a) Find a matrix \(X\) such that \(AX=A\). b) Verify by direct multiplication whether your matrix from part a also satisfies \(XA=A\). c) Use matrix dimensions to explain why the \(3\times3\) identity matrix \(I_3=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\) cannot serve as an identity matrix for \(A\).

Hints

- Identify the matrix analogue of multiplying a number by \(1\). - Use row-by-column multiplication to check the reverse product. - For a product to be defined, the inner dimensions must match.

Solution

1. Since \(A\) is \(2\times2\), choose the \(2\times2\) identity matrix: \(X=I_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). Then \(AX=A\). 2. Multiply in the reverse order: \(XA=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}3&-2\\4&1\end{pmatrix}=\begin{pmatrix}3&-2\\4&1\end{pmatrix}=A\). 3. The product \(AI_3\) is not defined because the \(2\) columns of \(A\) do not match the \(3\) rows of \(I_3\). Likewise, \(I_3A\) is not defined because the \(3\) columns of \(I_3\) do not match the \(2\) rows of \(A\).

Answer

a) \(X=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). b) Yes. \(XA=A\). c) Neither \(AI_3\) nor \(I_3A\) is defined because the inner dimensions do not match.

53074912

Matrices used for transition processes include several special matrix types. Investigate the additive identity matrix. 1. Find the entries of a \(2\times2\) matrix \(N=\begin{pmatrix}n_{11}&n_{12}\\n_{21}&n_{22}\end{pmatrix}\) such that \(M+N=M\) for every \(2\times2\) matrix \(M=\begin{pmatrix}m_{11}&m_{12}\\m_{21}&m_{22}\end{pmatrix}\). 2. Explain why an additive identity matrix exists for every matrix dimension \(m\times n\). 3. A square matrix is column-stochastic when all entries are between \(0\) and \(1\), inclusive, and every column sum is \(1\). Can the additive identity from part 1 be a column-stochastic transition matrix? Explain.

Hints

- An additive identity leaves an object unchanged when added. - Matrix addition is performed entry by entry. - Compare the column sums of the zero matrix with the required column sums of a transition matrix.

Solution

1. Matrix addition is entrywise. From \(M+N=M\), each entry must satisfy \(m_{ij}+n_{ij}=m_{ij}\). Therefore, \(n_{ij}=0\) for every position, so \(N=\begin{pmatrix}0&0\\0&0\end{pmatrix}\). 2. For any dimensions \(m\times n\), the matrix \(O_{m,n}\) with every entry equal to \(0\) satisfies \(A+O_{m,n}=A\) because \(a_{ij}+0=a_{ij}\) in every position. 3. Every column sum of a zero matrix is \(0\), not \(1\). Therefore, the additive identity cannot be a column-stochastic transition matrix.

Answer

1. \(N=\begin{pmatrix}0&0\\0&0\end{pmatrix}\). 2. The zero matrix \(O_{m,n}\) is the additive identity for all \(m\times n\) matrices. 3. No. Its column sums are \(0\), not \(1\).

53070612

A population transition matrix is \(M=\begin{pmatrix}0.7&0.2\\0.3&0.8\end{pmatrix}\). a) Suppose every entry is doubled proportionally. Find \(M_{\mathrm{new}}=2M\). b) Verify the distributive property by comparing \(3(M+M)\) with \(3M+3M\). c) Explain why matrix addition is commutative, using the commutative property of real-number addition. d) Explain why \(M_{\mathrm{new}}\) is not valid as a column-stochastic transition matrix.

Hints

- Scalar multiplication applies to every entry. - Carry out both sides of the distributive property entry by entry. - Check the sum of each column when interpreting a stochastic matrix.

Solution

1. Multiply every entry by \(2\): \(M_{\mathrm{new}}=\begin{pmatrix}1.4&0.4\\0.6&1.6\end{pmatrix}\). 2. Since \(M+M=\begin{pmatrix}1.4&0.4\\0.6&1.6\end{pmatrix}\), \(3(M+M)=\begin{pmatrix}4.2&1.2\\1.8&4.8\end{pmatrix}\). Also, \(3M=\begin{pmatrix}2.1&0.6\\0.9&2.4\end{pmatrix}\), so \(3M+3M=\begin{pmatrix}4.2&1.2\\1.8&4.8\end{pmatrix}\). 3. Matrix addition is entrywise. Because \(a_{ij}+b_{ij}=b_{ij}+a_{ij}\) for real numbers, corresponding entries are unchanged when the order of the matrices is reversed. 4. The columns of a column-stochastic matrix must each sum to \(1\). The columns of \(M_{\mathrm{new}}\) each sum to \(2\), so its entries cannot represent proportions distributed among the two states.

Answer

a) \(M_{\mathrm{new}}=\begin{pmatrix}1.4&0.4\\0.6&1.6\end{pmatrix}\). b) Both expressions equal \(\begin{pmatrix}4.2&1.2\\1.8&4.8\end{pmatrix}\). c) Matrix addition is commutative because corresponding real-number entries are added commutatively. d) Each column sums to \(2\), not \(1\), so it is not column-stochastic.

53070712

A sporting-goods company ships four running-shoe models—Air, Speed, Trail, and Comfort—to three distribution centers—West, South, and East. Rows represent the centers in that order, and columns represent the models in the order listed. April shipments are \(L_A=\begin{pmatrix}120&150&80&200\\100&120&110&180\\90&130&100&150\end{pmatrix}\). a) How many Trail shoes were shipped in total? Which center received the greatest total number of shoes? b) May, June, and July each have the same shipments as April. Find the matrix \(L_Q\) for the three-month total. c) August shipments are \(L_{\mathrm{Aug}}=\begin{pmatrix}140&160&90&210\\110&130&120&190\\100&140&110&160\end{pmatrix}\). Find the combined shipment matrix for July and August.

Hints

- Sum a column for a model total and a row for a center total. - Repeated identical monthly shipments can be represented by scalar multiplication. - Add the July and August matrices entrywise.

Solution

1. The Trail total is the sum of the third column: \(80+110+100=290\). The row totals are \(550\) for West, \(510\) for South, and \(470\) for East, so West received the most. 2. The three-month total is \(L_Q=3L_A=\begin{pmatrix}360&450&240&600\\300&360&330&540\\270&390&300&450\end{pmatrix}\). 3. July has the same matrix as April, so \(L_A+L_{\mathrm{Aug}}=\begin{pmatrix}260&310&170&410\\210&250&230&370\\190&270&210&310\end{pmatrix}\).

Answer

a) \(290\) Trail shoes; West received the most, with \(550\) shoes. b) \(L_Q=\begin{pmatrix}360&450&240&600\\300&360&330&540\\270&390&300&450\end{pmatrix}\). c) \(\begin{pmatrix}260&310&170&410\\210&250&230&370\\190&270&210&310\end{pmatrix}\).

53071112

Let \(A=\begin{pmatrix}4&-1&2\\0&3&-5\\1&2&1\end{pmatrix}\), \(B=\begin{pmatrix}-2&5&0\\1&-4&2\\3&0&-1\end{pmatrix}\), and \(C=\begin{pmatrix}1&2&-3\\-2&0&4\\0&1&5\end{pmatrix}\). Find: a) \(A+B-C\) b) \(2A-3B\) c) \(B-2(C-A)\)

Hints

- Perform each operation on corresponding entries. - Apply scalar multiplication before combining matrices. - Distribute the negative scalar carefully in part c).

Solution

1. Compute each entry of \(A+B-C\): \(A+B-C=\begin{pmatrix}1&2&5\\3&-1&-7\\4&1&-5\end{pmatrix}\). 2. Scale first, then subtract: \(2A-3B=\begin{pmatrix}14&-17&4\\-3&18&-16\\-7&4&5\end{pmatrix}\). 3. Distribute the scalar: \(B-2(C-A)=B-2C+2A\). Therefore, \(B-2(C-A)=\begin{pmatrix}4&-1&10\\5&2&-16\\5&2&-9\end{pmatrix}\).

Answer

a) \(\begin{pmatrix}1&2&5\\3&-1&-7\\4&1&-5\end{pmatrix}\) b) \(\begin{pmatrix}14&-17&4\\-3&18&-16\\-7&4&5\end{pmatrix}\) c) \(\begin{pmatrix}4&-1&10\\5&2&-16\\5&2&-9\end{pmatrix}\)

53071212

Let \(M=\begin{pmatrix}1.5&-2&4\\0&3.2&-1\end{pmatrix}\) and \(N=\begin{pmatrix}-0.5&4&0\\2&-1.6&3\end{pmatrix}\). Find: a) \(2M+4N\) b) \(M-0.5N\) c) the matrix \(X\) satisfying \(3N+X=M\).

Hints

- Multiply every entry by the scalar. - Rearrange a matrix equation using the same algebraic operations as for real-number equations. - Track decimal signs carefully.

Solution

1. Scale and add entrywise: \(2M+4N=\begin{pmatrix}1&12&8\\8&0&10\end{pmatrix}\). 2. Subtract the scaled matrix: \(M-0.5N=\begin{pmatrix}1.75&-4&4\\-1&4&-2.5\end{pmatrix}\). 3. Rearrange the matrix equation: \(X=M-3N\). Therefore, \(X=\begin{pmatrix}3&-14&4\\-6&8&-10\end{pmatrix}\).

Answer

a) \(\begin{pmatrix}1&12&8\\8&0&10\end{pmatrix}\) b) \(\begin{pmatrix}1.75&-4&4\\-1&4&-2.5\end{pmatrix}\) c) \(X=\begin{pmatrix}3&-14&4\\-6&8&-10\end{pmatrix}\)

53071512

A delivery company serves five locations, \(A, B, C, D, E\). The matrix \(D=(d_{ij})\) gives driving distances in miles from location \(i\) to location \(j\): \(D=\begin{pmatrix}\cdots&4.0&\cdots&7.0&\cdots\\\cdots&\cdots&3.0&\cdots&6.0\\5.0&\cdots&\cdots&8.0&\cdots\\\cdots&2.0&\cdots&\cdots&9.0\\1.0&\cdots&4.0&\cdots&\cdots\end{pmatrix}\). 1) Complete \(D\), assuming the distance is the same in both directions and the distance from a location to itself is \(0\). 2) State the mathematical properties that characterize this distance matrix. 3) Interpret a row vector in this context.

Hints

- Entries on the main diagonal represent travel from a location to itself. - Equal distances in both directions create matching entries across the main diagonal. - All entries in one row share the same starting location.

Solution

1. The diagonal entries are \(0\), and equal two-way distances imply \(d_{ij}=d_{ji}\). Therefore, \(D=\begin{pmatrix}0&4.0&5.0&7.0&1.0\\4.0&0&3.0&2.0&6.0\\5.0&3.0&0&8.0&4.0\\7.0&2.0&8.0&0&9.0\\1.0&6.0&4.0&9.0&0\end{pmatrix}\). 2. The matrix is square, has nonnegative entries, has zeros on its main diagonal, and is symmetric: \(D=D^T\). 3. Row \(i\) lists the distances from one fixed starting location \(i\) to all five destinations.

Answer

1) \(D=\begin{pmatrix}0&4.0&5.0&7.0&1.0\\4.0&0&3.0&2.0&6.0\\5.0&3.0&0&8.0&4.0\\7.0&2.0&8.0&0&9.0\\1.0&6.0&4.0&9.0&0\end{pmatrix}\). 2) \(D\) is square, symmetric, and nonnegative, with a zero main diagonal. 3) A row gives all distances from one starting location.

53071812

An exchange process among three states is represented by \(P=\begin{pmatrix}0.5&0.3&0.2\\x&0.4&y\\0.2&0.3&0.5\end{pmatrix}\). a) Find \(x\) and \(y\) so that \(P\) is symmetric. b) Using those values, determine whether \(P\) is row-stochastic, meaning that every row sum is \(1\). c) State the general component condition for a square matrix \(A\) to be symmetric.

Hints

- Compare entries reflected across the main diagonal. - Add each row after finding \(x\) and \(y\). - Express symmetry using indexed entries.

Solution

1. Symmetry requires corresponding entries across the main diagonal to match: \(x=p_{21}=p_{12}=0.3\) and \(y=p_{23}=p_{32}=0.3\). 2. The row sums are \(0.5+0.3+0.2=1\), \(0.3+0.4+0.3=1\), and \(0.2+0.3+0.5=1\). Therefore, \(P\) is row-stochastic. 3. A square matrix is symmetric when \(a_{ij}=a_{ji}\) for all \(i\) and \(j\), equivalently \(A=A^T\).

Answer

a) \(x=0.3\) and \(y=0.3\). b) Yes. Every row sum is \(1\). c) \(a_{ij}=a_{ji}\) for all \(i, j\), or \(A=A^T\).

53072512

A furniture manufacturer makes tables \(T\), chairs \(C\), and benches \(B\). The component requirements per item are \(M_{CP}=\begin{pmatrix}8&5&4\\24&16&12\\2&0&4\end{pmatrix}\), where rows represent wood panels, screws, and hinges, and columns represent \(T, C, B\). Two customer orders are shown below: <table> <tr> <td></td> <td>Customer 1</td> <td>Customer 2</td> </tr> <tr> <td>Tables (\(T\))</td> <td>10</td> <td>5</td> </tr> <tr> <td>Chairs (\(C\))</td> <td>40</td> <td>20</td> </tr> <tr> <td>Benches (\(B\))</td> <td>15</td> <td>10</td> </tr> </table> Find the component requirements for both orders and present the result as a matrix.

Hints

- Arrange products as rows of the order matrix so its dimensions match the component matrix. - Multiply row by column. - Interpret rows as components and columns as customers in the product matrix.

Solution

1. The order matrix is \(M_{PO}=\begin{pmatrix}10&5\\40&20\\15&10\end{pmatrix}\). 2. Multiply the component-per-product matrix by the order matrix: \(M_{CO}=M_{CP}M_{PO}\). For wood panels, the two totals are \(8\cdot10+5\cdot40+4\cdot15=340\) and \(8\cdot5+5\cdot20+4\cdot10=180\). For screws, the totals are \(24\cdot10+16\cdot40+12\cdot15=1060\) and \(24\cdot5+16\cdot20+12\cdot10=560\). For hinges, the totals are \(2\cdot10+0\cdot40+4\cdot15=80\) and \(2\cdot5+0\cdot20+4\cdot10=50\). 3. Therefore, \(M_{CO}=\begin{pmatrix}340&180\\1060&560\\80&50\end{pmatrix}\).

Answer

\(M_{CO}=\begin{pmatrix}340&180\\1060&560\\80&50\end{pmatrix}\). Customer 1 needs \(340\) wood panels, \(1060\) screws, and \(80\) hinges. Customer 2 needs \(180\) wood panels, \(560\) screws, and \(50\) hinges.

53072712

Let \(M=\begin{pmatrix}3&-1\\2&4\end{pmatrix}\), \(N=\begin{pmatrix}1&0&5\\-2&3&1\end{pmatrix}\), and \(P=\begin{pmatrix}2\\-3\\0\end{pmatrix}\). Find every defined product formed from two of these matrices, allowing the same matrix to be used twice.

Hints

- Write each matrix dimension before testing a product. - A product \(AB\) requires the number of columns of \(A\) to equal the number of rows of \(B\). - Check ordered products separately because matrix multiplication is not commutative.

Solution

1. The dimensions are \(M:2\times2\), \(N:2\times3\), and \(P:3\times1\). A product is defined when the inner dimensions match. 2. The product \(MN\) is defined: \(MN=\begin{pmatrix}5&-3&14\\-6&12&14\end{pmatrix}\). 3. The product \(NP\) is defined: \(NP=\begin{pmatrix}2\\-13\end{pmatrix}\). 4. The product \(M^2\) is defined: \(M^2=\begin{pmatrix}7&-7\\14&14\end{pmatrix}\). 5. The products \(MP\), \(NM\), \(N^2\), \(PM\), \(PN\), and \(P^2\) are not defined because their inner dimensions do not match.

Answer

\(MN=\begin{pmatrix}5&-3&14\\-6&12&14\end{pmatrix}\), \(NP=\begin{pmatrix}2\\-13\end{pmatrix}\), and \(M^2=\begin{pmatrix}7&-7\\14&14\end{pmatrix}\). No other two-matrix products are defined.

53072912

Let \(A=\begin{pmatrix}5&-2&0\\1&4&7\\-3&8&2\\6&0&-1\end{pmatrix}\) and \(P=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}\). a) Find \(AP\). b) Describe how right-multiplication by \(P\) changes the columns of \(A\). c) Find a matrix \(Q\) such that \(QA\) swaps the first and fourth rows of \(A\) while leaving the second and third rows unchanged.

Hints

- Each column of the product is selected by a column of the right-hand permutation matrix. - Right-multiplication changes columns; left-multiplication changes rows. - Begin with the identity matrix and swap the rows that should be exchanged.

Solution

1. The columns selected by \(P\) are the third, first, and second columns of \(A\), respectively. Therefore, \(AP=\begin{pmatrix}0&5&-2\\7&1&4\\2&-3&8\\-1&6&0\end{pmatrix}\). 2. Right-multiplication by \(P\) cyclically shifts the columns to the right: column 1 moves to column 2, column 2 moves to column 3, and column 3 moves to column 1. 3. Left-multiplication by a permutation matrix changes rows. To swap rows 1 and 4 while fixing rows 2 and 3, use \(Q=\begin{pmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{pmatrix}\).

Answer

a) \(AP=\begin{pmatrix}0&5&-2\\7&1&4\\2&-3&8\\-1&6&0\end{pmatrix}\). b) The columns shift cyclically to the right. c) \(Q=\begin{pmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{pmatrix}\).

53073612

Two consecutive changes in a dynamic system are represented by \(M=\begin{pmatrix}0.8&0.4\\0.2&0.6\end{pmatrix}\) and \(N=\begin{pmatrix}0.5&0\\0.5&1\end{pmatrix}\). Determine whether the order of the changes affects the overall result by finding \(MN\) and \(NM\).

Hints

- Compute each ordered product separately. - Matrix multiplication generally depends on order. - Compare corresponding entries of the two results.

Solution

1. Multiply in the order \(MN\): \(MN=\begin{pmatrix}0.6&0.4\\0.4&0.6\end{pmatrix}\). 2. Reverse the order: \(NM=\begin{pmatrix}0.4&0.2\\0.6&0.8\end{pmatrix}\). 3. Since \(MN\ne NM\), the order of the two changes affects the resulting transformation.

Answer

\(MN=\begin{pmatrix}0.6&0.4\\0.4&0.6\end{pmatrix}\) and \(NM=\begin{pmatrix}0.4&0.2\\0.6&0.8\end{pmatrix}\). The order matters.

53073912

A catering company delivers two lunch boxes, Box A and Box B, to three clients, Client X, Client Y, and Client Z. The material cost and preparation time for each box are shown below. <table> <tr><td></td><td>Box A</td><td>Box B</td></tr> <tr><td>Material cost (dollars)</td><td>\(5\)</td><td>\(8\)</td></tr> <tr><td>Preparation time (minutes)</td><td>\(12\)</td><td>\(15\)</td></tr> </table> The client orders for the first two weeks of the month are shown below. Week 1, \(B_1\): <table> <tr><td></td><td>Client X</td><td>Client Y</td><td>Client Z</td></tr> <tr><td>Box A</td><td>\(40\)</td><td>\(25\)</td><td>\(30\)</td></tr> <tr><td>Box B</td><td>\(20\)</td><td>\(30\)</td><td>\(10\)</td></tr> </table> Week 2, \(B_2\): <table> <tr><td></td><td>Client X</td><td>Client Y</td><td>Client Z</td></tr> <tr><td>Box A</td><td>\(20\)</td><td>\(35\)</td><td>\(10\)</td></tr> <tr><td>Box B</td><td>\(15\)</td><td>\(20\)</td><td>\(25\)</td></tr> </table> Find the total material cost and total preparation time for each client over both weeks in two different ways. State the matrix property that connects the methods.

Hints

- What information does adding the two weekly order matrices provide? - Interpret the product of the resource matrix and an order matrix. - For the second method, find the resources for each week separately and then combine them. - Compare whether addition occurs before or after multiplication in the two methods.

Solution

1. Write the matrices: \(R=\begin{pmatrix}5&8\\12&15\end{pmatrix}\), \(B_1=\begin{pmatrix}40&25&30\\20&30&10\end{pmatrix}\), and \(B_2=\begin{pmatrix}20&35&10\\15&20&25\end{pmatrix}\). 2. First add the order matrices: \(B_1+B_2=\begin{pmatrix}60&60&40\\35&50&35\end{pmatrix}\). Then multiply: \(R(B_1+B_2)=\begin{pmatrix}580&700&480\\1245&1470&1005\end{pmatrix}\). 3. Alternatively, calculate each week separately: \(RB_1=\begin{pmatrix}360&365&230\\780&750&510\end{pmatrix}\) and \(RB_2=\begin{pmatrix}220&335&250\\465&720&495\end{pmatrix}\). Adding gives \(RB_1+RB_2=\begin{pmatrix}580&700&480\\1245&1470&1005\end{pmatrix}\). 4. The equality \(R(B_1+B_2)=RB_1+RB_2\) is the distributive property of matrix multiplication over addition.

Answer

The total resource matrix is \(\begin{pmatrix}580&700&480\\1245&1470&1005\end{pmatrix}\). Client X: \(\$580\) and \(1245\) minutes. Client Y: \(\$700\) and \(1470\) minutes. Client Z: \(\$480\) and \(1005\) minutes. The matrix property is \(R(B_1+B_2)=RB_1+RB_2\).

53074412

A three-stage process is described by the transition matrices \(M_1=\begin{pmatrix}0.7&0.4\\0.3&0.6\end{pmatrix}\) from stage 1 to stage 2 and \(M_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}\) from stage 2 to stage 3. a) Find the matrix \(M_{\text{total}}\) for the direct transition from stage 1 to stage 3. b) Interpret the identity transition matrix \(M_2\) in the context of movement between two states. c) For an arbitrary matrix \(M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\), prove by multiplication that the identity matrix \(I=\begin{pmatrix}1&0\\0&1\end{pmatrix}\) commutes with \(M\).

Hints

- Recall what multiplication by an identity matrix does. - Interpret the diagonal \(1\)s and off-diagonal \(0\)s as transition proportions. - Use variables for the entries of a general matrix and compute both products.

Solution

1. The total transition is \(M_{\text{total}}=M_2M_1=IM_1=M_1=\begin{pmatrix}0.7&0.4\\0.3&0.6\end{pmatrix}\). 2. As a transition matrix, \(I\) means that every member remains in its current state. No movement occurs between the two states. 3. Multiply on the right: \(MI=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}=M\). Multiply on the left: \(IM=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}=M\). Therefore, \(MI=IM\).

Answer

a) \(M_{\text{total}}=\begin{pmatrix}0.7&0.4\\0.3&0.6\end{pmatrix}\). b) The identity transition leaves every state unchanged. c) \(MI=M\) and \(IM=M\), so \(MI=IM\).

53074812

A publisher sells two professional magazines, \(Z_1\) and \(Z_2\), in two regions, \(R_1\) and \(R_2\). The matrices \(G_1\) and \(G_2\) give the profit in dollars per copy during the first and second half of the year. Rows represent regions, and columns represent magazines. \(G_1=\begin{pmatrix}1.20&1.50\\1.10&1.40\end{pmatrix}\), \(G_2=\begin{pmatrix}1.30&1.60\\1.00&1.50\end{pmatrix}\). In each half-year, every region sells \(10{,}000\) copies of \(Z_1\) and \(8000\) copies of \(Z_2\), so the sales vector for both periods is \(\mathbf{v}=\begin{pmatrix}10{,}000\\8000\end{pmatrix}\). Find the total annual profit in each region. Verify \((G_1+G_2)\mathbf{v}=G_1\mathbf{v}+G_2\mathbf{v}\), and interpret the matrix \(G_{\text{year}}=G_1+G_2\) in context.

Hints

- Interpret the entries of \(G_1+G_2\) before multiplying. - Use the fact that the sales vector is the same in both half-years. - Check that multiplying dollars per copy by numbers of copies produces total dollars.

Solution

1. The first-half profits are \(G_1\mathbf{v}=\begin{pmatrix}1.20\cdot10{,}000+1.50\cdot8000\\1.10\cdot10{,}000+1.40\cdot8000\end{pmatrix}=\begin{pmatrix}24{,}000\\22{,}200\end{pmatrix}\). 2. The second-half profits are \(G_2\mathbf{v}=\begin{pmatrix}1.30\cdot10{,}000+1.60\cdot8000\\1.00\cdot10{,}000+1.50\cdot8000\end{pmatrix}=\begin{pmatrix}25{,}800\\22{,}000\end{pmatrix}\). Their sum is \(G_1\mathbf{v}+G_2\mathbf{v}=\begin{pmatrix}49{,}800\\44{,}200\end{pmatrix}\). 3. Add the matrices first: \(G_{\text{year}}=G_1+G_2=\begin{pmatrix}2.50&3.10\\2.10&2.90\end{pmatrix}\). Then \(G_{\text{year}}\mathbf{v}=\begin{pmatrix}2.50\cdot10{,}000+3.10\cdot8000\\2.10\cdot10{,}000+2.90\cdot8000\end{pmatrix}=\begin{pmatrix}49{,}800\\44{,}200\end{pmatrix}\). 4. Because the sales vectors are identical in both half-years, each entry of \(G_{\text{year}}\) is the combined profit from selling one copy in each half-year for the corresponding magazine and region.

Answer

The annual profit is \(\$49{,}800\) in region \(R_1\) and \(\$44{,}200\) in region \(R_2\). Both methods give \(\begin{pmatrix}49{,}800\\44{,}200\end{pmatrix}\), confirming \((G_1+G_2)\mathbf{v}=G_1\mathbf{v}+G_2\mathbf{v}\). \(G_{\text{year}}=\begin{pmatrix}2.50&3.10\\2.10&2.90\end{pmatrix}\) contains the combined per-copy profit margins for one copy sold in each half-year.

53075012

Consider multistep processes modeled by \(\mathbf{v}_{n+1}=M\mathbf{v}_n\), where \(M\) is a \(2\times2\) matrix. 1. Let \(A=\begin{pmatrix}0.7&0.4\\0.3&0.6\end{pmatrix}\). Find a matrix \(Z\) such that \(A+Z=A\). 2. Show by multiplication that \(AZ\ne A\). 3. The multiplicative identity is the identity matrix \(I\). Describe what happens to the process when \(M=I\), and compare it with the theoretical case \(M=Z\).

Hints

- The additive identity for matrices is the zero matrix. - Use row-by-column multiplication to find \(AZ\). - Compare what multiplication by \(1\) and multiplication by \(0\) do to each component of a state vector. - Recall what the identity matrix does to any compatible vector.

Solution

1. Since matrix addition is entrywise, \(A+Z=A\) requires every entry of \(Z\) to be \(0\). Thus, \(Z=\begin{pmatrix}0&0\\0&0\end{pmatrix}\). 2. Multiply: \(AZ=\begin{pmatrix}0.7&0.4\\0.3&0.6\end{pmatrix}\begin{pmatrix}0&0\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}\ne A\). 3. If \(M=I\), then \(\mathbf{v}_{n+1}=I\mathbf{v}_n=\mathbf{v}_n\), so the state remains unchanged. If \(M=Z\), then \(\mathbf{v}_{n+1}=Z\mathbf{v}_n=\begin{pmatrix}0\\0\end{pmatrix}\), so the state becomes the zero vector after one step.

Answer

1. \(Z=\begin{pmatrix}0&0\\0&0\end{pmatrix}\). 2. \(AZ=\begin{pmatrix}0&0\\0&0\end{pmatrix}\ne A\). 3. With \(M=I\), the state remains unchanged. With \(M=Z\), the state becomes the zero vector after one step.

53075312

Let \(A=\begin{pmatrix}1&2\\2&1\end{pmatrix}\). Find all real matrices \(B=\begin{pmatrix}a&b\\c&d\end{pmatrix}\) that commute with \(A\); that is, find all \(B\) such that \(AB=BA\).

Hints

- Multiply the matrices in both orders using variables for the entries of \(B\). - Equal matrices have equal corresponding entries. - Solve the resulting system for relationships among \(a\), \(b\), \(c\), and \(d\). - Identify which variables can remain free.

Solution

1. Compute both products: \(AB=\begin{pmatrix}a+2c&b+2d\\2a+c&2b+d\end{pmatrix}\) and \(BA=\begin{pmatrix}a+2b&2a+b\\c+2d&2c+d\end{pmatrix}\). 2. Equating corresponding entries gives \(a+2c=a+2b\), so \(c=b\), and \(b+2d=2a+b\), so \(d=a\). The other two entry equations give the same conditions. 3. Therefore, \(a\) and \(b\) are free real parameters, and every commuting matrix has the form \(B=\begin{pmatrix}a&b\\b&a\end{pmatrix}\), where \(a, b\in\mathbb{R}\).

Answer

All matrices \(B=\begin{pmatrix}a&b\\b&a\end{pmatrix}\), where \(a, b\in\mathbb{R}\).

53075612

A workshop produces two designer furniture pieces, \(F_1\) and \(F_2\), from three raw materials, \(M_1\), \(M_2\), and \(M_3\), through two intermediate production stages. The material flow is represented by \(A=\begin{pmatrix}1&0\\2&2\\0&3\end{pmatrix}\), which gives raw materials per assembly; \(B=\begin{pmatrix}2&1&0\\1&0&2\end{pmatrix}\), which gives assemblies per module; and \(C=\begin{pmatrix}1&2\\3&0\\0&1\end{pmatrix}\), which gives modules per furniture piece. Find the total-requirement matrix \(G\), whose entries give raw-material units per furniture piece. Then find the total number of units of raw material \(M_2\) needed for an order of \(10\) units of \(F_1\) and \(5\) units of \(F_2\).

Hints

- Combine the production stages by multiplying the matrices in the order represented by the process. - Check the dimensions before each multiplication. - Multiply the total-requirement matrix by the order vector. - Identify the row corresponding to raw material \(M_2\).

Solution

1. Combine the first two production stages: \(AB=\begin{pmatrix}1&0\\2&2\\0&3\end{pmatrix}\begin{pmatrix}2&1&0\\1&0&2\end{pmatrix}=\begin{pmatrix}2&1&0\\6&2&4\\3&0&6\end{pmatrix}\). 2. Include the final production stage: \(G=(AB)C=\begin{pmatrix}2&1&0\\6&2&4\\3&0&6\end{pmatrix}\begin{pmatrix}1&2\\3&0\\0&1\end{pmatrix}=\begin{pmatrix}5&4\\12&16\\3&12\end{pmatrix}\). 3. The second row of \(G\) gives the units of \(M_2\) needed for \(F_1\) and \(F_2\). Therefore, \(12\cdot10+16\cdot5=120+80=200\). Equivalently, \(G\begin{pmatrix}10\\5\end{pmatrix}=\begin{pmatrix}70\\200\\90\end{pmatrix}\).

Answer

\(G=\begin{pmatrix}5&4\\12&16\\3&12\end{pmatrix}\). The order requires \(200\) units of raw material \(M_2\).

53075812

Let \(M=\begin{pmatrix}2&0&1\\1&3&0\\0&1&2\end{pmatrix}\) and \(D=\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\). Find \(MD\) and \(DM\), and compare the results. What special effect does multiplication by a diagonal matrix have?

Hints

- Track which diagonal entry is paired with each column when the diagonal matrix is on the right. - Track which diagonal entry is paired with each row when the diagonal matrix is on the left. - Compare corresponding entries after computing both products.

Solution

1. Right-multiplication by \(D\) scales column 1 by \(1\), column 2 by \(2\), and column 3 by \(3\): \(MD=\begin{pmatrix}2&0&3\\1&6&0\\0&2&6\end{pmatrix}\). 2. Left-multiplication by \(D\) scales row 1 by \(1\), row 2 by \(2\), and row 3 by \(3\): \(DM=\begin{pmatrix}2&0&1\\2&6&0\\0&3&6\end{pmatrix}\). 3. Since the matrices are different, \(MD\ne DM\). Multiplication by a diagonal matrix on the right scales columns, while multiplication on the left scales rows.

Answer

\(MD=\begin{pmatrix}2&0&3\\1&6&0\\0&2&6\end{pmatrix}\) and \(DM=\begin{pmatrix}2&0&1\\2&6&0\\0&3&6\end{pmatrix}\). Right-multiplication by a diagonal matrix scales columns; left-multiplication scales rows.

53076212

A dynamic system is defined by \(x_{n+1}=x_n+y_n\) and \(y_{n+1}=x_n\), with \(x_0=1\) and \(y_0=0\). a) Write the system in the form \(\begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}=A\begin{pmatrix}x_n\\y_n\end{pmatrix}\), and give the matrix \(A\). b) Find the first six terms \(x_0\) through \(x_5\). Identify the sequence. c) Use repeated matrix multiplication to find \(x_6\), \(x_7\), and \(x_8\). Then calculate \(\frac{x_5}{x_4}\), \(\frac{x_6}{x_5}\), \(\frac{x_7}{x_6}\), and \(\frac{x_8}{x_7}\). Compare the ratios with \(\frac{1+\sqrt{5}}{2}\approx1.618\).

Hints

- Use the coefficients in the two recurrence equations as the rows of \(A\). - Apply the same matrix repeatedly to generate each new state vector. - Divide consecutive \(x\)-values and compare the decimal results with \(1.618\). - Look for a familiar recursive pattern in the \(x\)-values.

Solution

1. The coefficients of \(x_n\) and \(y_n\) give \(A=\begin{pmatrix}1&1\\1&0\end{pmatrix}\). 2. Starting with \(\begin{pmatrix}x_0\\y_0\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}\), repeated multiplication by \(A\) gives \(x_0=1\), \(x_1=1\), \(x_2=2\), \(x_3=3\), \(x_4=5\), and \(x_5=8\). This is the Fibonacci sequence beginning with \(1, 1\). 3. Continuing the process gives \(x_6=13\), \(x_7=21\), and \(x_8=34\). The ratios are \(\frac{x_5}{x_4}=\frac{8}{5}=1.6\), \(\frac{x_6}{x_5}=\frac{13}{8}=1.625\), \(\frac{x_7}{x_6}=\frac{21}{13}\approx1.615\), and \(\frac{x_8}{x_7}=\frac{34}{21}\approx1.619\). These values alternate around and approach \(\frac{1+\sqrt{5}}{2}\approx1.618\).

Answer

a) \(A=\begin{pmatrix}1&1\\1&0\end{pmatrix}\). b) \(1, 1, 2, 3, 5, 8\); this is the Fibonacci sequence. c) \(x_6=13\), \(x_7=21\), and \(x_8=34\). The ratios are \(1.6\), \(1.625\), approximately \(1.615\), and approximately \(1.619\), approaching \(\frac{1+\sqrt{5}}{2}\approx1.618\).

53077512

A commercial bakery makes three breads from wheat, rye, and spelt flour. The flour percentages are shown below. <table> <tr> <td></td> <th colspan="3">Flour percentage</th> </tr> <tr> <th>Bread</th> <th>Wheat</th> <th>Rye</th> <th>Spelt</th> </tr> <tr> <td>Harvest Wheat</td> <td>\(50\%\)</td> <td>\(40\%\)</td> <td>\(10\%\)</td> </tr> <tr> <td>Country Rye</td> <td>\(10\%\)</td> <td>\(80\%\)</td> <td>\(10\%\)</td> </tr> <tr> <td>Spelt Loaf</td> <td>\(5\%\)</td> <td>\(5\%\)</td> <td>\(90\%\)</td> </tr> </table> Two stores place the following weekend orders: Store A: \(120\) Harvest Wheat loaves, \(80\) Country Rye loaves, and \(50\) Spelt Loaves. Store B: \(150\) Harvest Wheat loaves, \(100\) Country Rye loaves, and \(40\) Spelt Loaves. Each loaf uses \(1.5\,\text{lb}\) of flour mixture. Use matrices to find the total pounds of each flour required.

Hints

- Write each percentage as a decimal in the flour-per-bread matrix. - Combine the two store orders before finding the flour amounts. - Account for the flour-mixture weight per loaf after applying the percentage matrix. - Check that the matrix dimensions are compatible.

Solution

1. Use rows for flour types and columns for bread types: \(M=\begin{pmatrix}0.50&0.10&0.05\\0.40&0.80&0.05\\0.10&0.10&0.90\end{pmatrix}\). The order matrix is \(B=\begin{pmatrix}120&150\\80&100\\50&40\end{pmatrix}\). 2. Add the store orders by multiplying by a vector of ones: \(B\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}270\\180\\90\end{pmatrix}\). 3. Find the flour proportions across all ordered loaves: \(M\begin{pmatrix}270\\180\\90\end{pmatrix}=\begin{pmatrix}157.5\\256.5\\126\end{pmatrix}\). 4. Each loaf uses \(1.5\,\text{lb}\) of flour mixture, so \(1.5\begin{pmatrix}157.5\\256.5\\126\end{pmatrix}=\begin{pmatrix}236.25\\384.75\\189\end{pmatrix}\). The entries correspond to wheat, rye, and spelt flour, respectively.

Answer

The bakery needs \(236.25\,\text{lb}\) of wheat flour, \(384.75\,\text{lb}\) of rye flour, and \(189\,\text{lb}\) of spelt flour.

53077612

An electronics manufacturer produces two control modules: audio modules, \(A\), and video modules, \(V\). The modules are made from resistors, capacitors, and transistors. The component requirements per module are \(M_{CM}=\begin{pmatrix}10&15\\5&10\\2&5\end{pmatrix}\), where the rows represent resistors, capacitors, and transistors, and the columns represent audio and video modules. Radios and televisions are assembled from the modules. The module requirements per finished product are \(M_{MP}=\begin{pmatrix}2&1\\0&3\end{pmatrix}\), where the columns represent radios and televisions. A wholesaler orders \(50\) radios and \(30\) televisions. a) Find the matrix \(M_{CP}\) that gives component requirements per finished product. b) Find the total numbers of resistors, capacitors, and transistors needed for the order.

Hints

- Follow the production stages from components to modules to finished products. - Interpret the units represented by each matrix before multiplying. - Keep the matrix factors in the order dictated by the production process. - Represent the order as a column vector.

Solution

1. Multiply the component-per-module matrix by the module-per-product matrix: \(M_{CP}=M_{CM}M_{MP}=\begin{pmatrix}10&15\\5&10\\2&5\end{pmatrix}\begin{pmatrix}2&1\\0&3\end{pmatrix}=\begin{pmatrix}20&55\\10&35\\4&17\end{pmatrix}\). 2. Represent the order by \(\mathbf{p}=\begin{pmatrix}50\\30\end{pmatrix}\). Then \(M_{CP}\mathbf{p}=\begin{pmatrix}20&55\\10&35\\4&17\end{pmatrix}\begin{pmatrix}50\\30\end{pmatrix}=\begin{pmatrix}2650\\1550\\710\end{pmatrix}\). The entries correspond to resistors, capacitors, and transistors.

Answer

a) \(M_{CP}=\begin{pmatrix}20&55\\10&35\\4&17\end{pmatrix}\). b) The order requires \(2650\) resistors, \(1550\) capacitors, and \(710\) transistors.

53077712

The line through the origin \(g:\mathbf{x}=\lambda\begin{pmatrix}3\\1\end{pmatrix}\), where \(\lambda\in\mathbb R\), is a line of reflection. a) Find the reflection of \(P=(-1, 3)\) across \(g\). b) The reflection is represented by a matrix \(M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\). Determine \(M\) by setting up and solving linear equations based on the images of a vector on \(g\) and the vector for \(P\).

Hints

- Check whether the vector for \(P\) is parallel or perpendicular to the reflection line. - A vector on the line is fixed by the reflection. - Apply the unknown matrix to two independent vectors and solve for its entries.

Solution

1. The vector for \(P\) is perpendicular to the direction vector of \(g\) because \((-1)\cdot3+3\cdot1=0\). A reflection across a line through the origin maps a perpendicular vector to its opposite. Therefore, \(P^{\prime}=(1, -3)\). 2. A vector on the reflection line is fixed, so \(M\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}\). Also, \(M\begin{pmatrix}-1\\3\end{pmatrix}=\begin{pmatrix}1\\-3\end{pmatrix}\). 3. For the first row of \(M\), \(3a+b=3\) and \(-a+3b=1\). Solving gives \(a=0.8\) and \(b=0.6\). 4. For the second row, \(3c+d=1\) and \(-c+3d=-3\). Solving gives \(c=0.6\) and \(d=-0.8\). Therefore, \(M=\begin{pmatrix}0.8&0.6\\0.6&-0.8\end{pmatrix}\).

Answer

a) \(P^{\prime}=(1, -3)\). b) \(M=\begin{pmatrix}0.8&0.6\\0.6&-0.8\end{pmatrix}\).

53077812

The matrix \(A=\begin{pmatrix}-0.6&0.8\\0.8&0.6\end{pmatrix}\) represents reflection across a line \(h\) through the origin. a) Verify that applying the reflection twice gives the identity transformation by showing that \(A^2=I\). b) The reflection line consists of the fixed points of the transformation. Find an equation of \(h\) by solving \((A-I)\mathbf{x}=\mathbf{0}\).

Hints

- Multiply using row-by-column products. - Fixed points satisfy \(A\mathbf{x}=\mathbf{x}\). - A dependent homogeneous system with one free variable describes a line through the origin.

Solution

1. Multiply the matrices: \(A^2=\begin{pmatrix}(-0.6)^2+(0.8)^2&(-0.6)\cdot0.8+0.8\cdot0.6\\0.8\cdot(-0.6)+0.6\cdot0.8&(0.8)^2+(0.6)^2\end{pmatrix}\). This simplifies to \(A^2=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I\). 2. Fixed points satisfy \(A\mathbf{x}=\mathbf{x}\), so \((A-I)\mathbf{x}=\mathbf{0}\). For \(\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix}\), the system is \(-1.6x+0.8y=0\) and \(0.8x-0.4y=0\). The equations are dependent. From the second equation, \(y=2x\). Therefore, the reflection line is \(h:y=2x\).

Answer

a) \(A^2=I\). b) \(h:y=2x\).

53073012

A four-state process is represented by a matrix \(M\). The states are cyclically relabeled so that old state 1 becomes new state 2, old state 2 becomes new state 3, old state 3 becomes new state 4, and old state 4 becomes new state 1. The relabeled matrix is \(M^{\prime}=SMS^T\), where \(S\) permutes rows and \(S^T\) permutes columns. a) Find \(S\). b) Find \(M^{\prime}\) when \(M=\begin{pmatrix}1&0&0&0\\0&2&0&0\\0&0&3&0\\0&0&0&4\end{pmatrix}\).

Hints

- Each row of \(S\) selects the old row that becomes a particular new row. - The transpose applies the matching permutation to columns. - For a diagonal matrix, track where each diagonal entry moves.

Solution

1. New row 1 must be old row 4, new row 2 must be old row 1, new row 3 must be old row 2, and new row 4 must be old row 3. Thus, \(S=\begin{pmatrix}0&0&0&1\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}\). 2. Its transpose is \(S^T=\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\end{pmatrix}\). 3. Left-multiplication moves the rows, and right-multiplication by \(S^T\) moves the corresponding columns. Therefore, \(M^{\prime}=SMS^T=\begin{pmatrix}4&0&0&0\\0&1&0&0\\0&0&2&0\\0&0&0&3\end{pmatrix}\).

Answer

a) \(S=\begin{pmatrix}0&0&0&1\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}\). b) \(M^{\prime}=\begin{pmatrix}4&0&0&0\\0&1&0&0\\0&0&2&0\\0&0&0&3\end{pmatrix}\).

53073712

A catering company creates three weekly meal plans, \(W_1\), \(W_2\), and \(W_3\), in several production stages. Three basic ingredients, \(I_1\), \(I_2\), and \(I_3\), are used to prepare two meals, \(M_1\) and \(M_2\). The matrix \(B=\begin{pmatrix}2&1\\1&2\\3&1\end{pmatrix}\) gives the number of ingredient units needed for each meal. The ingredients contain two nutrients, \(N_1\) and \(N_2\). The matrix \(A=\begin{pmatrix}12&15&10\\8&5&14\end{pmatrix}\) gives the nutrient units in one unit of each ingredient. The numbers of each meal included in the weekly plans are \(C=\begin{pmatrix}100&50&80\\40&120&60\end{pmatrix}\). Find the total nutrient requirements for the three weekly plans in two ways: 1. First find the nutrient content of each meal, and then find the requirements for the weekly plans. 2. First find the total ingredient requirements for the weekly plans, and then find the nutrient requirements. Compare the results and identify the matrix property illustrated.

Hints

- For the first method, determine the dimensions of the matrix that gives nutrient units per meal. - For the second method, combine the ingredient requirements for all meals in each weekly plan first. - Before multiplying, check that the number of columns in the left matrix equals the number of rows in the right matrix. - Interpret the rows and columns of the final matrix in context.

Solution

1. First calculate the nutrient content of each meal: \(AB=\begin{pmatrix}12&15&10\\8&5&14\end{pmatrix}\begin{pmatrix}2&1\\1&2\\3&1\end{pmatrix}=\begin{pmatrix}69&52\\63&32\end{pmatrix}\). Then multiply by the meal-plan matrix: \((AB)C=\begin{pmatrix}69&52\\63&32\end{pmatrix}\begin{pmatrix}100&50&80\\40&120&60\end{pmatrix}=\begin{pmatrix}8980&9690&8640\\7580&6990&6960\end{pmatrix}\). 2. First calculate the ingredient requirements: \(BC=\begin{pmatrix}240&220&220\\180&290&200\\340&270&300\end{pmatrix}\). Then calculate the nutrient requirements: \(A(BC)=\begin{pmatrix}12&15&10\\8&5&14\end{pmatrix}\begin{pmatrix}240&220&220\\180&290&200\\340&270&300\end{pmatrix}=\begin{pmatrix}8980&9690&8640\\7580&6990&6960\end{pmatrix}\). 3. Both methods give the same matrix, illustrating the associative property \((AB)C=A(BC)\).

Answer

1. \((AB)C=\begin{pmatrix}8980&9690&8640\\7580&6990&6960\end{pmatrix}\). 2. \(A(BC)=\begin{pmatrix}8980&9690&8640\\7580&6990&6960\end{pmatrix}\). The first row gives the amounts of \(N_1\) for \(W_1\), \(W_2\), and \(W_3\); the second row gives the corresponding amounts of \(N_2\). The equality illustrates \((AB)C=A(BC)\).

53073812

An electronics manufacturer produces three devices, \(G_1\), \(G_2\), and \(G_3\), from two modules, \(K_1\) and \(K_2\). The module requirements per device are \(M_2=\begin{pmatrix}3&1&2\\2&4&0\end{pmatrix}\). Two raw materials, \(R_1\) and \(R_2\), are needed to make the modules. Their requirements per module are \(M_1=\begin{pmatrix}4&2\\1&5\end{pmatrix}\). Two large orders, \(O_1\) and \(O_2\), are represented by \(D=\begin{pmatrix}10&5\\20&15\\15&10\end{pmatrix}\). a) Find \(M_{\text{total}}=M_1M_2\), and explain what its entries mean in context. b) Find the total raw-material requirements for both orders. Use the associative property and first calculate the module requirements for each order.

Hints

- Think about what happens when a matrix measured in units of raw material per module is multiplied by a matrix measured in modules per device. - In part b), follow the order of multiplication specified in the problem. - The associative property changes the grouping of three matrix factors without changing their order.

Solution

1. Multiply the raw-material-per-module matrix by the module-per-device matrix: \(M_{\text{total}}=M_1M_2=\begin{pmatrix}4&2\\1&5\end{pmatrix}\begin{pmatrix}3&1&2\\2&4&0\end{pmatrix}=\begin{pmatrix}16&12&8\\13&21&2\end{pmatrix}\). Each entry gives the units of a raw material required for one device. Rows correspond to \(R_1\) and \(R_2\), and columns correspond to \(G_1\), \(G_2\), and \(G_3\). 2. First find the module requirements for the orders: \(M_2D=\begin{pmatrix}3&1&2\\2&4&0\end{pmatrix}\begin{pmatrix}10&5\\20&15\\15&10\end{pmatrix}=\begin{pmatrix}80&50\\100&70\end{pmatrix}\). 3. Now find the raw-material requirements: \(M_1(M_2D)=\begin{pmatrix}4&2\\1&5\end{pmatrix}\begin{pmatrix}80&50\\100&70\end{pmatrix}=\begin{pmatrix}520&340\\580&400\end{pmatrix}\).

Answer

a) \(M_{\text{total}}=\begin{pmatrix}16&12&8\\13&21&2\end{pmatrix}\). Its entries give raw-material units per device. b) The total requirement matrix is \(\begin{pmatrix}520&340\\580&400\end{pmatrix}\). Order \(O_1\) requires \(520\) units of \(R_1\) and \(580\) units of \(R_2\). Order \(O_2\) requires \(340\) units of \(R_1\) and \(400\) units of \(R_2\).

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