Matrices · Lesson 5 of 8
Product Laws, Powers, and Applications
“Use product laws, recognise matrix power patterns, and model totals with labelled matrices.”
• Use associativity and both distributive laws while preserving factor order. • Explain how an identity matrix acts in a compatible product. • Calculate matrix powers and verify polynomial identities. • Prove a trigonometric matrix power pattern using induction. • Model costs, interest, revenue, and profit using matrix multiplication and consistent units.
Regrouping without rearranging
Matrix multiplication is generally not commutative, but it does obey useful laws. The distinction between regrouping factors and reversing them is central. Associativity changes which product we calculate first while leaving the factors in the same left-to-right order.
The (i,ℓ) entry on either side adds products aᵢⱼbⱼₖcₖℓ over all matching middle indices j and k. One grouping organises the arithmetic through AB and the other through BC. Both collect the same terms. This is the reason that A(BC) can be replaced by (AB)C, but cannot automatically be replaced by B(AC).
Problem
Use A = [1,1,−1;2,0,3;3,−1,2], B = [1,3;0,2;−1,4], C = [1,2,3,−4;2,0,−2,1] to verify (AB)C = A(BC).
- 1.The orders are 3 × 3, 3 × 2, and 2 × 4. AB is 3 × 2 and BC is 3 × 4; either complete grouping gives a 3 × 4 matrix.
- 2.Calculate AB = [2,1;−1,18;1,15]. Multiplying its first row by C gives [4,4,4,−7]. For instance, the final entry is 2(−4)+1(1) = −7.
- 3.The other rows give [35,−2,−39,22] and [31,2,−27,11]. Thus (AB)C = [4,4,4,−7;35,−2,−39,22;31,2,−27,11].
- 4.For the other grouping, first find BC = [7,2,−3,−1;4,0,−4,2;7,−2,−11,8].
- 5.Multiplying A by this result gives rows [7+4−7,2+0+2,−3−4+11,−1+2−8], [14+21,4−6,−6−33,−2+24], and [21−4+14,6−4,−9+4−22,−3−2+16].
- 6.Simplifying reproduces exactly [4,4,4,−7;35,−2,−39,22;31,2,−27,11], so the two groupings agree. No factors were interchanged.
Distribution and the identity matrix
A row-column total can be split into totals for separate contributions. This produces two distributive laws, one for a sum on the right and one for a sum on the left. The identity matrix gives another predictable product by selecting entries rather than changing them.
For the first law, the relevant row-column terms are aᵢⱼ(bⱼₖ+cⱼₖ) = aᵢⱼbⱼₖ+aᵢⱼcⱼₖ. Summing over j separates the total into the entries of AB and AC. The second law follows by splitting (aᵢⱼ+bᵢⱼ)cⱼₖ. The laws can therefore be used with compatible rectangular matrices as well as square ones.
Problem
For A = [0,6,7;−6,0,8;7,−8,0], B = [0,1,1;1,0,2;1,2,0], C = [2;−2;3], verify (A+B)C = AC+BC.
- 1.Add first: A+B = [0,7,8;−5,0,10;8,−6,0]. Multiplying by C gives [0−14+24;−10+0+30;16+12+0] = [10;20;28].
- 2.Calculate AC independently. Its entries are 0−12+21 = 9, −12+0+24 = 12, and 14+16+0 = 30. Thus AC = [9;12;30].
- 3.Similarly BC = [0−2+3;2+0+6;2−4+0] = [1;8;−2].
- 4.Adding these columns gives AC+BC = [10;20;28], equal to (A+B)C. Notice that C stays on the right in both terms.
In IₘA, a row of the identity has one 1 and all other entries 0, so it selects the matching row of A. In AIₙ, a column of the identity selects the matching column of A. For a rectangular matrix the left and right identity matrices have different orders, but each compatible product still returns A. Identity is a matrix, not the number 1 written in an array equation.
For square matrices, (A+B)² expands to A²+AB+BA+B². Combining the middle terms as 2AB requires the extra condition AB = BA. Associativity permits regrouping; it does not supply that condition.
Powers and polynomial expressions
A square matrix can be multiplied by itself repeatedly because every intermediate product has the same order. We write A² = AA and A³ = A²A = AA². Polynomial expressions combine these powers, scalar multiples, and an identity matrix of matching order.
The square of a matrix does not mean squaring each entry. Calculate A² by the row-column rule. In an expression such as A²−5A+7I, the scalar term 7 must be represented by 7I so that all terms have the same order. This lets us subtract matrices rather than attempt to subtract an ordinary number from an array.
Problem
For A = [1,2,3;3,−2,1;4,2,1], show that A³−23A−40I = O.
- 1.First calculate A². Its first row is [1+6+12,2−4+6,3+2+3] = [19,4,8]. The other rows are [1,12,8] and [14,6,15]. Thus A² = [19,4,8;1,12,8;14,6,15].
- 2.Multiply A² by A to obtain A³. Its first row is [19+12+32,38−8+16,57+4+8] = [63,46,69]. The other rows are [69,−6,23] and [92,46,63].
- 3.Calculate 23A = [23,46,69;69,−46,23;92,46,23]. Subtracting this from A³ gives [40,0,0;0,40,0;0,0,40] = 40I.
- 4.Therefore A³−23A−40I = O. The off-diagonal entries cancel before the diagonal 40s are removed by 40I.
- 5.This establishes the identity for the particular matrix supplied. It is not a formula asserting that every 3 × 3 matrix satisfies the same polynomial.
A trigonometric pattern and its proof
Some matrix powers follow a pattern that becomes visible after one multiplication. Trigonometric addition formulas explain one such pattern. Observing a few cases suggests the result; mathematical induction then establishes it for every positive integer.
For example, the top-left product entry is cos α cos β−sin α sin β = cos(α+β). The top-right entry is cos α sin β+sin α cos β = sin(α+β). The lower entries become −sin(α+β) and cos(α+β). Thus multiplying two matrices in this family adds their angle parameters. This is an algebraic identity supported by the entry calculations.
Problem
For A = R(θ), prove Aⁿ = [cos(nθ), sin(nθ); −sin(nθ), cos(nθ)] for every positive integer n.
- 1.Base case: for n = 1 the proposed expression is exactly the given matrix A, so the statement is true.
- 2.Induction assumption: suppose it is true for one positive integer k. Then Aᵏ = R(kθ). This is an assumption used to prove the next case, not a conclusion about every n yet.
- 3.By the definition of powers and associativity, Aᵏ⁺¹ = AᵏA = R(kθ)R(θ). The angle-addition product identity makes this R(kθ+θ) = R((k+1)θ).
- 4.Entrywise, the top-left becomes cos(kθ)cos θ−sin(kθ)sin θ, and the top-right becomes cos(kθ)sin θ+sin(kθ)cos θ. The lower row has the required negative sine and cosine.
- 5.Thus truth for k implies truth for k+1. Together with the base case, induction proves the formula for every positive integer n.
- 6.If θ = π/2, this predicts A² = −I and A⁴ = I. The power is governed by multiplying matrices, not by raising each trigonometric entry to the nth power.
Building a labelled application model
A practical matrix product begins with a choice of labels. Put quantities in rows and arrange the weights or prices in a matching column order. The product then forms one weighted total for each row. Its interpretation and units provide useful checks on the calculation.
| Contact method | Cost per contact in paise | City X contacts | City Y contacts |
|---|---|---|---|
| Telephone | 40 | 1000 | 3000 |
| House call | 100 | 500 | 1000 |
| Letter | 50 | 5000 | 10000 |
Problem
Use the contact data in the table to find the total amount spent in each city.
- 1.Write the city-by-method matrix B = [1000,500,5000;3000,1000,10000] and the matching cost column A = [40;100;50]. The product BA gives a 2 × 1 column, one total for each city.
- 2.For city X, the sum is 1000×40+500×100+5000×50 = 40000+50000+250000 = 340000 paise.
- 3.For city Y, the sum is 3000×40+1000×100+10000×50 = 120000+100000+500000 = 720000 paise.
- 4.Divide by 100 to convert paise to rupees. The totals are ₹3400 for city X and ₹7200 for city Y.
- 5.The product units are contacts times paise per contact, hence paise. Reading the raw numbers as rupees would overstate the cost by a factor of 100.
Revenue uses exactly the same structure: a sales-quantity matrix multiplied by a price column. A cost column gives total cost, and subtracting total cost from revenue gives gross profit in the simplified model. Bond allocation works in the other direction: unknown amounts x and y must have both a known sum and a specified weighted interest total. A product such as [0.05,0.07][x;y] records that interest condition compactly.
| Market | Product x units | Product y units | Product z units |
|---|---|---|---|
| I | 10000 | 2000 | 18000 |
| II | 6000 | 20000 | 8000 |
For the market table, the quantity matrix has order 2 × 3. Multiplying it by the 3 × 1 selling-price column [2.50;1.50;1.00] gives revenue for the two markets. Replacing that column by unit costs [2.00;1.00;0.50] gives cost. Alternatively, subtract the price and cost columns first, then multiply: the distributive law proves that this gives the same profit column. This connects the abstract law to a financial interpretation.
Quiz
Which operation is justified by associativity?
What is the valid expansion of (A+B)C?
A has order 2 × 3. Which identity product is valid?
What is always the correct expansion of (A+B)² for same-order square matrices?
If A = R(π/3) in the angle family defined above, what is A³?
A row of quantities multiplied by a column of prices per unit produces what?
If A² = A, what is (I+A)³−7A?
Practice Problems
- For F(x) = [cos x,−sin x,0; sin x,cos x,0;0,0,1], multiply F(x)F(y) entry by entry and show that it equals F(x+y). Explain the unchanged final row and column.
- For A = [2,0,1;2,1,3;1,−1,0], calculate A²−5A+6I. For B = [1,0,2;0,2,1;2,0,3], verify B³−6B²+7B+2I = O.
- For A = [3,−2;4,−2], find k such that A² = kA−2I. Use more than one entry to check the same value of k.
- Let T = [0,−tan(α/2);tan(α/2),0], for angles where tan(α/2) is defined. Verify I+T = (I−T)[cos α,−sin α;sin α,cos α] using half-angle identities.
- A fund must divide ₹30000 between bonds yielding 5% and 7% per year. Use a capital equation and a matrix interest product to find the allocations giving (a) ₹1800 annual interest and (b) ₹2000 annual interest.
- A bookshop sells 10 dozen chemistry books at ₹80 each, 8 dozen physics books at ₹60 each, and 10 dozen economics books at ₹40 each. Write a quantity row and price column and calculate the total receipts. Convert dozens to individual books first.
- Find x in [1,2,1][1,2,0;2,0,1;1,0,2][0;2;x] = O. State the order of every intermediate product.
- For A = [3,1;−1,2], verify A²−5A+7I = O. Explain why 7I is used rather than an ordinary scalar 7.
- Find every real x satisfying [x,−5,−1][1,0,2;0,2,1;2,0,3][x;4;1] = O. Reduce the product to a scalar equation and check all roots.
- Use the two-market table above, prices [2.50;1.50;1.00], and costs [2.00;1.00;0.50] to find revenue in each market, cost in each market, profit in each market, and total gross profit. Verify both distributive approaches.
- If A² = A, simplify A³ and A⁴, then expand (I+A)³−7A. Explain why the mixed terms can be combined here even though arbitrary matrices do not commute.
Key Takeaways
• Associativity changes grouping while preserving factor order. • Both distributive laws hold for compatible matrices; factors remain on their original sides. • Identity matrices select rows or columns, and their dimensions must fit the product. • Calculate powers by matrix multiplication and use I for constant terms in matrix polynomials. • A pattern suggested by examples needs reasoning; angle addition and induction establish the trigonometric power formula. • Quantity-by-weight products model costs, interest, and revenue; labels and units are essential checks.