Add entry by entry
Two matrices of the same order are added by adding the entries in the same place. Take A = (3 1; 0 5) and B = (1 4; 6 2), where the semicolon separates the first row from the second.
The top left entries give 3 + 1 = 4, the top right 1 + 4 = 5, the bottom left 0 + 6 = 6, and the bottom right 5 + 2 = 7. So A + B = (4 5; 6 7), another 2 × 2 matrix.
The rule makes sense because each place holds the same kind of quantity in both matrices. If A and B are two weeks of sales, with one row for each shop and one column for each item, the top right entries are the same shop’s sales of the same item, and adding them gives that shop’s total for the item.
The top right entry of A + B comes from the top right entries of A and B: 1 + 4 = 5.
Subtracting
Subtraction works the same way: subtract each entry of B from the entry in the same place in A. That gives 3 − 1 = 2, 1 − 4 = −3, 0 − 6 = −6 and 5 − 2 = 3, so A − B = (2 −3; −6 3).
The order of a subtraction matters, as it does for numbers. B − A has every entry with the opposite sign: B − A = (−2 3; 6 −3).
Check a difference by adding it back on. B + (A − B) should give A, and it does: 1 + 2 = 3, 4 + (−3) = 1, 6 + (−6) = 0 and 2 + 3 = 5.
The bottom left entry of A − B is 0 − 6 = −6. A difference can have negative entries even when both matrices are positive.
The orders must match
A sum pairs every entry with the entry in the same place, so both matrices must have the same order. A 2 × 2 and a 2 × 3 cannot be added: the third column of the 2 × 3 has nothing to be added to.
Holding the same number of entries is not enough. A 2 × 3 and a 3 × 2 each hold 6 entries, but the places do not line up: the 2 × 3 has a row 1, column 3 and the 3 × 2 does not. Writing one of them with its rows as columns does not rescue the sum, because that makes a different matrix.
A 2 × 3 and a 3 × 2 each hold six entries, but their places do not match, so this sum cannot be done.
Either order for a sum
Each entry of A + B is a sum of two numbers, and addition of numbers is commutative, so A + B = B + A for any two matrices of the same order. In the same way (A + B) + C = A + (B + C): matrix addition is associative.
Multiplying by a scalar
A single number multiplying a matrix is called a scalar. It multiplies every entry: 3 × (2 1; 0 4) = (6 3; 0 12), because 3 × 2 = 6, 3 × 1 = 3, 3 × 0 = 0 and 3 × 4 = 12. The order does not change.
This agrees with addition. Write M = (2 1; 0 4). Then 3M means M + M + M, and adding three copies of M adds each entry to itself three times: the top left becomes 2 + 2 + 2 = 6.
Three copies of M = (2 1; 0 4) added together give 3M. The bottom right entry is 4 + 4 + 4 = 3 × 4 = 12.
Fractions, negatives and combinations
A scalar can be a fraction or a decimal: ½ × (4 6; 2 8) = (2 3; 1 4). It can be negative too, and −1 × B changes the sign of every entry, so A − B is the same as A + (−1)B.
Scalar multiples and sums combine. With A = (3 1; 0 5) and B = (1 4; 6 2), 2A = (6 2; 0 10). Subtracting B entry by entry gives 6 − 1 = 5, 2 − 4 = −2, 0 − 6 = −6 and 10 − 2 = 8, so 2A − B = (5 −2; −6 8).
The usual mistakes
Adding entries from different places. The top right of A + B is 1 + 4, from the top right of each matrix, not 1 + 6, which takes the 6 from the bottom left of B.
Adding the scalar instead of multiplying. 3 × (2 1; 0 4) has 6 in the top left, not 2 + 3 = 5.
Multiplying only one entry by the scalar. Every entry is multiplied, so a 0 stays 0 and every other entry changes.
Adding matrices of different orders, or turning one on its side to make the orders agree. Only matrices of the same order can be added.
Sales and prices
In the first application below, two weeks of café sales are added for the total and subtracted for the change, where a negative entry is a fall. In the second, a price rise and then a discount are each a scalar, and the two scalars multiply into one.
Worked example: Two Weeks of Café Sales Added, and the Change from One Week to the Next
Question A café has two branches, Station and Park. The cups of coffee, tea and juice sold in two weeks are given by the matrices below, with one row for each branch: week 1 is A = 1204530956025 and week 2 is B = 11050351055530. (a) Find A + B, and say what the entry in row 2, column 1 means. (b) Find B − A. Which drinks sold fewer cups in week 2 than in week 1, and at which branch?
1.Both matrices have order 2 × 3, with the branches in the rows and coffee, tea and juice in the columns, so they can be added and subtracted entry by entry.
Both matrices are 2 × 3, with the same branch in each row and the same drink in each column, so they add entry by entry. 2.Add each entry to the one in the same place: A + B = 120 + 11045 + 5030 + 3595 + 10560 + 5525 + 30 = 230956520011555.
A + B = 230956520011555. 3.(a) The entry in row 2, column 1 is 200. Row 2 is Park and column 1 is coffee, so the Park branch sold 200 cups of coffee over the two weeks.
(a) Row 2, column 1 is Park and coffee: the Park branch sold 200 cups of coffee over the two weeks. 4.Subtract entry by entry: B − A = 110 − 12050 − 4535 − 30105 − 9555 − 6030 − 25 = −105510−55.
B − A = −105510−55. 5.(b) A negative entry is a fall. Station sold 10 fewer cups of coffee and Park sold 5 fewer cups of tea; every other entry is positive, so those sales rose. Check: A + (B − A) gives B again, for example 120 + (−10) = 110.
(b) The negative entries are the falls: coffee at Station, down 10 cups, and tea at Park, down 5 cups.
Answer: (a) A + B = 230956520011555; the 200 is the cups of coffee sold at Park over the two weeks; (b) B − A = −105510−55: coffee at Station fell by 10 cups and tea at Park by 5 cups
Common mistakes
- Subtracting the wrong way round and finding A − B. The change from week 1 to week 2 is week 2 minus week 1, so that a fall comes out negative.
- Trying to add A to a matrix of a different order, such as the same figures written with one row for each drink. Matrices can be added only when they have the same order, so that each entry has a partner in the same place.
Worked example: A Price Rise and Then a Discount Card Applied to a Price List
Question Two shops, A and B, sell school shirts, trousers and jackets. Their prices in dollars are the matrix P = 203550253040, with one row for each shop. Next term every price rises by 20%. (a) Find the matrix of new prices. (b) A parent has a card that takes 25% off every new price. Write the prices the parent pays as a single number times P, find them, and say how each compares with the price before the rise.
1.A rise of 20% multiplies every price by 1 + 0.2 = 1.2, so the new prices are the scalar multiple 1.2P: every entry of P is multiplied by 1.2.
A rise of 20% multiplies every price by 1.2: the new prices are the scalar multiple 1.2P. 2.(a) 1.2P = 1.2 × 201.2 × 351.2 × 501.2 × 251.2 × 301.2 × 40 = 244260303648 dollars.
(a) 1.2P = 244260303648 dollars. 3.Taking 25% off leaves 75%, so the card multiplies each new price by 0.75. The parent pays 0.75(1.2P), and the two scalars multiply together: 0.75 × 1.2 = 0.9, so the parent pays 0.9P.
The card leaves 75% of each new price, so the parent pays 0.75(1.2P) = 0.9P. 4.0.9P = 0.9 × 200.9 × 350.9 × 500.9 × 250.9 × 300.9 × 40 = 1831.54522.52736 dollars: for example, the trousers at shop A cost the parent $31.50.
0.9P = 1831.54522.52736 dollars. 5.(b) The parent pays 0.9P, so every price is 10% below what it was before the rise. Check on one entry: the jacket at shop B rises to 1.2 × 40 = $48, and 75% of $48 is $36, which is 0.9 × 40.
(b) The parent pays 0.9P: every price is 10% below the price before the rise.
Answer: (a) 1.2P = 244260303648 dollars; (b) 0.9P = 1831.54522.52736 dollars, every price 10% below the price before the rise
Common mistakes
- Combining the percentages as 20% − 25% = −5% and multiplying by 0.95. The discount is taken off the new, higher price, so the scalars multiply: 1.2 × 0.75 = 0.9, which is 10% off.
- Adding 20 to every entry. A 20% rise is a different amount on each item, $4 on a $20 shirt and $10 on a $50 jacket, so every entry is multiplied by 1.2 instead.