Lesson 12 – 2 Multiplication of Matrices Pre-calculus

Learning Objective To solve quadratic systems

Matrix Matrix Multiplication The product of two matrices, 𝐴 𝑚 𝑥 𝑝 and 𝐵 𝑝 𝑥 𝑛 , is the matrix 𝐴𝐵 with dimensions 𝑚 𝑥 𝑛. Any element in the 𝑖 𝑡ℎ row and 𝑗 𝑡ℎ column of this product matrix is the sum of the products of the corresponding elements of the 𝑖 𝑡ℎ row of 𝐴 and the 𝑗 𝑡ℎ column of 𝐵 Don’t Write. You have it! When you multiply matrices, they need to be conformable, for multiplication. Yes, write it Extra  This means: # of columns in 1st matrix = # of rows in 2nd matrix

Matrix 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 ∙ 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 2 x 3 3 x 2 MUST MATCH TO MULTIPLY Multiplying Matrices – order MATTERS Matrix 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 ∙ 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 2 x 3 3 x 2 MUST MATCH TO MULTIPLY Answer will be a 2 x 2

Matrix 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 ∙ 𝑎 𝑏 𝑐 𝑑 3 x 2 2 x 2 MUST MATCH TO MULTIPLY 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 ∙ 𝑎 𝑏 𝑐 𝑑 3 x 2 2 x 2 MUST MATCH TO MULTIPLY Answer will be a 3 x 2

Matrix 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 ∙ 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 2 x 3 2 x 3 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 ∙ 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 2 x 3 2 x 3 Don’t match, CANNOT MULTIPLY

Matrix 1. Find 𝐴𝐵: 𝐴= 4 5 7 2 1 3 and 𝐵= 2 3 5 6 3 x 2 2 x 2 match dimensions of product 3 × 2 This is the first row, first column so we take 1st row of A × 1st column of B write yourself how to get this element 𝑎 𝑑 𝑏 𝑒 𝑐 𝑓 4 2 +5(5) 4 3 +5(6) 7 2 +2(5) 7 3 +2(6) 1 2 +3(5) 1 3 +3(6) 𝐴𝐵= 33 42 24 33 17 21

2. Find 𝐴 2 : 𝐴= 4 5 7 2 1 3 Matrix 𝐴 2 = 4 5 7 2 1 3 4 5 7 2 1 3 =𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 3 x 2 3 x 2 Wait! You Can’t!

Matrix We can solve for unknown elements in a matrix equation 3. Solve for x & y 3 1 −2 0 𝑥 4 −4 5𝑦 = 2 −3 −4 −8 3x – 4 = 2 12 + 5y = –3 x = 2 y = –3

Matrix The Identity Matrix The identity matrix is the equivalent to the algebraic 1. Multiplying by it does not change the original. Yes, write it  𝐼 3 𝑥 3 = 1 0 0 0 1 0 0 0 1 𝐼 2 𝑥 2 = 1 0 0 1 Pattern: 1’s along the diagonal & 0’s everywhere else *If the product of two matrices is 𝐼, then they are inverses of each other. You can also multiply by a 0 matrix to get a 0 matrix.

Matrix Properties of Matrix Multiplication for Square Matrices If 𝐴, 𝐵, 𝑎𝑛𝑑 𝐶 are 𝑛 𝑥 𝑛 matrices, then 𝐴𝐵 is an 𝑛 𝑥 𝑛 matrix Closure Don’t Write. You have it! 𝐴𝐵 𝐶=𝐴(𝐵𝐶) Associative 𝐼 𝑛 𝑥 𝑛 𝐴=𝐴 𝐼 𝑛 𝑥 𝑛 =𝐴 Multiplicative Identity 𝑂 𝑛 𝑥 𝑛 𝐴=𝐴 𝑂 𝑛 𝑥 𝑛 = 𝑂 𝑛 𝑥 𝑛 Multiplicative Property of the Zero Matrix 𝐴 −1 is the multiplicative inverse of 𝐴 if 𝐴 −1 is defined and 𝐴 𝐴 −1 = 𝐴 −1 𝐴= 𝐼 𝑛 𝑥 𝑛 Multiplicative Inverse

Matrix Properties of Matrix Multiplication for Square Matrices If 𝐴, 𝐵, 𝑎𝑛𝑑 𝐶 are 𝑛 𝑥 𝑛 matrices, then Don’t Write. You have it! 𝐴 𝐵+𝐶 =𝐴𝐵+𝐴𝐶 Distributive Properties 𝐵+𝐶 𝐴=𝐵𝐴+𝐶𝐴 What properties are not here?? Yes, write it Extra  When we “store” information in matrices, we may have to transpose them (switch rows & columns) to make them conformable for multiplication. Per 3 Per 4 Boys Girls Ex. 𝐴= 𝐵𝑜𝑦𝑠 𝐺𝑖𝑟𝑙𝑠 18 20 17 14 𝐴′= 𝑃𝑒𝑟 3 𝑃𝑒𝑟 4 18 17 20 14 It’s still the same!

Matrix Don’t Write. You have it! Cost Yes, write it  4. A fruit stand owner packages fruit in three different ways for gift packages. Economy package, E, has 6 apples, 3 oranges, and 3 pears. Standard package, S has 5 apples, 4 oranges, and 4 pears. Luxury package, L, has 6 types of each fruit. The costs are $0.50 for an apple, $1.10 for an orange, and $0.80 for a pear. What is the total cost of preparing each package of fruit? Matrix Don’t Write. You have it! Cost Apple Orange Pear 𝐴=𝑐𝑜𝑠𝑡 [ $0.50 $1.10 $0.80 ] Yes, write it  Number of Items Apple Orange Pear 𝐵= 𝐸 𝑆 𝐿 6 3 3 5 4 4 6 6 6

Matrix Cost Apple Orange Pear 𝐴=𝑐𝑜𝑠𝑡 [ $0.50 $1.10 $0.80 ] Number of Items Apple Orange Pear 𝐵= 𝐸 𝑆 𝐿 6 3 3 5 4 4 6 6 6 If we multiply in this state … the labels don’t match up Cost per fruit package per fruit Cost per fruit fruit per package

𝐴 𝐵 𝑡 = $0.50 $1.10 $0.80 6 5 6 3 4 6 3 4 6 Matrix E S L =cost $8.70 $10.10 $14.40

Assignment Pg. 608 #1, 3, 5, 8, 15, 16, 18, 19, 20, 29, 31, 34, 41, 42

Lesson 12 – 2 Multiplication of Matrices Pre-calculuS SUPPLEMENTAL NOTES

Multiplying Matrices 5. 𝐴= 2 8 −3 0 and 𝐵= −2 −8 3 0 find 𝐴𝐵 2 8 −3 0 −2 −8 3 0 = 2 −2 +8(3) 2 −8 +8(0) −3 −2 +0(3) −3 −8 +0(0) = 2 −2 +8(3) 2 −8 +8(0) −3 −2 +0(3) = 2 −2 +8(3) 2 −8 +8(0) = 2 −2 +8(3) = 20 −16 6 24

Multiplying Matrices 6. 𝐴= 2 −1 3 4 and 𝐵= −3 1 0 2 find 𝐴𝐵 2 −1 3 4 −3 1 0 2 = 2 −3 +(−1)(0) 2 1 +(−1)(2) 3 −3 +4(0) 3 1 +4(2) = 2 −3 +(−1)(0) 2 1 +(−1)(2) 3 −3 +4(0) = 2 −3 +(−1)(0) 2 1 +(−1)(2) = 2 −3 +(−1)(0) = −6 0 −9 11

Multiplying Matrices 7. 𝐴= 2 −1 3 4 and 𝐵= −3 1 0 2 find 𝐵𝐴 −3 1 0 2 2 −1 3 4 = −3 2+(1)(3) −3 (−1)+(1)(4) 0 2 +2(3) 0 −1 +2(4) = −3 2+(1)(3) −3 (−1)+(1)(4) 0 2 +2(3) = −3 2+(1)(3) = −3 2+(1)(3) −3 (−1)+(1)(4) = −3 7 6 8 ***NOT the same as AB (#6)!!!!

Multiplying Matrices 8. −1 1 1 4 −3 5 (1 x 2)(2 x 2) = 1 x 2 8. −1 1 1 4 −3 5 (1 x 2)(2 x 2) = 1 x 2 = −1 1 +1(−3) −1 4 +1(5) = −1 1 +1(−3) = −4 1

Multiplying Matrices 9. 1 4 −3 5 −1 1 (2 x 2)(1 x 2) Can’t multiply!

Multiplying Matrices 10. −1 −2 0 2 3 1 −1 2 −3 2 0 1 (3 x 2)(2 x 3) 10. −1 −2 0 2 3 1 −1 2 −3 2 0 1 (3 x 2)(2 x 3) = 3 x 3 = −1 −1 +(−2)(2) −1 2 +(−2)(0) −1 −3 +(−2)(1) 0 −1 +2(2) 0 2 +2(0) 0 −3 +2(1) = −1 −1 +(−2)(2) −1 2 +(−2)(0) −1 −3 +(−2)(1) 0 −1 +2(2) 0 2 +2(0) 0 −3 +2(1) 3 −1 +1(2) = −1 −1 +(−2)(2) −1 2 +(−2)(0) −1 −3 +(−2)(1) 0 −1 +2(2) 0 2 +2(0) = −1 −1 +(−2)(2) −1 2 +(−2)(0) −1 −3 +(−2)(1) 0 −1 +2(2) 0 2 +2(0) 0 −3 +2(1) 3 −1 +1(2) 3 2 +1(0) 3 −3 +1(1) = −1 −1 +(−2)(2) −1 2 +(−2)(0) −1 −3 +(−2)(1) 0 −1 +2(2) 0 2 +2(0) 0 −3 +2(1) 3 −1 +1(2) 3 2 +1(0) = −1 −1 +(−2)(2) −1 2 +(−2)(0) −1 −3 +(−2)(1) = −1 −1 +(−2)(2) −1 2 +(−2)(0) −1 −3 +(−2)(1) 0 −1 +2(2) = −1 −1 +(−2)(2) −1 2 +(−2)(0) = −1 −1 +(−2)(2) = −3 −2 1 4 0 2 −1 6 −8

Multiplying Matrices 11. −1 2 −3 2 0 1 −1 −2 0 2 3 1 (2 x 3)(3 x 2) 11. −1 2 −3 2 0 1 −1 −2 0 2 3 1 (2 x 3)(3 x 2) = 2 x 2 = −1 −1 +2 0 +(−3)(3) = −1 −1 +2 0 +(−3)(3) −1 −2 +2 2 +(−3)(1) = −1 −1 +2 0 +(−3)(3) −1 −2 +2 2 +(−3)(1) 2 −1 +0 0 +1(3) = −1 −1 +2 0 +(−3)(3) −1 −2 +2 2 +(−3)(1) 2 −1 +0 0 +1(3) 2 −2 +0 2 +1(1) = −8 3 1 −3