1. Algebra 2 Chapter 4 Notes Matrices & Determinants Algebra 2 Chapter 4 Notes Matrices & Determinants

2. MATRIX a MATRIX is a rectangular arrangement of rows & columns DIMENSIONS are the number of rows (horizontal) by the number of columns (vertical) Matrix A Matrix A = This matrix has 2 rows x 3 columns, so it is a 2 x 3 matrix. 62─1 ─205 To add or subtract matrices, just add or subtract the corresponding entries. NOTE: you can add or subtract only if each matrix has the same dimensions. 3 ─4 7 1 0 3 + = 4 10 83 40 2─7 6─1 - = 610 ─21 REMEMBER: “Rows x Columns” when describing a Matrix. Matrix Operations 4.1

3. Multiplying Scalars In matrix algebra, a real number is often called a scalar To multiply a matrix by a scalar, you multiply each entry in the matrix by the scalar. This process is called “Scalar Multiplication.”” + = ─20 47 ─60 1221 = 3 1─2 03 ─45 ─2 ─45 0─8 ─26 4 0─6 8─10 + ─45 0─8 ─26 = ─69 6─14 6─4 4.1 Multiplying Scalars

4. REMEMBER: “Rows x Columns” when describing a Matrix. It is NOT “ Columns x Rows” 4.2 Properties of Matrix Operations

5. The product of 2 matrices A and B is defined provided the number of columns in matrix A = the number of rows in matrix B A x B = AB m x n n x p m x p dimensions of AB Example 1: 3 x 2 x 2 x 4 = 3 x 4 6 entries 8 entries 12 entries 1234 5678 9101112 12 34 56 1234 5678 Multiplying Matrices 4.2

6. To multiply each matrix, multiply EACH Row by EACH column Properties of Matrix Mulltiplication Associative Property of x( A B ) C = A ( B C ) Left distributiveA (B + C) = AB + AC Right distributive ( A + B ) C = AC + BC Associative Property of a scalar C ( A B ) = (CA) B = A (CB) Properties of Matrix Multiplications 4.2

7. 12 34 56 78 910 = 1 x 7 + 2 x 91 x 8 + 2 x 10 3 x 7 + 4 x 93 x 8 + 4 x 10 5 x 7 + 6 x 95 x 8 + 6 x 10 = 7 + 18 = 258 + 20 = 28 21 + 36 = 5724 + 40 = 64 35 + 54 = 8940 + 60 = 100 = 2528 5764 89100 A B AB 3 x 2 x 2 x 2 = 3 x 2 REMEMBER: you multiply each ENTIRE row by each ENTIRE column. You multiple and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix. Multiplying Matrices 4.2

8. 12 34 56 78 910 = 1 x 7 + 2 x 91 x 8 + 2 x 10 3 x 7 + 4 x 93 x 8 + 4 x 10 5 x 7 + 6 x 95 x 8 + 6 x 10 = 7 + 18 = 258 + 20 = 28 21 + 36 = 5724 + 40 = 64 35 + 54 = 8940 + 60 = 100 = 2528 5764 89100 A B AB 3 x 2 x 2 x 2 = 3 x 2 REMEMBER: you multiply each ENTIRE row by each ENTIRE column and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix. Multiplying Matrices 4.2

9. 12 34 56 78 910 = 1 x 7 + 2 x 91 x 8 + 2 x 10 3 x 7 + 4 x 93 x 8 + 4 x 10 5 x 7 + 6 x 95 x 8 + 6 x 10 = 7 + 18 = 258 + 20 = 28 21 + 36 = 5724 + 40 = 64 35 + 54 = 8940 + 60 = 100 = 2528 5764 89100 A B AB 3 x 2 x 2 x 2 = 3 x 2 REMEMBER: you multiply each ENTIRE row by each ENTIRE column and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix. Multiplying Matrices 4.2

10. 12 34 56 78 910 = 1 x 7 + 2 x 91 x 8 + 2 x 10 3 x 7 + 4 x 93 x 8 + 4 x 10 5 x 7 + 6 x 95 x 8 + 6 x 10 = 7 + 18 = 258 + 20 = 28 21 + 36 = 5724 + 40 = 64 35 + 54 = 8940 + 60 = 100 = 2528 5764 89100 A B AB 3 x 2 x 2 x 2 = 3 x 2 REMEMBER: you multiply each ENTIRE row by each ENTIRE column and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix. Multiplying Matrices 4.2

11. 12 34 56 78 910 = 1 x 7 + 2 x 91 x 8 + 2 x 10 3 x 7 + 4 x 93 x 8 + 4 x 10 5 x 7 + 6 x 95 x 8 + 6 x 10 = 7 + 18 = 258 + 20 = 28 21 + 36 = 5724 + 40 = 64 35 + 54 = 8940 + 60 = 100 = 2528 5764 89100 A B AB 3 x 2 x 2 x 2 = 3 x 2 REMEMBER: you multiply each ENTIRE row by each ENTIRE column and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix. Multiplying Matrices 4.2

12. 12 34 56 78 910 = 1 x 7 + 2 x 91 x 8 + 2 x 10 3 x 7 + 4 x 93 x 8 + 4 x 10 5 x 7 + 6 x 95 x 8 + 6 x 10 = 7 + 18 = 258 + 20 = 28 21 + 36 = 5724 + 40 = 64 35 + 54 = 8940 + 60 = 100 = 2528 5764 89100 A B AB 3 x 2 x 2 x 2 = 3 x 2 REMEMBER: you multiply each ENTIRE row by each ENTIRE column and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix. Multiplying Matrices 4.2

13. 12 34 56 78 910 = 1 x 7 + 2 x 91 x 8 + 2 x 10 3 x 7 + 4 x 93 x 8 + 4 x 10 5 x 7 + 6 x 95 x 8 + 6 x 10 = 7 + 18 = 258 + 20 = 28 21 + 36 = 5724 + 40 = 64 35 + 54 = 8940 + 60 = 100 = 2528 5764 89100 A B AB 3 x 2 x 2 x 2 = 3 x 2 REMEMBER: you multiply each ENTIRE row by each ENTIRE column and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix. Multiplying Matrices 4.2

14. ─23 1 ─4─4 60 ─1 3─2 4 = ─2 x ─1 + 3 x ─2─2 x 3 + 3 x 4 1 x ─1 + ─ 4 x ─21 x 3 + ─ 4 x 4 6 x ─1 + 0 x ─26 x 3 + 0 x 4 = 2 + ─ 6 = ─ 4 ─ 6 + 12 = 6 ─ 1 + 8 = 73 ─ 16 = ─ 13 ─ 6 + 0 = ─ 618 + 0 = 18 = ─ 46 7 ─ 13 ─ 618 A B AB 3 x 2 x 2 x 2 = 3 x 2 REMEMBER: you multiply each row by each column. That is why the number of columns of the first matrix must equal the number of rows of the second matrix. Multiplying Matrices 4.2

15. det ab cd = Determinants and Cramer’s Rule Associated with each square matrix is a real # called it’s determinant, denoted by det A or │ A │. ab cd =ad ─ bc The determinant of a 2 x 2 matrix is the difference of the products of the entries on the diagonals det Example 1 13 25 = 13 25 =1 5 ─ 2 3= 5 ─ 6 = ─ 1 Determinants and Cramer’s Rule 4.3

16. det = 2 steps to get the determinant of a 3 x 3 matrix: 1.Repeat first 2 columns to the right of the matrix 2.Select the sum of the products the products in red from the sum of the products in blue. = det Example 1 = = abc def ghi abc def ghi ab de gh (aei + bfg + cdh ) ─ ( gec + hfa + idb ) 2 ─ 1 3 ─ 2 0f 124 2 ─ 1 3 ─ 2 0f 124 2 ─ 1 ─ 2 0 12 (0 + ─ 1 + ─ 12 ) ─ ( 0 + 4 + 8 ) = ─ 13 ─ 12 = ─ 25 Determinants and Cramer’s Rule 4.3

17. A = ± ½ A Determinant can find the area of a triangle whose vertices are points in a coordinate plane. Area of a Triangle with vertices, ( x 1, y 1 ), ( x 2, y 2 ), ( x 3, y 3 ) A = ± ½ Example 1 = ± ½ x1x1 y1y1 1 x2x2 y2y2 1 x3x3 y3y3 1 1 2 1 4 01 621 (0 + 12 + 8 ) ─ ( 0 + 2 + 8 ) = ½ ( 20 ─ 10) = 5 Where ± indicates the appropriate sign should be chosen to yield a positive sign. ( x 1, y 1 ) ( x 2, y 2 ) ( x 3, y 3 ) 1 2 4 0 62 Determinants and Cramer’s Rule 4.3

18. Cramer’s Rule for a 2 x 2 matrix: Let A be the coefficient matrix of this linear system. Linear System: ax + by = e cx + dy = f Coefficient Matrix: ab cd =det A If det A ≠ 0, then the system has exactly one solution, the solution is: eb fd x = det A ae cf y = and det A Determinants and Cramer’s Rule 4.3

19. Cramer’s Rule for a 2 x 2 matrix: Let A be the coefficient matrix of this linear system. 8 x + 5 y = 2 2 x ─ 4 y = ─ 10 Coefficient Matrix: 85 2 ─ 4 = det A If det A ≠ 0, then the system has exactly one solution, the solution is: eb fd x = det A ae cf y = det A Linear System: a x + b y = e c x + d y = f = (8) ( ─ 4 ) ─ (2) (5 ) ( ─ 32 ) ─ (10) ( ─ 42 ) = 25 ─ 10 ─ 4─ 4 ─ 42 = 82 2 ─ 10 ─ 42 (2) ( ─ 4 ) ─ ( ─ 10) (5 ) =( ─ 8 ) + (50)==42 ─ 42 = (8) ( ─ 10 ) ─ (2) (2 ) ─ 42 = = ─ 1 ( ─ 80 ) ─ (4) ─ 42 = ─ 84 ─ 42 = 2 ( x, y ) ( ─ 1, 2 ) Determinants and Cramer’s Rule 4.3

20. Cramer’s Rule for a 3 x 3 matrix: Let A be the coefficient matrix of this linear system. Coefficient Matrix: = det A x = det A y = det A Linear System: a x + b y + c z = j d x + e y + f z = k g x + h y + i z = l ( x, y, z ) abc def ghi ab de gh = ( aei +bfg + cdh ) – ( ceg + afh + bdi ) If det A ≠ 0, then the system has exactly one solution, the solution is: jbc kef lHi ajc dkf gli z = abj dek gHl det A 4.3

21. Cramer’s Rule for a 3 x 3 matrix: Let A be the coefficient matrix of this linear system. Coefficient Matrix: = det A x = – 10 y = – 10 Linear System: x + 4 y = 16 3 x + 8 y + 3 z = 92 2 y + z = 18 ( x, y, z ) ( 12, 1, 16 ) 140 383 02I 14 38 02 = ( 8 + 0 + 0 ) – ( 0 + 6 + 12 ) = – 10 If det A ≠ 0, then the system has exactly one solution, the solution is: 1640 9283 1821 1160 3923 0181 z = 1416 3892 0218 164 928 182 116 392 018 14 38 02 = ( 128 + 216 + 0 ) – ( 0 + 96 + 368) – 10 = ( 92 + 0 + 0 ) – ( 0 + 54 + 48) – 10 = ( 144 + 0 + 96 ) – ( 0 + 184 + 216) – 10 = – 120 – 10 = – 10 – 10 = – 160 – 10 = 12 = 1 = 16 – 10 4.3

22. Identity and Inverse Matrices The number 1 is the multiplicative identity for real #’s because 1 a = a 1 = a For matrices, the n x n Identity matrix is the matrix that has 1’s on the main diagonal and 0’s elsewhere. I = 100 010 00I 10 01 2 x 2 Identity matrix 3 x 3 Identity matrix If A is any n x n matrix and I is the Identity matrix, then I A = A and A I = A Two n x n matrices are inverses of each other if their products (in both orders) is the n x n identity matrix. Remember: 3/4 4/3 = 1 AB = BA = 3–1 –52 21 53 21 53 3–1 –52 10 01 10 01 = = = I Identity and Inverse Matricies 4.4

23. Inverse Matrices The inverse of a 2 x 2 matrix A = is ab cd A –1 = 1 │ A │ d ––b––b ––c––ca = 1 ad–cbad–cb provided ad–cb ≠ 0 d ––b––b ––c––ca 1 32–41 A –1 = 2 ––1––1 ––4––43 Remember : │ A │ and ad–cb are the determinant of matrix A. 31 42 A = Example: = 1 6–4 2 ––1––1 ––4––43 = 1 2 2 ––1––1 ––4––43 = 1 ––12––12 ––2––23 332332 Identity and Inverse Matricies 4.4

24. Inverse Matrices ab cd A –1 = 1 ad–cb d ––b––b ––c––ca A =  Switch a and d  Change b and c to opposite signs  Multiply by 1 divided by determinant of matrix A  Switch a and d  Change b and c to opposite signs  Multiply by 1 divided by determinant of matrix A 4.4

25. Example 2: Solve matrix equation, A X = B for 2 x 2 matrix X 4–1 –31 8–5 –63 X = Solution: Find inverse of A A –1 = 1 4 – 3 11 34 = 11 34 4–1 –31 11 34 X = 8–5 –63 11 34 10 01 X = 2–2 0–3 X = 2–2 0–3 AB A X = B 1 A X = 1 B A 1 X = B A X = B A Solving a Matrix Equation 4.4

26. Solving systems using inverse matricies 5–4 12 Remember: Row x Column = Coefficient matrix Matrix of variables 8 6 x y Matrix of constants 5 x – 4 y = 8 1 x + 2 y = 6 Example 1: Write a system of linear equations as a matrix equation – 3 x + 4 y = 5 2 x – 1 y = – 10 –34 2 – 1 x y = 5 – 10 Example 2: Use matricies to solve the linear system A –1 = 1 3 – 8 – 1 –4 –2–3 = 1515 4545 2525 3535 A –1 = 1 – 5 – 1 –4 –2–3 x y = 5 – 10 1515 4545 2525 3535 = 1 ─ 8 2 – 10 = ─ 7 – 4 ( X, Y ) ( ─ 7, – 4 ) Solving Systems using Inverse Matricies 4.4

27. 62─1 ─205 83 4─1 3 ─2 7 14 000 000 14 0 00 62─1 1234 5678 9101112 1234 5678 1 0