Syllabus Objectives: 4. The student will organize data using matrices. 4.2 The student will simplify matrix expressions using the properties of matrices. Matrix: a rectangular arrangement of numbers (called entries) in rows and columns. Plural of matrix: matrices Dimensions (Order) of a Matrix: Number of Rows (by) Number of Columns Square Matrix: a matrix with the same number of rows and columns Organizing Data in a Matrix Ex: Store A sells 550 DVDs, 420 video games, and 90 CDs on average every week. Store B sells 405 DVDs, 300 video games, and 00 CDs on average every week. Use a matrix to organize this information. State the dimensions of this matrix. DVDs VGs CDs Solution: Store A 550 420 90 The dimensions (order) of the matrix: 2 3 Store B 405 300 00 Adding & Subtracting Matrices: In order to add or subtract matrices, the dimensions must be the same. Ex: Add the matrices: [ 4 3 ] + [ 8 5]. Step One: Determine if the matrices can be added by checking the dimensions. The matrices both have dimensions 3, so they can be added. Step Two: Add the corresponding entries. The sum of the matrices will have the same dimensions as the original matrices. [ 4 3 ] + [ 8 5] = [ 4+ 8 3+ + 5] = [ 2 4 4] Ex: Subtract the matrices: 3 0 2 5 9 5 8 6 0 3 7 Step One: Determine if the matrices can be added by checking the dimensions. The matrices both have dimensions 2 3, so they can be subtracted. 3 0 2 5 9 Step Two: Change to an addition problem (add the opposite). + 5 8 6 0 3 7 Step Three: Add the matrices. 3 0 2 5 9 2 9 5 8 6 + = 0 3 7 5 2 Page of 3
Scalar: another name for a real number Scalar Multiplication Ex: Find the product. 6 0 2 3 3 4 2 5 Multiply every entry in the matrix by the scalar. 6 8 3 0 2 0 6 3 = 3 4 9 2 2 5 6 5 Equal Matrices: two matrices are equal if their dimensions are the same, and all of the corresponding entries are equal. Ex: Solve the equation for x and y. 4 2x 5 9 3 + = 8 5y 2 2 3 Step One: Add the matrices on the left side of the equation. 4+ 5 2x+ 9 2x+ = 8+ 5y 2 8+ 5y 3 Step Two: Set the corresponding entries equal to each other and solve for the variables. 2x + = 3 8+ 5y = 2 2x = 2 5y = 20 x = 6 y = 4 You Try: Perform the indicated operations. 7 0 2 0 8 6 3 2 4 6 3 5 QOD: Is matrix addition commutative? Explain your answer. Page 2 of 3
Syllabus Objective: 4.2 The student will simplify matrix expressions using the properties of matrices. Matrix Multiplication: In order to multiply matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. The product matrix will have the same number of rows of the first matrix and the same number of columns of the second matrix. Let A be an m nmatrix, and B be an n pmatrix. The product A B exists, and will have the dimensions m p. Ex: Find the product AB (if possible). 2 3 A = 5 0 2 4 B = 3 6 Step One: Determine if the product exists. If it does, find its dimensions. Matrix A is a 3 2 matrix. Matrix B is a 2 2 matrix. The number of columns in matrix A equals the number of rows in matrix B. Therefore, the product exists and will be a 3 2matrix. Step Two: Multiply each entry in the rows of matrix A to each entry in the columns of matrix B. Then find the sum of these products. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 4 + 3 3 2 + 3 6 20 3 5 6 9 29 + + = 04+ 2 3 0+ 26 6 2 Ex: Find the product BA (if possible) of the matrices from the previous example. Step One: Determine if the product exists. If it does, find its dimensions. Matrix B is a 2 2 matrix. Matrix A is a 3 2 matrix. The number of columns in Matrix B does not equal the number of rows in matrix A. Therefore, the product BA does not exist. Page 3 of 3
Ex: Find the product AC if 2 7 3 3 A = 2 0 8 4 and 6 C = 8 2 (if possible). Matrix A is a 4 3 matrix. Matrix C is a 3 matrix. The number of columns in matrix A equals the number of rows in matrix C. Therefore, the product AC exists, and has the dimensions 4. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 6 + 2 8+ 72 + 6+ 4 3 6 + 3 8+ 32 6 24 6 36 AC = = = 46+ 5 8+ 22 24 40 + 4 2 06+ 8 8+ 42 0 64 8 72 Application Problem Multiplying Matrices Ex: A school is selling tickets to a school play. They sell tickets for $5 for balcony seating and $7 for floor seating. The school sells 60 balcony seats and 25 floor seats to parents and relatives. They also sell 40 balcony seats and 50 floor seats to students. Set up the school s profit as a product of two matrices. Seat Matrix Ticket Price Matrix = Profit Seat Matrix: Non-Student Tickets 60 25 Student Tickets 40 50 Ticket Price Matrix: Balcony Price 5 Floor Price 7 Multiply the matrices: ( ) + ( ) ( ) + ( ) 60 25 5 60 5 25 7 Non-Student Profit 205 40 50 7 = = 40 5 50 7 Student Profit 550 You Try: Find the product CA using the matrices 5 6 A = 4 and 4 3 C = 2. 0 8 QOD: Is matrix multiplication commutative? Explain why or why not. Page 4 of 3
Syllabus Objective: 4.3 The student will find the determinant of a matrix with and without technology. 4.5 The student will solve systems of equations using matrices. Determinant of a Matrix Notation: Determinant of Matrix A = det A = A Evaluating the Determinant of a a b ad bc c d = 2 2Matrix Ex: What is the determinant of the matrix A? 3 A = 7 9 det A = ( 3)( 9) ( )( 7) = 27 7 = 20 Evaluating the Determinant of a 3 3 Matrix a b c a b c a b det d e f = d e f d e = ( a e i + b f g + c d h) ( b d i + a f h + c e g ) g h i g h i g h (Sum of the products of (Sum of the products of the diagonals left to right) the diagonals right to left) Ex: Evaluate the following: 5 2 0 3 4 4 2 5 2 5 ( ) ( ) 0 3 4 0 3 = 5 3 2+ 4 + 2 0 4 0 2+ 5 4 4+ 2 3 4 2 4 =( 30 4 + 0) ( 0 80 6) = 26 ( 86) = 26 + 86 = 2 Application: Finding the Area of a Triangle Using Determinants The area of a triangle with vertices (, ),(, ), and (, ) x y x y x y is 2 2 3 3 x y ± x2 y2. 2 x y 3 3 Note: The ± sign indicates that we need to choose which sign will make this a positive number. Page 5 of 3
Ex: Find the area of a triangle whose vertices are the points (,2 ),( 3, 4 ), and ( 0,6). Area = 2 2 2 ± 3 4 = ± 3 4 3 4 2 2 0 6 0 6 0 6 =± ( 4 2 0 3 6) ( 2 3 6 4 0) 2 + + + + =± ( 4 + 0 + 8) ( 6 6 + 0) =± ( 22 0) =± ( 22) = 2 2 2 Solution: Because the determinant is positive, we will use the positive answer for the area of the triangle, sq. units. Solving a Linear System Using Cramer s Rule Linear System: ax + by = e cx + dy = f Coefficient Matrix: a c b d Constant Matrix: e f Cramer s Rule Solution of a Linear System: e b a e f d c f a b x = and y =, if 0 a b a b c d c d c d Note: Constant matrix replaces the column of the coefficients of the variable being solved for. Ex: Solve the system of equations using Cramer s Rule. x+ 7y = 3 3x 5y = 7 3 7 ( 3 5) ( 7 7) ( ) ( ) 7 5 5 9 04 x = = = = = 4 7 5 7 3 5 2 26 3 5 3 ( 7) ( 33) 3 7 7 + 9 26 y = = = = = 7 26 26 26 3 5 x+ 7y = 3: ( 4) + 7( ) = 3 3= 3 Check: Solution: ( 4, ) 3x 5y = 7 : 3( 4) 5( ) = 7 2 + 5 = 7 7 = 7 Page 6 of 3
Solving a System of Three Equations and Three Unknowns with Cramer s Rule Ex: Solve the system using Cramer s Rule. x+ 3y z = 2x 6y+ z = 3 3x+ 5y 2z = 4 Note: We will need the determinant of the coefficient matrix to find the value of each variable, so we will calculate this first. 3 3 3 ( ) ( ) 2 6 = 2 6 2 6 = 2 + 9 + 0 2 + 5 + 8 = 3 35 = 4 3 5 2 3 5 2 3 5 To find x, replace the x-column of the coefficient matrix with the constant matrix. 3 3 3 3 6 3 6 3 6 ( ) ( ) 2 2 2 + 2 + 5 8 + 5 + 24 39 47 8 x = = = = = 2 4 4 4 4 4 = To find y, replace the y-column of the coefficient matrix with the constant matrix. 2 3 2 3 2 3 ( ) ( ) 3 4 2 3 4 2 3 4 6+ 3+ 8 4+ 4+ 9 7 7 0 y = = = = = 0 4 4 4 4 4 = To find z, replace the z-column of the coefficient matrix with the constant matrix. 3 3 3 2 6 3 2 6 3 2 6 ( ) ( ) ( ) 3 5 4 3 5 4 3 5 24 27 0 24 5 8 6 57 4 z = = = = = 4 4 4 4 4 = Solution: (2,0,) (Check this on your own in the original three equations.) Note: We could also use Cramer s Rule to find x and y, then substitute these values into one of the original equations to find z. Page 7 of 3
Graphing Calculator Evaluating Determinants 6 2 5 Ex: Evaluate det 0 3. 9 4 8 Step One: Entering in a Matrix In the Matrix Menu, choose Edit, then Matrix A. Our matrix is a 3 3, so enter in the dimensions. Then enter in the entries of the matrix. Keystrokes: Step Two: Evaluating the Determinant In the Home Screen, go to the Matrix Menu, then MATH, and choose det(. Then go back to the Matrix Menu and choose Matrix A under NAMES. Keystrokes: Note: You can perform all operations with matrices on the graphing calculator. You Try:. Use Cramer s Rule to solve the system 2x+ 5y+ 3z = 0 3x y+ 4z = 8. 5x 2y+ 7z = 2 2. Find the area of the triangle with vertices shown in the coordinate grid. Use your graphing calculator. y 0 5-0 -5 5 0 x -5-0 QOD: In order to use Cramer s Rule, what must be true about the determinant of the coefficient matrix, and why? Page 8 of 3
Syllabus Objective: 4.4 The student will find inverse matrices with or without technology. Recall (let a be a real number): Identity of Multiplication = a = a = a ( ) Identity Matrix 2 2 = 0 = I 0 a b 0 a + b 0 a 0+ b a b c d 0 = = c + d 0 c 0+ d c d Now verify on your own that 0 a b a b 0 = c d c d ( ) Identity Matrix 3 3 = 0 0 0 0 = I 0 0 Recall: The product of a number and its multiplicative inverse is the identity. a = a The inverse of matrix A is denoted by A, and A A A A I = =. Ex: Verify that if 3 A =, then 4 2 A 2 =. 3 2 2 Find the product: 3 3 0 = 4 2 = = 3 3 0 2 4 + 2 2 4 + 2 2 2 2 3 2 3 A A 2 + 2 + 2 Now verify on your own that the product 0 A A= 0. Finding the Inverse of a 2 2. a b If A =, then c d d b A = ad bc c a Note: ( ad bc 0) Ex: Find the inverse of 5 4 A = 4 4. A 4 4 4 4 = ( 54) ( 4 4 ) = 4 =.25 Page 9 of 3
Solving a Matrix Equation Recall: To solve the equation ax = b, we multiply both sides by the multiplicative inverse of a to isolate b the variable. ax = b x = a a a To solve the matrix equation equation by the inverse of A. AX = B for X, where A and B are matrices, multiply both sides of the A AX = A B X = A B Ex: Solve the equation AX = B if 0 0 6 8 A and B = = 6 2 4 2 2. Step One: Write the equation. 0 0 6 8 6 2 X = 4 2 2 Step Two: To solve for X, multiply both sides of the equation by the inverse of A. A AX = A B A 2 0 2 0 = = ( ) ( ) 0 2 0 6 6 2 6 = 3 2 0 0 0 0 6 8 X = 3 6 2 3 4 2 2 2 2 0 + 0 4 6 + 02 8 + 0 2 0 6 8 X = = 30 + 4 36 + 2 38 + 2 32 24 25 2 2 2 Finding Inverse Matrices on the Graphing Calculator Ex: Find the inverse of the matrix 3 2 5 6. 7 2 6 Step One: Enter the matrix into Matrix A. Step Two: On the Home Screen, bring up Matrix A. Then use the key to find the inverse. To make the entries fractions, go to the MATH Menu and choose Frac. Keystrokes: Page 0 of 3
Solving a Two-Step Matrix Equation Ex: Solve the equation 7 9 3 4 9 X + = 4 3 6 6 for X. Step One: Isolate the X term by subtracting the matrix 3 4 4 3 from both sides. 7 9 3 4 3 4 9 3 4 X + 4 3 4 3 = 6 6 4 3 7 9 2 5 X = 2 3 Step Two: Isolate X by multiplying both sides by the inverse of 7 9. 7 9 Inverse of = 5 9 5 9 5 9 = = 7 5 9 4 4 7 35 + 36 4 7 4 7 ( ) ( ) 5 9 7 9 5 9 2 5 4 7 X = 4 7 2 3 5 2+ 9 2 5 5+ 9 3 8 2 X = = 4 2+ 7 2 4 5+ 7 3 6 7 5 8 2 9 3 You Try: Solve the matrix equation X. 4 3 = 6 6 2 QOD: What is the special relationship between a matrix and its inverse? Page of 3
Syllabus Objective: 4.5 The student will solve systems of equations using matrices. Solving Systems of Equations Using Inverse Matrices Ex: Solve the system 2x+ 7y = 3 3x 8y = 23 using an inverse matrix. Step One: Rewrite the system of equations as a matrix equation AX matrix, X is the variable matrix, and B is the matrix of constants. = B, where A is the coefficient 2 7 x 3 = 3 8 y 23 Note: Use matrix multiplication to show that this represents the original system. Step Two: Find the inverse of Matrix A. 2 7 A = 3 8 8 7 8 7 A = = ( 2 8) ( 7 3 ) 3 2 37 3 2 Note: Do not multiply by the scalar. Step Three: Multiply both sides of the equation by A. 8 7 2 7 x 8 7 3 37 3 2 3 8 y = 37 3 2 23 x 5 Therefore, x 8 3 + 7 23 85 5 y = = = = y 37 3 3+ 2 23 37 37 Solution: ( 5,) Check the solution in the original equations. 3 2( ) ( ) = 23 3( ) ( ) 2x+ 7y = 5 + 7 = 3 0+ 7= 3 3= 3 3x 8y 5 8 = 23 5 8 = 23 23 = 23 Now try the same problem on the graphing calculator: Enter the coefficient matrix into Matrix A and the constant matrix into matrix B. Find the product A Bon the home screen. Page 2 of 3
Application Problem System of Three Equations Ex: Lawrence has $25 to spend on picking 2 pounds of three different types of apples in an orchard. Empire apples cost $.40 per pound, Red Delicious apples cost $.0 per pound, and Golden Delicious apples cost $.30 per pound. He wants twice as many Red Delicious apples as the other two kinds combined. How many pounds of each type of apple should Lawrence pick? Let E = # of pounds of Empire apples, R = # of pounds of Red Delicious apples, and G = # of pounds of Golden Delicious apples Total Pounds: E+ R+ G = 2 Total Cost:.4 E+.R+.3G = 25 Other Information: ( ) E 2.4..3 R = 25 2 2 G 0 R = 2 E+ G 2E R+ 2G =0 Now we have a system of three equations with three variables. Rewrite the system as a matrix equation. Enter the coefficient matrix as Matrix A and the constant matrix as Matrix B in the graphing calculator. On the Home Screen, multiply the Inverse of Matrix A by Matrix B. Solution: Lawrence should pick 5 pounds of Empire apples, 4 pounds of Red Delicious apples, and 2 pounds of Golden Delicious apples. You Try: Solve the system the graphing calculator. x+ y = 2x 3y = 2 using an inverse matrix (by hand). Check your answer using QOD: Explain why the solution of AX = B is not X = BA. Page 3 of 3