LU11 Matrices.ppt - Matrices Definition Equality of Matrices Addition and Subtraction of Matrices Multiplication of Matrices Identitiy Matrices

LU11 Matrices.ppt - Matrices Definition Equality of...

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Unformatted text preview: Matrices: Definition & Equality of Matrices Addition and Subtraction of Matrices Multiplication of Matrices Identitiy Matrices Determinant of Matrices Inverse of Matrix Solving System Equations: Inverse of Matrices Method & Cramer’s Rule Method 1 A matrix is an ordered rectangular array of numbers. A matrix with m rows and n columns has size m ☓ n. The entry in the ith row and jth column is denoted by aij. 2 Definition of Matrices A rectangular array of numbers written within brackets 1 3 7 A 5 0 4 2 row: 3 column 2 x 3 matrix 5 0 H 2 3 4 11 1 6 2 8 0 1 4 row: 3 column 4 x 3 matrix Definition of Matrices n x n = Square Matrix of order n 0.5 0.2 1.0 0.0 0.3 0.5 0.7 0.0 0.2 Square Matrix of order 3 3 x 3 matrix 3 2 1 0 Column Matrix 4 x 1 matrix 2 1 2 2 0 3 Row Matrix 1 x 4 matrix Definition of Matrices Notation: aij; i = row and j = column 1 3 7 A 5 0 4 a12 = “a sub one two” a11 = 1 a12 = -3 a13 = 7 a21 = 5 a22 = 0 a23 = -4 Two matrices are equal if they have the same size and their corresponding entries are equal. 6 Solve the following matrix equation for x, y, and z: x 3 1 4 z 1 2 y 1 2 2 1 2 Solution Since the corresponding elements of the two matrices must be equal, we find that x = 4, z = 3, and y – 1 = 1, or y = 2. 7 If A and B are two matrices of the same size, then: 1. The sum A + B is the matrix obtained by adding the corresponding entries in the two matrices. 2. The difference A – B is the matrix obtained by subtracting the corresponding entries in B from those in A. 8 If A, B, and C are matrices of the same size, then 1. A + B = B + A Commutative law 2. (A + B) + C = A + (B + C) Associative law 9 The total output of Acrosonic for May is Model A Model B Model C Model D Location I 320 280 460 280 Location II 480 360 580 0 Location III 540 420 200 880 The total output of Acrosonic for June is Model A Model B Model C Model D Location I 210 180 330 180 Location II 400 300 450 40 Location III 420 280 180 740 Find the total output of the company for May and June. 10 Solution Expressing the output for May and June as matrices: The total output of Acrosonic for May is 320 A 480 540 280 360 420 460 280 580 0 200 880 The total output of Acrosonic for June is 210 B 400 420 180 330 180 300 450 40 280 180 740 11 Solution The total output of the company for May and June is given by the matrix 320 A B 480 540 280 360 420 530 880 960 460 660 700 460 280 210 580 0 400 200 880 420 790 460 1030 40 380 1620 180 300 280 330 180 450 40 180 740 12 The Acrosonic Company manufactures four different loudspeaker systems at three separate locations. The company’s May output is as follows: Model A Model B Model C Model D Location I 320 280 460 280 Location II 480 360 580 0 Location III 540 420 200 880 If we agree to preserve the relative location 460we 280 320 summarize of each entry in the280 table, can 360 580 0 the set of data 480 as follows: 540 420 200 880 13 We have Acrosonic’s May output expressed as a matrix: 320 280 460 280 P 480 360 580 0 540 420 200 880 a. What is the size of the matrix P? Solution Matrix P has three rows and four columns and hence has size 3 ☓ 4. 14 We have Acrosonic’s May output expressed as a matrix: 320 280 460 280 P 480 360 580 0 540 420 200 880 b. Find a24 (the entry in row 2 and column 4 of the matrix P) and give an interpretation of this number. Solution The required entry lies in row 2 and column 4, and is the number 0. This means that no model D loudspeaker system was manufactured at location II in May. 15 We have Acrosonic’s May output expressed as a matrix: 320 280 460 280 P 480 360 580 0 540 420 200 880 c. Find the sum of the entries that make up row 1 of P and interpret the result. Solution The required sum is given by 320 + 280 + 460 + 280 = 1340 which gives the total number of loudspeaker systems manufactured at location I in May as 1340 units. 16 We have Acrosonic’s May output expressed as a matrix: 320 280 460 280 P 480 360 580 0 540 420 200 880 d. Find the sum of the entries that make up column 4 of P and interpret the result. Solution The required sum is given by 280 + 0 + 880 = 1160 giving the output of Model D loudspeaker systems at all locations in May as 1160 units. 17 If A is an m ☓ n matrix with elements aij, then the transpose of A is the n m matrix AT with elements aji. ☓ 18 Find the transpose of the matrix 1 A 4 7 2 5 8 3 6 9 Solution The transpose of the matrix A is 1 AT 2 3 4 5 6 7 8 9 19 If A is a matrix and c is a real number, then the scalar product cA is the matrix obtained by multiplying each entry of A by c. 20 Given 3 4 A 1 2 and 3 2 B 1 2 find the matrix X that satisfies 2X + B = 3A Solution 2 X B 3 A 2 X 3 A B 3 3 1 4 2 3 1 2 2 9 12 3 2 6 10 2 4 3 6 1 2 1 6 10 3 5 X 2 2 4 1 2 21 The management of Acrosonic has decided to increase its July production of loudspeaker systems by 10% (over June output). Find a matrix giving the targeted production for July. Solution We have seen that 180 330 180total output for 210 Acrosonic’s 400 300 450 June may be represented by the B 40 matrix 420 280 180 740 22 The management of Acrosonic has decided to increase its July production of loudspeaker systems by 10% (over June output). Find a matrix giving the targeted production for July. Solution 210 180 330 180 The required matrix is given by (1.1) B 1.1 400 300 450 40 420 280 180 740 231 440 462 198 330 308 363 198 495 44 198 814 23 If we have a row matrix of sizeA 1 ☓[an, a2 1 a3 an ] b1 b And a column matrix of size n ☓ 1, 2 B b3 bn Then we may define the matrix product of A and B, written AB, by AB [a1 a2 a3 b1 b 2 an ] b3 a1b1 a2b2 a3b3 anbn 24 bn Let A [1 2 3 5] and 2 3 B 0 1 Find the matrix product AB. Solution 2 3 AB [1 2 3 5] (1)(2) ( 2)(3) (3)(0) (5)( 1) 9 0 1 25 Note from the last example that for the multiplication to be feasible, the number of columns of the row matrix A must be equal to the number of rows of the column matrix B. From last example, note that the product matrix AB has size 1 ☓ 1. This has to do with the fact that we are multiplying a row matrix with a column matrix. Size of AB (1 ☓ 1) We can establish the dimensions of a product matrix schematically: Size of A (1 ☓ n) (n ☓ 1) Size of B Same 26 More generally, if A is a matrix of size m ☓ n and B is a matrix of size n ☓ p, then the matrix product of A and B, AB, is defined and is a matrix of size m ☓ p. Schematically: Size of AB (m ☓ p) Size of A (m ☓ n) (n ☓ p) Size of B Same The number of columns of A must be the same as the number of rows of B for the multiplication to be feasible. 27 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a a a 23 21 22 b11 B b21 b31 b12 b13 b22 b23 b32 b33 b14 b24 b34 From the schematic Size of A (2 ☓ 3) Size of AB (2 ☓ 4) (3 ☓ 4) Size of B Same we see that the matrix product C = AB is feasible (since the number of columns of A equals the number of rows of B) and has size 2 ☓ 4. 28 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a a a 23 21 22 Thus, b11 B b21 b31 b12 b13 b22 b23 b32 b33 b14 b24 b34 c11 c12 c13 c14 C c c c c 21 22 23 24 To see how to calculate the entries of C consider entry c11: c11 [a11 a12 b11 a13 ] b21 a11b11 a12b21 a13b31 b31 29 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a a a 23 21 22 Thus, b11 B b21 b31 b12 b13 b22 b23 b32 b33 b14 b24 b34 c11 c12 c13 c14 C c c c c 21 22 23 24 Now consider calculating the entry c12: c12 [a11 a12 b12 a13 ] b22 a11b12 a12b22 a13b32 b32 30 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a a a 23 21 22 Thus, b11 B b21 b31 b12 b13 b22 b23 b32 b33 b14 b24 b34 c11 c12 c13 c14 C c c c c 21 22 23 24 Now consider calculating the entry c21: c21 [a21 a22 b11 a23 ] b21 a21b11 a22b21 a23b31 b31 31 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a a a 23 21 22 Thus, b11 B b21 b31 b12 b13 b22 b23 b32 b33 b14 b24 b34 c11 c12 c13 c14 C c c c c 21 22 23 24 Other entries are computed in a similar manner. 32 Let 3 1 4 A 1 2 3 1 3 3 B 4 1 2 2 4 1 Compute AB. Solution Since the number of columns of A is equal to the number of rows of B, the matrix product C = AB is defined. The size of C is 2 ☓ 3. 33 3 1 4 A 1 2 3 Let 1 3 3 B 4 1 2 2 4 1 Compute AB. Solution Thus, 1 3 3 c11 c12 c13 3 1 4 C AB 4 1 2 c c c 1 2 3 2 4 1 21 22 23 Calculate all entries for C: 1 c11 [3 1 4] 4 (3)(1) (1)(4) (4)(2) 15 2 34 3 1 4 A 1 2 3 Let 1 3 3 B 4 1 2 2 4 1 Compute AB. Solution Thus, 1 3 3 15 c12 c13 3 1 4 C AB 4 1 2 c c c 1 2 3 2 4 1 21 22 23 Calculate all entries for C: 3 c12 [3 1 4] 1 (3)(3) (1)( 1) (4)(4) 24 4 35 3 1 4 A 1 2 3 Let 1 3 3 B 4 1 2 2 4 1 Compute AB. Solution Thus, 1 3 3 15 24 c13 3 1 4 C AB 4 1 2 c c c 1 2 3 2 4 1 21 22 23 Calculate all entries for C: 3 c13 [3 1 4] 2 (3)( 3) (1)(2) (4)(1) 3 1 36 3 1 4 A 1 2 3 Let 1 3 3 B 4 1 2 2 4 1 Compute AB. Solution Thus, 1 3 3 15 24 3 3 1 4 C AB 4 1 2 c c c 1 2 3 2 4 1 21 22 23 Calculate all entries for C: 1 c21 [ 1 2 3] 4 ( 1)(1) (2)(4) (3)(2) 13 2 37 3 1 4 A 1 2 3 Let 1 3 3 B 4 1 2 2 4 1 Compute AB. Solution Thus, 1 3 3 15 24 3 3 1 4 C AB 4 1 2 13 c c 1 2 3 2 4 1 22 23 Calculate all entries for C: 3 c22 [ 1 2 3] 1 ( 1)(3) (2)( 1) (3)(4) 7 4 38 3 1 4 A 1 2 3 Let 1 3 3 B 4 1 2 2 4 1 Compute AB. Solution Thus, 1 3 3 15 24 3 3 1 4 C AB 4 1 2 13 7 c 1 2 3 2 4 1 23 Calculate all entries for C: 3 c23 [ 1 2 3] 2 ( 1)( 3) (2)(2) (3)(1) 10 1 39 Let 3 1 4 A 1 2 3 1 3 3 B 4 1 2 2 4 1 Compute AB. Solution Thus, 1 3 3 3 1 4 15 24 3 C AB 4 1 2 1 2 3 13 7 10 2 4 1 40 If the products and sums are defined for the matrices A, B, and C, then 1. (AB)C = A(BC) Associative law 2. A(B + C) = AB + AC Distributive law 41 Definition of Matrices Diagonal Matrix A square matrix where all elements are zero except elements in the principal diagonal a11 0 : 0 0 a22 : 0 .. 0 .. 0 : : 0 anm Definition of Matrices Identity Matrix A diagonal matrix with all elements in the principal are 1. Denoted by I 1 0 0 1 1 0 0 0 1 0 0 0 1 The identity matrix of size n is given by 1 0 I n 0 0 1 0 0 0 n rows 1 n columns 44 The identity matrix has the properties that In A = A for any n ☓ r matrix A. BIn = B for any s ☓ n matrix B. In particular, if A is a square matrix of size n, Ithen n A AI n A 45 Let 1 3 1 A 4 3 2 1 0 1 Then 1 0 0 1 3 1 1 3 1 I 3 A 0 1 0 4 3 2 4 3 2 A 0 0 1 1 0 1 1 0 1 1 3 1 1 0 0 1 3 1 AI 3 4 3 2 0 1 0 4 3 2 A 1 0 1 0 0 1 1 0 1 So, I3A = AI3 = A. 46 Determinant of Matrices Determinant of 2 x 2 Matrix Second Order Determinant a11 A a21 a12 a11 a22 a21a12 a22 a c b ad bc d 1 2 Det ( A) ( 1)( 4) (3)(2) 4 6 2 3 4 Minor of Matrix Determinant of matrix obtained by deleting the ith row and jth row column of A 2 3 1 A 4 6 5 9 8 7 Minor of element A21 Minor of element A23 2 3 1 A 4 6 5 9 8 7 3 1 8 7 3(7) 8(1) 13 2 3 1 A 4 6 5 9 8 7 2 3 9 8 2(8) 9(3) 11 Cofactor Matrix Cofactor Cofactor Theminor minor of of aaparticular particular element element The togetherwith with its itsassociated associatedsign sign together a13 = + a32 = - Cofactor Matrix 2 3 1 A 4 6 5 9 8 7 Cofactor of A21 3 1 8 7 [3(7) 8(1)] 13 Cofactor of A23 2 3 9 8 [2(8) 9(3)] 11 Determinants of Matrices Third-Order Determinant (I) – Cofactor Method a22 A a11 a32 a11 A a21 a12 a22 a13 a23 a31 a32 a33 a23 a21 a12 a33 a31 a23 a21 a22 a13 a33 a31 a32 Determinants of Matrices Third-Order Determinant (I) – Cofactor Method 2 1 3 4 0 1 1 2 3 EXAMPLE 4 2 0 1 2 3 1 4 1 1 3 ( 3) 4 0 1 2 2[ 0(3) 2(1) ] 1[ 4(3) 1(1) ] 3[ 4(2) 1(0) ] 2( 2) 1(11) 3(8) 4 11 24 39 Determinants of Matrices Third-Order Determinant (II) – Expansion Method a11 a 21 a31 P4 a12 a 22 a32 P5 a13 a11 a12 a 23 a 21 a 22 a33 a31 a32 P6 P1 P2 P3 ( P1 P2 P3 ) ( P4 P5 P6 ) P1 P2 P3 P4 P5 P6 a11a22 a33 a12 a23a31 a13a21a32 a13a22 a31 a11a23a32 a12 a21a33 Determinants of Matrices Third-Order Determinant (II) – Expansion Method 2 1 3 EXAMPLE 5 2 1 3 4 0 1 2 4 1 0 1 2 1 2 3 4 0 1 1 2 3 [2(0)(3) 1(1)(1) ( 3)(4)(2)] [1(0)( 3) 2(1)(2) 3(4)(1)] [0 1 24] [0 4 12] 23 16 39 Evaluate the determinant of the following: a)Cofactor method EXAMPLE 6 b)Expansion method 6 1 2 A 5 3 7 4 2 1 2 A 5 4 6 3 2 1 7 1 Cofactor Method EXAMPLE 6 SOLUTION 2 3 7 2 1 6 5 7 4 1 ( 1) 5 3 4 2 2[ 3(1) ( 7)( 2) ] 6[ 5(1) ( 7)( 4) ] ( 1)[ 5( 2) (3)( 4) ] 2( 11) 6( 23) ( 1)( 2) 22 138 2 114 Expansion Method 2 6 1 2 6 5 3 7 5 3 4 2 1 4 2 [2(3)(1) 6( 7)( 4) ( 1)(5)( 2)] [ 1(3)( 4) 2( 7)( 2) 6(5)(1)] [6 168 10] [12 28 30] 184 70 114 Matrix Equation Linear equations can be represented by using matrix multiplication x1 4 x2 2 x3 4 2 x1 3 x2 x3 3 A system of linear equations can be expressed in the form of an equation of matrices. Consider 2 x the 4 y system z 6 3x 6 y 5z 1 x 3y 7z 0 The coefficients on the left-hand side of the equation can be expressed as matrix A below, the variables as matrix X, and the constants 4 of 1 the equation 6 2 side x on right-hand as matrix A 3 6 5 X y B 1 B: 1 3 7 z 0 58 A system of linear equations can be expressed in the form of an equation of matrices. Consider 2 x the 4 y system z 6 3x 6 y 5z 1 x 3y 7z 0 The matrix representation of the system of linear equations 4 1 by 2is given x AX 6= B, or 3 6 5 y 1 1 3 7 z 0 59 A system of linear equations can be expressed in the form of an equation of matrices. Consider the system 2x 4 y z 6 3x 6 y 5z 1 x 3y 7z 0 To confirm this, we can multiply the two matrices on the left-hand side of the equation, obtaining 2 x 4 y z 6 3x 6 y 5z 1 x 3 y 7 z 0 which, by matrix equality, is easily seen to be equivalent to the given system of linear equations. 60 Let A be a square matrix of size n. A square matrix A–1 of size n such that A 1 A AA 1 I n is called the inverse of A. Not every matrix has an inverse. A square matrix that has an inverse is said to be nonsingular. A square matrix that does not have an inverse is said to be singular. 61 Inverse Matrix [2x2] Step Details 1 Find the determinant of the matrix. 2 Find the adjoint matrix. 3 Find inverse matrix: 1 A Adjoin t A 1 Inverse Matrix a A c b d d ad bc 1 A c ad bc b ad bc 1 d a A c ad bc Determinant b a Adjoint Matrix A ad bc; provided A 0 • If A= 0, it is impossible to calculate the inverse, because we cannot divide by zero. • If the matrix has zero determinant it is said to be non-singular (inverse does not exist), otherwise it is said to be singular (inverse exist). Inverse Matrix [2x2] 3 1 A 4 2 EXAMPLE 7 A 3(2) 4(1) 6 4 2 d b 1 ad bc c a 1 1 1 2 1 1 2 A 2 4 3 2 11 2 Inverse Matrix [2x2] EXAMPLE 8 3 0 A 2 1 2 3 B 1 0 6 3 C 2 1 Given the above A, B and C matrix, find the inverse for each matrix of the above. 3 A 2 0 1 Step 1: Find the determinant of the matrix. EXAMPLE 8 SOLUTION 3 A 2 0 (3)(1) ( 2)(0) 3 0 3 1 Step 2: Find the adjoint matrix. 3 Adjoin t of A 2 0 1 1 2 Step 3: Find inverse matrix A 1 A 1 1 Adjoin t A 1 1 3 2 0 1 3 3 2 3 0 1 0 3 1 2 0 3 2 B 1 3 0 Step 1: Find the determinant of the matrix. EXAMPLE 8 SOLUTION 2 B 1 3 ( 2)(0) (1)(3) 0 3 3 0 Step 2: Find the adjoint matrix. 2 3 Adjoin t of B 1 0 0 3 1 2 Step 3: Find inverse matrix 1 B Adjoin t B 1 1 1 0 3 0 B 1 2 1 2 3 3 3 1 6 3 C 2 1 EXAMPLE 8 SOLUTION Step 1: Find the determinant of the matrix. 6 3 C (6)(1) ( 2)(3) 6 6 0 2 1 Answer: Unable to find the inverse matrix as the determinant is 0 1 2 The matrixA 3 4 1 2 A 3 has a matrix 1 2 2 1 as its inverse. This can be demonstrated by multiplying them: 1 1 0 1 2 2 AA I 3 1 3 4 2 2 0 1 1 1 1 2 1 0 2 A A 3 I 1 2 2 3 4 0 1 1 69 0 1 The matrixB 0 0 does not have an inverse. a b where c d 1 If B had an inverse given B by a, b, c, and d are some appropriate numbers, then by definition of an inverse we would have BB–1 = I. 0 1 a b 1 0 That is 0 0 c d 0 1 c d 1 0 0 0 0 1 implying that 0 = 1, which is impossible! 70 Finding the Inverse of a Square Matrix Inverse Matrix [3x3] Step Details 1 Find the determinant of the matrix. 2 Find the minors of the matrix. 3 Find the cofactor of the matrix: 4 Find the adjoint matrix. 5 Find the inverse matrix. Inverse Matrix [3x3] Find the inverse matrix of the following. EXAMPLE 9 2 1 3 A 0 3 1 1 0 2 Find the inverse matrix of the following. 2 1 3 A 0 3 1 1 0 2 EXAMPLE 9 SOLUTION Step 1: Find the determinant of the matrix. 2 1 3 2 1 A0 3 1 0 3 1 0 2 1 0 [( 2)(3)( 2) ( 1)(1)(1) (3)(0)(0)] [(1)(3)(3) (0)(1)( 2) (2)(0)( 1)] [12 1 0] [9 0 0] 11 9 2 Step 2: Find the minor of the matrix. EXAMPLE 9 2 1 3 3 1 M 11 0 3 1 6 0 6 0 2 1 0 2 2 1 3 1 3 M 21 0 3 1 ( 2 0) 2 0 2 1 0 2 2 1 3 1 3 M 31 0 3 1 1 9 10 3 1 1 0 2 SOLUTION M 12 2 0 1 M 22 2 0 1 M 32 2 0 1 1 3 0 3 1 1 0 2 1 (0 1) 1 2 1 3 2 3 1 1 0 2 3 4 3 1 2 1 3 2 3 3 1 (2 0) 2 0 1 0 2 M 13 2 1 3 0 3 0 3 1 0 3 3 1 0 1 0 2 M 23 2 1 3 2 1 0 3 1 (0 1) 1 1 0 1 0 2 M 33 2 1 3 2 1 0 3 1 6 0 6 0 3 1 0 2 The minor of the matrix 1 3 6 1 2 1 10 2 6 Step 3: Find the cofactor of the matrix. EXAMPLE 9 SOLUTION 1 3 6 1 3 6 1 2 1 1 2 1 10 2 10 2 6 6 The minor of the matrix Adjoint Matrix a11 a12 A a21 a22 a 31 a32 a13 a23 a33 11 12 13 21 22 23 32 33 31 Matrix of Cofactor T 11 12 13 11 21 31 adj A 21 22 23 12 22 32 32 33 23 33 31 13 Step 4: Find the adjoint matrix. EXAMPLE 9 SOLUTION 6 Cofactor matrix 2 10 1 1 2 6 Adj A 2 10 T 1 1 2 3 1 6 Step 5: Find the inverse matrix. 3 1 6 6 1 3 2 1 1 A 1 2 10 3 6 1 1 A 1 1 2 1 2 2 3 1 6 3 2 10 2 6 1 adj A A 1 5 1 1 2 1 3 2 If AX = B is a linear system of n equations in n unknowns and if A–1 exists, then X = A–1B is the unique solution of the system. 78 Solving Equation (Inverse Matrix [2x2]) Step Details 1 Convert to matrix equation. 2 Find the determinant of the matrix. 3 Find the adjoint matrix. 4 Find inverse matrix: 5 Solve the equation. 1 A Adjoin t A 1 Using Inverse to Solve a System x1 2 x2 5 3 x1 7 x2 18 EXAMPLE Step 1: Convert to matrix equation. 10 1 2 x1 5 3 7 x 18 2 A X B Ax = B Step 2: Find the determinant of the matrix. A EXAMPLE 10 1 2 3 7 (1)(7) (3)(2) 7 6 1 Step 3: Find the adjoint matrix. 1 2 7 2 Adjoin t of A 3 7 3 1 Step 4: Find inverse matrix A 1 EXAMPLE 10 1 Adjoin t A 1 7 2 7 2 A 1 3 1 3 1 1 Step 5: Solve the matrix equation. x=A B -1 1 3 2 x1 5 7 x 2 18 A X B x1 7 2 5 35 36 1 x 3 1 18 15 18 3 2 Therefore x1 = -1 and x2 = 3 Ax = B Solving Equation (Inverse Matrix [3x3]) Step Details 1 Convert to matrix equation. 2 Find the determinant of the matrix. 3 Find the minors of the matrix. 4 Find the cofactor of the matrix: 5 Find the adjoint matrix. 6 Find the inverse matrix. 7 Solve the equation. Solving Equation (Inverse Matrix [3x3]) x y z 6 x y z 2 EXAMPLE 12 2 x y 3 z 6 Step 1: Convert to matrix equation. 1 1 1 x 6 1 1 1 y 2 2 1 3 z 6 A X B Ax = B Step 2: Find the determinant of the matrix. EXAMPLE 12 1 1 1 1 1 A1 1 1 1 1 2 1 3 2 1 [(1)( 1)(3) (1)(1)(2) (1)(1)( 1)] [(2)( 1)(1) ( 1)(1)(1) (3)(1)(1)] [ 3 2 1] [ 2 1 3] 2 0 2 Step 3: Find the minor of the matrix. EXAMPL E 12 1 1 1 1 1 M 11 1 1 1 3 ( 1) 2 1 3 2 1 3 1 1 M 13 1 1 1 2 1 1 1 M 21 1 1 1 3 ( 1) 4 1 3 2 1 3 M 23 1 2 1 1 1 1 1 M 31 1 1 1 1 ( 1) 2 1 1 2 1 3 1 1 2 M 12 1 1 2 1 1 1 1 1 1 2 3 1 3 2 1 3 M 22 1 1 2 1 1 1 1 1 1 2 3 1 3 2 1 3 M 32 1 1 2 1 1 1 1 1 1 1 1 1 0 1 1 3 M 33 1 1 1 1 1 2 1 3 1 1 1 1 1 2 1 3 1 1 ( 2) 1 1 1 1 2 3 1 1 1 1 1 1 1 1 1 2 1 1 1 3 The minor of the matrix 2 1 1 4 1 3 2 0 2 Step 4: Find the cofactor of the matrix. EXAMPLE 12 . 2 1 1 2 1 1 3 4 1 3 4 1 2 0 2 2 0 2 Step 5: Find the adjoint matrix. T 2 1 1 2 4 2 adj A 4 1 3 1 1 0 2 1 0 2 3 2 EXAMPLE 12 2 4 2 adj A 1 1 0 1 3 2 1 A adj A A 1 Step 6: Find the inverse matrix. 2 2 4 2 1 1 1 A 0 1 1 1 1 2 2 2 3 2 1 3 1 2 2 1 0 1 Step 7: Solve the equation. X = A-1 B EXAMPLE 12 2 1 6 6 4 ( 6) 4 x 1 y 1 2 1 0 2 3 1 0 2 2 3 z 1 2 2 1 6 3 3 6 0 Therefore x = 4, y = 2 and z = 0 Solve the system of linear equations 2x y z 1 3x 2 y z 2 2 x y 2 z 1 a)Write the system of equations in the form AX =B b)Solve the matrices equation using inverse of matrices method 90 Solve the system of linear equations 2x y z 1 3x 2 y z 2 2 x y 2 z 1 Solution Write the system of equations in the form AX = B where 2 A 3 2 1 2 1 1 1 2 x X y z 1 B 2 1 91 Solve the system of linear equations 2x y z ...
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