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6. Solve 6. Solve (a) 4x + 3y = 28 2x + 5y = 42 the system Ax = d by matrix inversion, where (b) 4x1 + x2 - 5x3 = 8 -2x1 + 3x2 + x3 = 12 3X1 X2 + 4x3 - 7. Is it possible for a matrix to be its own inverse? = 5

6. Solve 6. Solve (a) 4x + 3y = 28 2x + 5y = 42 the system Ax = d by matrix inversion, where (b) 4x1 + x2 - 5x3 = 8 -2x1 + 3x2 + x3 = 12 3X1 X2 + 4x3 - 7. Is it possible for a matrix to be its own inverse? = 5

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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5. Find the inverse of 4 1 -5 -[ 3 1 3 -1 4 A = -2 Chapter 5 Linear Models and Matrix Algebra (Continued) 103 6. Solve the system Ax = d by matrix inversion, where (a) 4x + 3y = 28 (b) 2x + 5y = 42 5.5 Cramer's Rule 7. Is it possible for a matrix to be its own inverse? 4x1 + x2 - 5x3 = 8 -2x₁ + 3x2 + x3 = 12 3x1 - x2 + 4x3 = 5 The method of matrix inversion discussed in Sec. 5.4 enables us to derive a practical, if not always efficient, way of solving a linear-equation system, known as Cramer's rule. Derivation of the Rule Given an equation system Ax = d, where A is n x n, the solution can be written as 1 [by (5.16)] 1-¹d- - (adj A)d
Transcribed Image Text:5. Find the inverse of 4 1 -5 -[ 3 1 3 -1 4 A = -2 Chapter 5 Linear Models and Matrix Algebra (Continued) 103 6. Solve the system Ax = d by matrix inversion, where (a) 4x + 3y = 28 (b) 2x + 5y = 42 5.5 Cramer's Rule 7. Is it possible for a matrix to be its own inverse? 4x1 + x2 - 5x3 = 8 -2x₁ + 3x2 + x3 = 12 3x1 - x2 + 4x3 = 5 The method of matrix inversion discussed in Sec. 5.4 enables us to derive a practical, if not always efficient, way of solving a linear-equation system, known as Cramer's rule. Derivation of the Rule Given an equation system Ax = d, where A is n x n, the solution can be written as 1 [by (5.16)] 1-¹d- - (adj A)d
© 198 STEP 1: Find A x B Exercise 5.प A = 1274 98 니 911 42 433 9a3 952 (4) (3) (4) - (40) (-1) ५ -9604 98X 98 23 3-14 9.3921932 913922931 + (-3) (0) (0) -(3) () () +/- 14=98 98 - 5 31 -14 STEP 3: Find A*1 = 98 (C) -1078 2) (₁ 98 1 १४ -98 798 X ( 48 ) (4+ * 44 ) (++ * ) X 98 98 98x 98 - 588 98% 98 -১। उप ) 11 -5/4-5 443 4 उन 11 -514 -51 4 1 311-2 14-51 -23 STEP 2: Find Cofactor of each element 98 4/3 -686 98 * = 9604 Xx98 -23 31 X 1 X> X 3 912 925 931 911921933 +013 - 024) 5 d= 48 +4+3 +8 -10 + 45= 98 1274 9604 -98 9604 1 1-14-6 14 - 9604 9604 8 12 13-11-7 -1 1 -1078 9604 १४ 9604 n -588 9604 = CM -686 9604 98 9604 9604 9604 " " A -1 Exercise 5.4 Step4 A1 (1) 1274 1604 -98 9604 -9604 9604 -391 -1078 4604 9604 48 9604 -588 9604 97,883,968 72,236,816 - 7,529,536 72,236,816 737,894,528 92,236,816 + 98 9604 1604 9604 8 -45,177,216 92,236, 816 12 124, 237,344 92,136, 816 11, 294, 304 92,236,816 [76,832 67,765,824 92.736,816 4604 115,248 9604 48,020 4504 18 323 340 13,136,816 4,705, 360 2-235,816 461,124.080 72.26.816
Transcribed Image Text:© 198 STEP 1: Find A x B Exercise 5.प A = 1274 98 니 911 42 433 9a3 952 (4) (3) (4) - (40) (-1) ५ -9604 98X 98 23 3-14 9.3921932 913922931 + (-3) (0) (0) -(3) () () +/- 14=98 98 - 5 31 -14 STEP 3: Find A*1 = 98 (C) -1078 2) (₁ 98 1 १४ -98 798 X ( 48 ) (4+ * 44 ) (++ * ) X 98 98 98x 98 - 588 98% 98 -১। उप ) 11 -5/4-5 443 4 उन 11 -514 -51 4 1 311-2 14-51 -23 STEP 2: Find Cofactor of each element 98 4/3 -686 98 * = 9604 Xx98 -23 31 X 1 X> X 3 912 925 931 911921933 +013 - 024) 5 d= 48 +4+3 +8 -10 + 45= 98 1274 9604 -98 9604 1 1-14-6 14 - 9604 9604 8 12 13-11-7 -1 1 -1078 9604 १४ 9604 n -588 9604 = CM -686 9604 98 9604 9604 9604 " " A -1 Exercise 5.4 Step4 A1 (1) 1274 1604 -98 9604 -9604 9604 -391 -1078 4604 9604 48 9604 -588 9604 97,883,968 72,236,816 - 7,529,536 72,236,816 737,894,528 92,236,816 + 98 9604 1604 9604 8 -45,177,216 92,236, 816 12 124, 237,344 92,136, 816 11, 294, 304 92,236,816 [76,832 67,765,824 92.736,816 4604 115,248 9604 48,020 4504 18 323 340 13,136,816 4,705, 360 2-235,816 461,124.080 72.26.816
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