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Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Where did I go wrong on this problem?
![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](https://content.bartleby.com/qna-images/question/18def6ce-9aea-4bab-b8da-98b00690cf31/599142bd-ecc1-48f7-a971-726d064191a4/4q412vw_thumbnail.jpeg)
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

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
Expert Solution
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