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1 1 A = (1 2 A = 1 a) g) A = m) -1 3 5 -4 -3 -2 -3 A = 3 b) h) A= n) A = 3 3 4 -3 -4) 10 1 1 1 1 c) A =0 1 1 i) A = 6 5 4 o) A=6 3 1 1 1 13 10 8 5 -2 -3) (1 2 -3 1. 3 1 3 1 10 d) A = 0 1 j) A =3 2 5 P) A= 2 5 4 0 0 1 2 2 -1 -3 1 2. 2.

1 1 A = (1 2 A = 1 a) g) A = m) -1 3 5 -4 -3 -2 -3 A = 3 b) h) A= n) A = 3 3 4 -3 -4) 10 1 1 1 1 c) A =0 1 1 i) A = 6 5 4 o) A=6 3 1 1 1 13 10 8 5 -2 -3) (1 2 -3 1. 3 1 3 1 10 d) A = 0 1 j) A =3 2 5 P) A= 2 5 4 0 0 1 2 2 -1 -3 1 2. 2.

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter7: Systems Of Equations And Inequalities
Section7.8: Solving Systems With Cramer's Rule
Problem 3SE: Explain what it means in terms of an inverse for a matrix to have a 0 determinant.
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Q3 - Find the inverse of the matrix? 1 2 A = 35 1 1 A = 1-1 A = -4 a) g) m) -3 3 -2 -1 b) A = h) A = n) A = 4 -3 -4 (1 1 1 1 c) A =0 1 1 i) A = 6 5 o) A =6 4 1 1 1 13 10 8 5 - 2 -3 1 2 -3 (1 3 1 1 10 d) A =0 1 j) A = 3 2 5 P) A= 2 4 0 0 1 2 2 -1 - 3 1 3.
Transcribed Image Text:Q3 - Find the inverse of the matrix? 1 2 A = 35 1 1 A = 1-1 A = -4 a) g) m) -3 3 -2 -1 b) A = h) A = n) A = 4 -3 -4 (1 1 1 1 c) A =0 1 1 i) A = 6 5 o) A =6 4 1 1 1 13 10 8 5 - 2 -3 1 2 -3 (1 3 1 1 10 d) A =0 1 j) A = 3 2 5 P) A= 2 4 0 0 1 2 2 -1 - 3 1 3.
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