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you may assume that matrix multiplication is associative and that The determinant | | = ad - bc If ad- bc 0, с -1 (a b)² = 1 (a b) ad-bc For matrices A and B in M₂(R) the determinant of a product is the product of determinats involved, i.e. |AB| = |A||B| Is Sl₂ (R) = {(ab): : ad – bc = 1} a ་ subgroup of Gl₂(R) ? Write a proof if the answer is YES, an example if the answer is NO.

you may assume that matrix multiplication is associative and that The determinant | | = ad - bc If ad- bc 0, с -1 (a b)² = 1 (a b) ad-bc For matrices A and B in M₂(R) the determinant of a product is the product of determinats involved, i.e. |AB| = |A||B| Is Sl₂ (R) = {(ab): : ad – bc = 1} a ་ subgroup of Gl₂(R) ? Write a proof if the answer is YES, an example if the answer is NO.

Algebra for College Students
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter11: Systems Of Equations
Section11.4: Determinants
Problem 45PS
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you may assume that matrix multiplication is associative and that The determinant | | = ad - bc If ad- bc 0, с -1 (a b)² = 1 (a b) ad-bc For matrices A and B in M₂(R) the determinant of a product is the product of determinats involved, i.e. |AB| = |A||B| Is Sl₂ (R) = {(ab): : ad – bc = 1} a ་ subgroup of Gl₂(R) ? Write a proof if the answer is YES, an example if the answer is NO.
Transcribed Image Text:you may assume that matrix multiplication is associative and that The determinant | | = ad - bc If ad- bc 0, с -1 (a b)² = 1 (a b) ad-bc For matrices A and B in M₂(R) the determinant of a product is the product of determinats involved, i.e. |AB| = |A||B| Is Sl₂ (R) = {(ab): : ad – bc = 1} a ་ subgroup of Gl₂(R) ? Write a proof if the answer is YES, an example if the answer is NO.
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