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In Exercises 31–36, mention an appropriate theorem in your explanation.Show that if $A$ is invertible, then $\operatorname{det} A^{-1}=\frac{1}{\operatorname{det} A}$

see the proof

Algebra

Chapter 3

Determinants

Section 2

Properties of Determinants

Introduction to Matrices

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Lectures

01:32

In mathematics, the absolu…

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In Exercises 31–36, mentio…

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Use matrix algcbra to show…

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If $A$ is an invertible $n…

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Find $\operatorname{det}(A…

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If $A$ and $S$ are $n \tim…

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Show that if $A B$ is inve…

were given a statement and were asked to prove this statement using the theorems from this section. The statement is if a is in vertebral than the determinant of a inverse is equal to one over the determinant of a now A is in vertebral right away. This tells us that the determinant of a is non zero. This is important because otherwise one over the determinant of a is not defined. And also by the're, um six. We have that the determinant of a times the determinant of a inverse will notice also that a inverse is also in vertebral. So this makes sense so by this theory, and we have the determinant eight times the determined today in verse is equal to determinant of a times a inverse or I which is equal to one. And therefore it follows that the determinant of a inverse is equal to one over the determinant today

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