For A a square matrix such that ATA = I show that det A = ±1.   If A is square and A5 = 0 show that A is singular.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.8: Determinants
Problem 34E
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For A a square matrix such that ATA = I show that det A = ±1.

 

If A is square and A5 = 0 show that A is singular.

For A a square matrix such that ATA = I show that det A = +1.
If A is square and A5 = 0 show that A is singular.
Transcribed Image Text:For A a square matrix such that ATA = I show that det A = +1. If A is square and A5 = 0 show that A is singular.
Expert Solution
Step 1

Given,

ATA=IdetATA=detIdetATdetA=1  Since, detI=1detA·detA=1    Since, detAT=AdetA2=1detA=±1

This completes the proof.

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