• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Linear Algebra and its Applications
Found in: Page 165
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

Answers without the blur.

Just sign up for free and you're in.


Short Answer

Question:In Exercises 31–36, mention an appropriate theorem in your explanation.

36. Let U be a square matrix such that \({U^T}U = I\). Show that\(det{\rm{ }}U = \pm 1\).

It is proved that \(\det U = \pm 1\).

See the step by step solution

Step by Step Solution

Step 1: Write the multiplicative property

According to theorem 6, if A andB are square matrices, then the determinant of the product matrix AB is equal to the product of the determinant of A and the determinant of B.

\(\det AB = \left( {\det A} \right)\left( {\det B} \right)\)

If matrices A and B are the same, then the general form is \(\det {A^n} = {\left( {\det A} \right)^n}\).

According to theorem 5, the determinant of square matrix A is equal to the determinant of the transpose matrix A.

\(\det {A^T} = \det A\)

Step 2: Prove the statement

Apply the theorem \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\) by substituting \(A = {U^T}\)and \(B = U\), as shown below:

\(\det \left( {{U^T}U} \right) = \left( {\det {U^T}} \right)\left( {\det U} \right)\)

Apply the theorem \(\det {A^T} = \det A\).

\(\begin{aligned}{}\det \left( {{U^T}U} \right) &= \left( {\det U} \right)\left( {\det U} \right)\\ &= {\left( {\det U} \right)^2}\end{aligned}\)

It is given that \({U^T}U = I\). By using this,

\(\begin{aligned}{}\det \left( {{U^T}U} \right) &= {\left( {\det U} \right)^2}\\{\left( {\det U} \right)^2} &= \det \left( I \right)\\{\left( {\det U} \right)^2} &= 1\\\det U &= \pm 1.\end{aligned}\)

Hence, it is proved that \(\det U = \pm 1\).

Most popular questions for Math Textbooks


Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.