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Q35E

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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

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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 \(\det U = \pm 1\).

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