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Q35E

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Found in: Page 165

### 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

# 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 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 $$\det U = \pm 1$$.