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Q41E

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Linear Algebra With Applications

Found in: Page 309

Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

If all the diagonal entries of an $n \times n$ matrix $A$ are odd integers and all the other entries are even integers, then $A$ must be an invertible matrix.

Therefore, A is invertible. So, the given statement is true.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Matrix Definition

Matrix is a set of numbers arranged in rows and columns so as to form a rectangular array.

The numbers are called the elements, or entries, of the matrix.

If there are $m$ rows and $n$ columns, the matrix is said to be an “ $m$ by $n$ ” matrix, written “ $m \times n$ .”

Step 2: To check whether the given condition is true or false

Using the definition of $d e t A$ , the product in the addend corresponding to the diagonal pattern is odd, while all other addends are even.

Thus, $d e t A$ is odd, so it must be different than 0.

Thus, $A$ is invertible. So, the given statement is true.

Most popular questions for Math Textbooks

There exists a real $3 \times 3$ matrix $A$ such that $A^{2} = - l_{3}$ .

If an $n \times n$ matrix A is invertible, then there must be an $(n - 1) \times (n - 1)$ sub matrix of (obtained by deleting a row and a column of ) that is invertible as well.

There exist invertible $2 \times 2$ matrices A and B such that $d e t (A + B) = d e t A + d e t B$ .

In Exercises 62 through 64, consider a function D from $ℝ^{2 \times 2}$ to $ℝ$ that is linear in both columns and alternating on the columns. See Examples 4 and 6 and the subsequent discussions. Assume that $D (l_{2}) = 1$ .

62. Show that $D (A) = 0$ for any $2 \times 2$ matrix whose two columns are equal.

For every nonzero $2 \times 2$ matrix A there exists a $2 \times 2$ matrix B such that $d e t (A + B) \neq d e t A + d e t B$ .

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