Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

The matrix $[\begin{matrix} k^{2} & 1 & 4 \\ k & - 1 & - 2 \\ 1 & 1 & 1 \end{matrix}]$ is invertible for all positive constants k.

Therefore, the given condition is true.

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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: Given.

Given Matrix,

$[\begin{matrix} k^{2} & 1 & 4 \\ k & - 1 & - 2 \\ 1 & 1 & 1 \end{matrix}]$

Step 3: To check whether the given condition is true or false.

We compute,

$\begin{array}{l} d e t A = - k^{2} - 2 + 4 k + 4 + 2 k^{2} - k \\ d e t A = k^{2} + 3 k + 2 \end{array}$

which turns out to be 0 if and only if $k = - 2$ or $k = - 1$ , so for all positive constants $k$ , the matrix is regular.

Therefore, the given condition is true.

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Linear Algebra With Applications

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Short Answer

The matrix $[\begin{matrix} k^{2} & 1 & 4 \\ k & - 1 & - 2 \\ 1 & 1 & 1 \end{matrix}]$ is invertible for all positive constants k.

Step by Step Solution

Step 1: Matrix Definition.

Step 2: Given.

Step 3: To check whether the given condition is true or false.

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Linear Algebra With Applications

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The matrix [k214k-1-2111] is invertible for all positive constants k.

Step by Step Solution

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The matrix $[\begin{matrix} k^{2} & 1 & 4 \\ k & - 1 & - 2 \\ 1 & 1 & 1 \end{matrix}]$ is invertible for all positive constants k.