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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

In Exercises 27 and 28, A and B are \[n \times n\] matrices. Mark each statement True or False. Justify each answer.

27. a. A row replacement operation does not affect the determinant of a matrix.

b. The determinant of A is the product of the pivots in any echelon form U of A, multiplied by \({\left( { - {\bf{1}}} \right)^r}\), where r is the number of row interchanges made during row reduction from A to U.

c. If the columns of A are linearly dependent, then \(det\left( A \right) = 0\).

d. \(det\left( {A + B} \right) = det{\rm{ }}A + det{\rm{ }}B\).

  1. The given statement is true.
  2. The given statement is false.
  3. The given statement is true.
  4. The given statement is false.
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

(a) Step 1: Mark the first statement true or false

Recall Theorem 3:

Consider a square matrix A. If any row of matrix A has a multiple that is added to another row to give a new matrix B, then the determinant of matrix A must be equal to the determinant of matrix B.

\(\det \left( A \right) = \det \left( B \right)\)

Therefore, statement (a) is true.

(b) Step 2: Mark the second statement true or false

Recall that when A is invertible, then the determinant of matrix A can be written as shown below:

\(\det {\rm{ }}A = {\left( { - 1} \right)^r} \cdot \left( {{\rm{product of pivots in }}U} \right)\)

If it is not invertible, then the determinant of the matrix is 0.

Also, scaling a row by a constant (non-zero) during conversion to an echelon form may affect the value of the determinant.

Therefore, statement (b) is false.

(c) Step 3: Mark the third statement true or false

The set of vectors \({{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3},...,{{\bf{b}}_n}\) is said to be linearly independent if the determinant of the matrix \(\left[ {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}& \cdots &{{{\bf{b}}_n}}\end{aligned}} \right]\) is 0 (or \(\left| {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}& \cdots &{{{\bf{b}}_n}}\end{aligned}} \right| = 0\)).

So, the columns of matrix A are linearly independent, and the determinant of the matrix is zero.

Therefore, statement (c) is true.

(d) Step 4: Mark the fourth statement true or false

Check the equality \(\det \left( {A + B} \right) = \det {\rm{ }}A + \det {\rm{ }}B\).

Let the matrices be \(A = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\) and \(B = \left[ {\begin{aligned}{*{20}{c}}u&v\\w&x\end{aligned}} \right]\).

Obtain the determinant of each matrix.

\(\begin{aligned}{c}\det {\rm{ }}A = \left| {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right|\\ = ad - bc\end{aligned}\)

And

\(\begin{aligned}{c}\det {\rm{ }}B = \left| {\begin{aligned}{*{20}{c}}u&v\\w&x\end{aligned}} \right|\\ = ux - vw\end{aligned}\)

Add both the determinants.

\(\begin{aligned}{c}\det {\rm{ }}A + \det {\rm{ }}B = \left( {ad - bc} \right) + \left( {ux - vw} \right)\\ = ad - bc + ux - vw\end{aligned}\)

Thus, \(\det {\rm{ }}A + \det {\rm{ }}B = ad - bc + ux - vw\).

Add both the matrices as shown below:

\(\begin{aligned}{c}A + B = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right] + \left[ {\begin{aligned}{*{20}{c}}u&v\\w&x\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}{a + u}&{b + v}\\{c + w}&{d + x}\end{aligned}} \right]\end{aligned}\)

Obtain the determinant of \(A + B = \left[ {\begin{aligned}{*{20}{c}}{a + u}&{b + v}\\{c + w}&{d + x}\end{aligned}} \right]\).

\(\begin{aligned}{c}\det \left( {A + B} \right) = \left| {\begin{aligned}{*{20}{c}}{a + u}&{b + v}\\{c + w}&{d + x}\end{aligned}} \right|\\ = \left( {a + u} \right)\left( {d + x} \right) - \left( {c + w} \right)\left( {b + v} \right)\\ = ad + ax + ud + ux - cb - cv - bw - vw\end{aligned}\)

It is observed that \(\det \left( {A + B} \right) \ne \det {\rm{ }}A + \det {\rm{ }}B\).

Therefore, it is not true that \(\det \left( {A + B} \right) = \det {\rm{ }}A + \det {\rm{ }}B\).

Most popular questions for Math Textbooks

Compute the determinant in Exercise 9 by cofactor expansions. At each step, choose a row or column that involves the least amount of computation.

9. \(\left| {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{0}}&{\bf{0}}&{\bf{5}}\\{\bf{1}}&{\bf{7}}&{\bf{2}}&{ - {\bf{5}}}\\{\bf{3}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{8}}&{\bf{3}}&{\bf{1}}&{\bf{7}}\end{array}} \right|\)

Question: In Exercise 15, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.

15. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{0}}&{\bf{0}}\\{ - {\bf{1}}}&{\bf{1}}&{\bf{0}}\\{ - {\bf{2}}}&{\bf{3}}&{ - {\bf{1}}}\end{array}} \right)\)

Question: Use Cramer’s rule to compute the solutions of the systems in Exercises1-6.

2. \(\begin{array}{l}4{x_1} + {x_2} = 6\\3{x_1} + 2{x_2} = 7\end{array}\)

Compute the determinants of the elementary matrices given in Exercise 25-30.

29. \(\left[ {\begin{aligned}{*{20}{c}}1&0&0\\0&k&0\\0&0&1\end{aligned}} \right]\).

The expansion of a \({\bf{3}} \times {\bf{3}}\) determinant can be remembered by the following device. Write a second type of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals.

atr

Add the downward diagonal products and subtract the upward products. Use this method to compute the determinants in Exercises 15-18. Warning: This trick does not generalize in any reasonable way to \({\bf{4}} \times {\bf{4}}\) or larger matrices.

\(\left| {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{3}}&{\bf{4}}\\{\bf{2}}&{\bf{3}}&{\bf{1}}\\{\bf{3}}&{\bf{3}}&{\bf{2}}\end{aligned}} \right|\)
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