• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Deutsch (DE)
Deutsch (UK)
Deutsch (US)

Americas

Europe

Answers without the blur. Sign up and see all textbooks for free! → Illustration

Q8E

Expert-verified
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

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later
Illustration

Short Answer

Compute the determinant in Exercise 8 using a cofactor expansion across the first row.

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

\(\left| {\begin{array}{*{20}{c}}4&1&2\\4&0&3\\3&{ - 2}&5\end{array}} \right| = - 3\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the determinant formula

The determinant computed by cofactor expansion across the ith row is

\(\det A = {a_{i1}}{C_{i1}} + {a_{i2}}{C_{i2}} + \cdots + {a_{in}}{C_{in}}\).

Here, A is an \(n \times n\) matrix, and \({C_{ij}} = {\left( { - 1} \right)^{i + j}}{A_{ij}}\).

Step 2: Use cofactor expansion across the first row

\(\begin{array}{c}\left| {\begin{array}{*{20}{c}}4&1&2\\4&0&3\\3&{ - 2}&5\end{array}} \right| = {a_{11}}{C_{11}} + {a_{12}}{C_{12}} + {a_{13}}{C_{13}}\\ = {a_{11}}{\left( { - 1} \right)^{1 + 1}}\det {A_{11}} + {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{13}}{\left( { - 1} \right)^{1 + 3}}\det {A_{13}}\\ = 4\left| {\begin{array}{*{20}{c}}0&3\\{ - 2}&5\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}4&3\\3&5\end{array}} \right| + 2\left| {\begin{array}{*{20}{c}}4&0\\3&{ - 2}\end{array}} \right|\\ = 4\left( 6 \right) - 1\left( {11} \right) + 2\left( { - 8} \right)\\ = 24 - 11 - 16\\ = - 3\end{array}\)

Step 3: Draw a conclusion

The determinant obtained by cofactor expansion across the first row is \( - 3\).

Most popular questions for Math Textbooks

Find the determinant in Exercise 20, where \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{\bf{d}}&{\bf{e}}&{\bf{f}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right| = {\bf{7}}\].

20. \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{{\bf{d}} + {\bf{3g}}}&{{\bf{e}} + {\bf{3h}}}&{{\bf{f}} + {\bf{3i}}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right|\]

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.

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

In Exercises 24–26, use determinants to decide if the set of vectors is linearly independent.

24. \(\left( {\begin{aligned}{*{20}{c}}4\\6\\2\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{ - 7}\\0\\7\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{ - 3}\\{ - 5}\\{ - 2}\end{aligned}} \right)\)

Compute the determinants in Exercises 9-14 by cofactor expnasions. At each step, choose a row or column that involves the least amount of computation.

\(\left| {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{5}}&{ - {\bf{6}}}&{\bf{4}}\\{\bf{0}}&{ - {\bf{2}}}&{\bf{3}}&{ - {\bf{3}}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{5}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{3}}\end{aligned}} \right|\)

Compute the determinants in Exercises 9-14 by cofactor expnasions. At each step, choose a row or column that involves the least amount of computation.

\(\left| {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{0}}&{ - {\bf{7}}}&{\bf{3}}&{ - {\bf{5}}}\\{\bf{0}}&{\bf{0}}&{\bf{2}}&{\bf{0}}&{\bf{0}}\\{\bf{7}}&{\bf{3}}&{ - {\bf{6}}}&{\bf{4}}&{ - {\bf{8}}}\\{\bf{5}}&{\bf{0}}&{\bf{5}}&{\bf{2}}&{ - {\bf{3}}}\\{\bf{0}}&{\bf{0}}&{\bf{9}}&{ - {\bf{1}}}&{\bf{2}}\end{array}} \right|\)
Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

Decision Maths

Read explanation

Calculus

Read explanation

Probability and Statistics

Read explanation

Statistics

Read explanation

Mechanics Maths

Read explanation

Geometry

Read explanation

Pure Maths

Read explanation

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.