/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q17SE Question: 17. Let A be the origi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Q17SE (page 165)

Question: 17. Let A be the original matrix given in Exercise 16 i.e., \(A = \left( {\begin{array}{*{20}{c}}a&b&b& \cdots &b\\b&a&b& \cdots &b\\b&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&b&b& \cdots &a\end{array}} \right)\), and let

\(B = \left( {\begin{array}{*{20}{c}}{a - b}&b&b& \cdots &b\\{\bf{0}}&a&b& \cdots &b\\{\bf{0}}&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\{\bf{0}}&b&b& \cdots &a\end{array}} \right)\)

\(C = \left( {\begin{array}{*{20}{c}}b&b&b& \cdots &b\\b&a&b& \cdots &b\\b&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&b&b& \cdots &a\end{array}} \right)\)

Notice that A, B, and C are nearly the same except that the first column of A equals the sum of the first columns of B and C. A linearity property of the determinant function, discussed in Section 3.2, says that \({\bf{det}}\,A = {\bf{det}}\,B + {\bf{det}}\,C\). Use this fact to prove the formula in Exercise 16 i.e., \(det\,A = {\left( {a - b} \right)^{n - {\bf{1}}}}\left( {a + \left( {n - {\bf{1}}} \right)b} \right)\)by induction on size of matrix A.

Short Answer

Expert verified

Hence, \(\det A = {\left( {a - b} \right)^{n - 1}}\left( {a + \left( {n - 1} \right)b} \right)\). This is exactly what needs to be shown. Thus, the formula is proved by mathematical induction.

Step by step solution

01

Consider the case \(n = 2\)

In this case, \(B = \left( {\begin{array}{*{20}{c}}{a - b}&b\\0&a\end{array}} \right)\), and \(C = \left( {\begin{array}{*{20}{c}}b&b\\b&a\end{array}} \right)\). Then,

\(\begin{array}{c}\det B = \left| {\begin{array}{*{20}{c}}{a - b}&b\\0&a\end{array}} \right|\\ = a\left( {a - b} \right) - 0\\\det B = a\left( {a - b} \right)\end{array}\)

\(\begin{array}{c}\det \,C = \left| {\begin{array}{*{20}{c}}b&b\\b&a\end{array}} \right|\\ = ab - {b^2}\\\det C = b\left( {a - b} \right)\end{array}\)

Hence,

\(\begin{array}{c}\det A = \det B + \det C\\ = a\left( {a - b} \right) + b\left( {a - b} \right)\\ = \left( {a - b} \right)\left( {a + b} \right)\\\det A = {\left( {a - b} \right)^{2 - 1}}\left( {a + \left( {2 - 1} \right)b} \right)\end{array}\)

Thus, the formula holds for \(n = 2\).

02

Use the induction

Now, assume that the formula holds for \(n = k - 1\). Use this to prove that the formula holds for \(n = k\). In this case,

\(\begin{array}{c}B = \left( {\begin{array}{*{20}{c}}{a - b}&b&b& \cdots &b\\0&a&b& \cdots &b\\0&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&b&b& \cdots &a\end{array}} \right)\\\det B = \left| {\begin{array}{*{20}{c}}{a - b}&b&b& \cdots &b\\0&a&b& \cdots &b\\0&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&b&b& \cdots &a\end{array}} \right|\\ = \left( {a - b} \right)\left| {\begin{array}{*{20}{c}}a&b& \cdots &b\\b&a& \cdots &b\\ \vdots & \vdots & \ddots & \vdots \\b&b& \cdots &a\end{array}} \right|\\ = \left( {a - b} \right){\left( {a - b} \right)^{\left( {k - 1} \right) - 1}}\left( {a + \left( {k - 1 - 1} \right)b} \right)\\ = {\left( {a - b} \right)^{k - 1}}\left( {a + \left( {k - 2} \right)b} \right)\end{array}\)

Also, \(C = \left( {\begin{array}{*{20}{c}}b&b&b& \cdots &b\\b&a&b& \cdots &b\\b&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&b&b& \cdots &a\end{array}} \right)\). So, \(\det C = \left| {\begin{array}{*{20}{c}}b&b&b& \cdots &b\\b&a&b& \cdots &b\\b&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&b&b& \cdots &a\end{array}} \right|\). At row 2, subtract row 1 from row 2; at row 3, subtract row 1 from row 3, and so on. Therefore,

\(\det C = \left| {\begin{array}{*{20}{c}}b&b&b& \cdots &b\\0&{a - b}&0& \cdots &0\\0&0&{a - b}& \cdots &0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&0&0& \cdots &{a - b}\end{array}} \right|\)

This is a triangular matrix. Hence, its determinant is given by the product of its diagonal entries.Therefore,

\(\begin{array}{c}\det C = b\left( {a - b} \right)\left( {a - b} \right) \cdots \left( {a - b} \right)\\ = b{\left( {a - b} \right)^{k - 1}}\end{array}\)

03

Conclusion

\(\begin{array}{c}\det A = \det B + \det A\\ = {\left( {a - b} \right)^{k - 1}}\left( {a + \left( {k - 2} \right)b} \right) + b{\left( {a - b} \right)^{k - 1}}\\ = {\left( {a - b} \right)^{k - 1}}\left( {a + \left( {k - 2} \right)b + b} \right)\\\det A = {\left( {a - b} \right)^{k - 1}}\left[ {a + \left( {k - 1} \right)b} \right)\end{array}\)

Here \(n = k\). Hence, \(\det A = {\left( {a - b} \right)^{n - 1}}\left( {a + \left( {n - 1} \right)b} \right)\).

This is exactly what needs to be shown. Thus, the formula is proved by mathematical induction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: In Exercise 8, determine the values of the parameter s for which the system has a unique solution, and describe the solution.

8.

\(\begin{array}{c}{\bf{3}}s{x_{\bf{1}}} + {\bf{5}}{x_{\bf{2}}} = {\bf{3}}\\12{x_{\bf{1}}} + {\bf{5}}s{x_{\bf{2}}} = {\bf{2}}\end{array}\)

In Exercise 33-36, verify that \(\det EA = \left( {\det E} \right)\left( {\det A} \right)\)where E is the elementary matrix shown and \(A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\).

35. \(\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]\)

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: Use Cramer鈥檚 rule to compute the solutions of the systems in Exercises1-6.

3. \(\begin{array}{c}3{x_1} - 2{x_2} = 3\\ - 4{x_1} + 6{x_2} = - 5\end{array}\)

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.

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.

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

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.