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Q16E

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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

Question: In Exercise 16, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.

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

The adjugate matrix is \(\left( {\begin{array}{*{20}{c}}6&4&{14}\\0&{ - 2}&{ - 1}\\0&0&{ - 3}\end{array}} \right)\), and the inverse matrix is

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: First, find the determinant

Let \(A = \left( {\begin{array}{*{20}{c}}1&2&4\\0&{ - 3}&1\\0&0&{ - 2}\end{array}} \right)\). Then,

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&2&4\\0&{ - 3}&1\\0&0&{ - 2}\end{array}} \right|\\ = 1\left| {\begin{array}{*{20}{c}}{ - 3}&1\\0&{ - 2}\end{array}} \right| + 0 + 0\\\det A = 6 \ne 0\end{array}\)

Here, \(\det A \ne 0\). Hence the inverse of A exists.

Step 2: Compute the adjugate matrix  

The nine cofactors are:

\(\begin{array}{c}{C_{11}} = {\left( { - 1} \right)^2}\left| {\begin{array}{*{20}{c}}{ - 3}&1\\0&{ - 2}\end{array}} \right|\\ = 6\end{array}\)

\(\begin{array}{c}{C_{12}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}}0&1\\0&{ - 2}\end{array}} \right|\\ = 0\end{array}\)

\(\begin{array}{c}{C_{13}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}0&{ - 3}\\0&0\end{array}} \right|\\ = 0\end{array}\)

\(\begin{array}{c}{C_{21}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}}2&4\\0&{ - 2}\end{array}} \right|\\ = - \left( { - 4} \right)\\ = 4\end{array}\)

\(\begin{array}{c}{C_{22}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}1&4\\0&{ - 2}\end{array}} \right|\\ = - 2\end{array}\)

\(\begin{array}{c}{C_{23}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}}1&2\\0&0\end{array}} \right|\\ = 0\end{array}\)

\(\begin{array}{c}{C_{31}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}2&4\\{ - 3}&1\end{array}} \right|\\ = 14\end{array}\)

\(\begin{array}{c}{C_{32}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}}1&4\\0&1\end{array}} \right|\\ = - 1\end{array}\)

\(\begin{array}{c}{C_{33}} = {\left( { - 1} \right)^6}\left| {\begin{array}{*{20}{c}}1&2\\0&{ - 3}\end{array}} \right|\\ = - 3\end{array}\)

The adjugate matrix is the transpose of the matrix of cofactors. Hence,

\(\begin{array}{c}{\rm{adj}}\,A = \left( {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{21}}}&{{C_{31}}}\\{{C_{12}}}&{{C_{22}}}&{{C_{32}}}\\{{C_{13}}}&{{C_{23}}}&{{C_{33}}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}6&4&{14}\\0&{ - 2}&{ - 1}\\0&0&{ - 3}\end{array}} \right)\end{array}\)

Step 3: Use Theorem 8 to find \({A^{ - {\bf{1}}}}\)

By Theorem 8,

Most popular questions for Math Textbooks

Compute the determinant in Exercise 7 using a cofactor expansion across the first row.

7. \[\left| {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{3}}&{\bf{0}}\\{\bf{6}}&{\bf{5}}&{\bf{2}}\\{\bf{9}}&{\bf{7}}&{\bf{3}}\end{array}} \right|\]

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

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

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: 17. Show that if A is \({\bf{2}} \times {\bf{2}}\), then Theorem 8 gives the same formula for \({A^{ - {\bf{1}}}}\) as that given by theorem 4 in Section 2.2.

In Exercise 19-24, explore the effect of an elementary row operation on the determinant of a matrix. In each case, state the row operation and describe how it affects the determinant.

\(\left[ {\begin{array}{*{20}{c}}a&b&c\\{\bf{3}}&{\bf{2}}&{\bf{1}}\\{\bf{4}}&{\bf{5}}&{\bf{6}}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{\bf{3}}&{\bf{2}}&{\bf{1}}\\a&b&c\\{\bf{4}}&{\bf{5}}&{\bf{6}}\end{array}} \right]\)
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