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Q6E

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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: 6. Use Cramer’s rule to compute the solution of the following system.

\(\begin{array}{c}{x_{\bf{1}}} + {\bf{3}}{x_{\bf{2}}} + \,{x_{\bf{3}}} = {\bf{4}}\\ - {x_{\bf{1}}} + \,\,\,\,\,\,\,\,\,\,{\bf{2}}{x_{\bf{3}}} = {\bf{2}}\\{\bf{3}}{x_{\bf{1}}} + \,{x_{\bf{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\, = {\bf{2}}\end{array}\)

The solution is \({x_1} = \frac{2}{5}\),\({x_2} = \frac{4}{5}\), and \({x_3} = \frac{6}{5}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the matrix form

The matrix form of the given system is:

\(\left( {\begin{array}{*{20}{c}}1&3&1\\{ - 1}&0&2\\3&1&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4\\2\\2\end{array}} \right)\)

Thus, \(Ax = b\), where \(A = \left( {\begin{array}{*{20}{c}}1&3&1\\{ - 1}&0&2\\3&1&0\end{array}} \right)\), \(b = \left( {\begin{array}{*{20}{c}}4\\2\\2\end{array}} \right)\), and \(x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\).

Then, \({A_1}\left( b \right) = \left( {\begin{array}{*{20}{c}}4&3&1\\2&0&2\\2&1&0\end{array}} \right)\),\({A_2}\left( b \right) = \left( {\begin{array}{*{20}{c}}1&4&1\\{ - 1}&2&2\\3&2&0\end{array}} \right)\), and \({A_3}\left( b \right) = \left( {\begin{array}{*{20}{c}}1&3&4\\{ - 1}&0&2\\3&1&2\end{array}} \right)\).

Step 2: Find the determinants

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&3&1\\{ - 1}&0&2\\3&1&0\end{array}} \right|\\ = - 3\left| {\begin{array}{*{20}{c}}{ - 1}&2\\3&0\end{array}} \right| + 0 - 1\left| {\begin{array}{*{20}{c}}1&1\\{ - 1}&2\end{array}} \right|\\ = - 3\left( { - 6} \right) - \left( 3 \right)\\\det A = 15\end{array}\)

\(\begin{array}{c}\det {A_1}\left( b \right) = \left| {\begin{array}{*{20}{c}}4&3&1\\2&0&2\\2&1&0\end{array}} \right|\\ = - 3\left| {\begin{array}{*{20}{c}}2&2\\2&0\end{array}} \right| + 0 - 1\left| {\begin{array}{*{20}{c}}4&1\\2&2\end{array}} \right|\\ = - 3\left( { - 4} \right) - 6\\ = 12 - 6\\\det {A_1}\left( b \right) = 6\end{array}\)

\(\begin{array}{c}\det {A_2}\left( b \right) = \left| {\begin{array}{*{20}{c}}1&4&1\\{ - 1}&2&2\\3&2&0\end{array}} \right|\\ = 1\left| {\begin{array}{*{20}{c}}{ - 1}&2\\3&2\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}1&4\\3&2\end{array}} \right| + 0\\ = - 8 - 2\left( { - 10} \right)\\ = - 8 + 20\\\det {A_2}\left( b \right) = 12\end{array}\)

\(\begin{array}{c}\det {A_3}\left( b \right) = \left| {\begin{array}{*{20}{c}}1&3&4\\{ - 1}&0&2\\3&1&2\end{array}} \right|\\ = - 3\left| {\begin{array}{*{20}{c}}{ - 1}&2\\3&2\end{array}} \right| + 0 - 1\left| {\begin{array}{*{20}{c}}1&4\\{ - 1}&2\end{array}} \right|\\ = - 3\left( { - 8} \right) - 6\\ = 24 - 6\\\det {A_3}\left( b \right) = 18\end{array}\)

Step 3: Use Cramer’s rule

By Cramer’s rule, \({x_i} = \frac{{\det {A_i}\left( b \right)}}{{\det A}}\), \(i = 1,2,3\). Hence,

\(\begin{array}{c}{x_1} = \frac{{\det {A_1}\left( b \right)}}{{\det A}}\\ = \frac{6}{{15}}\\{x_1} = \frac{2}{5}\end{array}\)

\(\begin{array}{c}{x_2} = \frac{{\det {A_2}\left( b \right)}}{{\det A}}\\ = \frac{{12}}{{15}}\\{x_2} = \frac{4}{5}\end{array}\)

\(\begin{array}{c}{x_3} = \frac{{\det {A_3}\left( b \right)}}{{\det A}}\\ = \frac{{18}}{{15}}\\{x_3} = \frac{6}{5}\end{array}\)

Most popular questions for Math Textbooks

Let \(A = \left[ {\begin{array}{*{20}{c}}3&1\\4&2\end{array}} \right]\). Write \(5A\). Is \(\det 5A = 5\det A\)?

Question: 12. Use the concept of area of a parallelogram to write a statement about a \(2 \times 2\) matrix A that is true if and only if A is invertible.

Find the determinants in Exercises 5-10 by row reduction to echelon form.

\(\left| {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{3}}&{\bf{2}}&{ - {\bf{4}}}\\{\bf{0}}&{\bf{1}}&{\bf{2}}&{ - {\bf{5}}}\\{\bf{2}}&{\bf{7}}&{\bf{6}}&{ - {\bf{3}}}\\{ - {\bf{3}}}&{ - {\bf{10}}}&{ - {\bf{7}}}&{\bf{2}}\end{aligned}} \right|\)

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

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

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

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