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Chapter 3: Q6E (page 165)

Question: 6. Use Cramer鈥檚 rule to compute the solution of the following system.

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

Short Answer

Expert verified

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

Step by step solution

01

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

02

Find the determinants

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

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

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

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

03

Use Cramer’s rule

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

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

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

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

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Most popular questions from this chapter

In Exercises 21鈥�23, use determinants to find out if the matrix is invertible.

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

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

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

Each equation in Exercises 1-4 illustrates a property of determinants. State the property

\(\left| {\begin{array}{*{20}{c}}{\bf{3}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{3}}&{\bf{5}}&{ - {\bf{5}}}\\{\bf{1}}&{\bf{3}}&{\bf{3}}\end{array}} \right| = {\bf{3}}\left| {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{2}}}&{\bf{3}}\\{\bf{3}}&{\bf{5}}&{ - {\bf{5}}}\\{\bf{1}}&{\bf{3}}&{\bf{3}}\end{array}} \right|\)

Find the determinant in Exercise 19, where \[\left| {\begin{aligned}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{\bf{d}}&{\bf{e}}&{\bf{f}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{aligned}} \right| = {\bf{7}}\].

19. \[\left| {\begin{aligned}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{{\bf{2d}} + {\bf{a}}}&{{\bf{2e}} + {\bf{b}}}&{{\bf{2f}} + {\bf{c}}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{aligned}} \right|\]

Combine the methods of row reduction and cofactor expansion to compute the determinant in Exercise 11.

11. \(\left| {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{4}}&{ - {\bf{3}}}&{ - {\bf{1}}}\\{\bf{3}}&{\bf{0}}&{\bf{1}}&{ - {\bf{3}}}\\{ - {\bf{6}}}&{\bf{0}}&{ - {\bf{4}}}&{\bf{3}}\\{\bf{6}}&{\bf{8}}&{ - {\bf{4}}}&{ - {\bf{1}}}\end{aligned}} \right|\)
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