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Found in: Page 165

### 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

# 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 $${x_3} = \frac{6}{5}$$.

See the step by step solution

## 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,

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

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

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