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### 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

# In Exercises 10, write a vector equation that is equivalent to the given system of equations.10. $$4{x_1} + {x_2} + 3{x_3} = 9$$ $$\begin{array}{c}{x_1} - 7{x_2} - 2{x_3} = 2\\8{x_1} + 6{x_2} - 5{x_3} = 15\end{array}$$

The system of equations in the vector equation form is$${x_1}\left[ {\begin{array}{*{20}{c}}4\\1\\8\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\6\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}3\\{ - 2}\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\2\\{15}\end{array}} \right]$$.

See the step by step solution

## Step 1: Write the system of equations in a single vector

Consider the system of equations as shown below:

$$\begin{array}{c}4{x_1} + {x_2} + 3{x_3} = 9\\{x_1} - 7{x_2} - 2{x_3} = 2\\8{x_1} + 6{x_2} - 5{x_3} = 15\end{array}$$

Write the left-hand part and the right-hand part of the system of equations in the vector form as shown below:

$$\left[ \begin{array}{c}4{x_1} + {x_2} + 3{x_3}\\{x_1} - 7{x_2} - 2{x_3}\\8{x_1} + 6{x_2} - 5{x_3}\end{array} \right] = \left[ {\begin{array}{*{20}{c}}9\\2\\{15}\end{array}} \right]$$

## Step 2: Separate the left-hand side of the system of equations

The left-hand side of the vector equation contains $$\left[ \begin{array}{c}4{x_1} + {x_2} + 3{x_3}\\{x_1} - 7{x_2} - 2{x_3}\\8{x_1} + 6{x_2} - 5{x_3}\end{array} \right]$$. Write it as sums of vectors in terms of unknowns $${x_1}$$, $${x_2}$$, and $${x_3}$$.

$$\left[ {\begin{array}{*{20}{c}}{4{x_1}}\\{{x_1}}\\{8{x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{x_2}}\\{ - 7{x_2}}\\{6{x_2}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{3{x_3}}\\{ - 2{x_3}}\\{ - 5{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\2\\{15}\end{array}} \right]$$

## Step 3: Take the unknowns’ common from each vector

It is observed that $${x_1}$$ is the common in the vector $$\left[ {\begin{array}{*{20}{c}}{4{x_1}}\\{{x_1}}\\{8{x_1}}\end{array}} \right]$$. Take $${x_1}$$ as the common from the first vector as shown below:

$${x_1}\left[ {\begin{array}{*{20}{c}}4\\1\\8\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{x_2}}\\{ - 7{x_2}}\\{6{x_2}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{3{x_3}}\\{ - 2{x_3}}\\{ - 5{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\2\\{15}\end{array}} \right]$$

## Step 4: Take the unknowns’ common from each vector

Similarly, take $${x_2}$$ as the common from the vector $$\left[ {\begin{array}{*{20}{c}}{{x_2}}\\{ - 7{x_2}}\\{6{x_2}}\end{array}} \right]$$, and $${x_3}$$ as the common from the vector $$\left[ {\begin{array}{*{20}{c}}{3{x_3}}\\{ - 2{x_3}}\\{ - 5{x_3}}\end{array}} \right]$$. Take $${x_2}$$ and $${x_3}$$ as commons from each vector as shown below:

$${x_1}\left[ {\begin{array}{*{20}{c}}4\\1\\8\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\6\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}3\\{ - 2}\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\2\\{15}\end{array}} \right]$$

Thus, the system of equations in the vector equation form is:

$${x_1}\left[ {\begin{array}{*{20}{c}}4\\1\\8\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\6\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}3\\{ - 2}\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\2\\{15}\end{array}} \right]$$