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Q6E

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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 6, write a system of equations that is equivalent to the given vector equation.6. $${x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]$$

The system of equations is

$$\begin{array}{*{20}{c}}{ - 2{x_1} + 8{x_2} + {x_3} = 0}\\{3{x_1} + 5{x_2} - 6{x_3} = 0}\end{array}$$

See the step by step solution

Step 1: Write the conditions for the vector addition and scalar multiple of a vector by a constant

From the given vector equation, it can be observed that the vectors contain two entries. So, the vectors can be denoted as $${\mathbb{R}^2}$$.

Add the corresponding terms of the vectors to obtain the sum.

Multiply each entry of a $${\mathbb{R}^2}$$ vector by the unknowns $${x_1}$$, $${x_2}$$, and $${x_3}$$ to obtain the scalar multiple of a vector by a scalar.

Step 2: Compute the scalar multiplication

Consider the vector equations $${x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]$$.

Obtain the scalar multiplication of the vector $$\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right]$$ by the unknown $${x_1}$$; obtain the scalar multiplication of the vector $$\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right]$$ by the unknown $${x_2}$$; and $$\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right]$$ with $${x_3}$$.

$$\left[ {\begin{array}{*{20}{c}}{ - 2{x_1}}\\{3{x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{8{x_2}}\\{5{x_2}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{x_3}}\\{ - 6{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]$$

Now, add the vectors $$\left[ {\begin{array}{*{20}{c}}{ - 2{x_1}}\\{3{x_1}}\end{array}} \right]$$, $$\left[ {\begin{array}{*{20}{c}}{8{x_2}}\\{5{x_2}}\end{array}} \right]$$, and $$\left[ {\begin{array}{*{20}{c}}{{x_3}}\\{ - 6{x_3}}\end{array}} \right]$$ on the left-hand side of the equation.

$$\left[ {\begin{array}{*{20}{c}}{ - 2{x_1} + 8{x_2} + {x_3}}\\{3{x_1} + 5{x_2} - 6{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]$$

Step 4: Equate the vectors and write in the equation form

The unknowns $${x_1}$$, $${x_2}$$, and $${x_3}$$ must satisfy the system of equations to make the equation $$\left[ {\begin{array}{*{20}{c}}{ - 2{x_1} + 8{x_2} + {x_3}}\\{3{x_1} + 5{x_2} - 6{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]$$ true. So, equate the vectors as shown below:

$$\begin{array}{*{20}{c}}{ - 2{x_1} + 8{x_2} + {x_3} = 0}\\{3{x_1} + 5{x_2} - 6{x_3} = 0}\end{array}$$

Thus, the system of equations is:

$$\begin{array}{*{20}{c}}{ - 2{x_1} + 8{x_2} + {x_3} = 0}\\{3{x_1} + 5{x_2} - 6{x_3} = 0}\end{array}$$