Question

In Exercises 6, write a system of equations that is equivalent to the given vector equation.

6. \({x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\)

Short answer

The system of equations is

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

Step-by-step solution

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 ofthe 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.

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 \({x_3}\).

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

Add the vectors

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.

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

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:

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