In Exercises 5, write a system of equations that is equivalent to the given vector equation.

5. \({x_1}\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\)

From the given vector equation, it can be observed that the vectors contain three entries. So, the vectors can be denoted as \({\mathbb{R}^3}\).

Add the corresponding terms ofthe vectors to obtain the sum.

Multiply each entry of a \({\mathbb{R}^3}\) vector by the unknowns \({x_1}\) and \({x_2}\) to obtain the scalar multiple of a vector by a scalar.

Consider the vector equations \({x_1}\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\).

Obtain the scalar multiplication of the vector \(\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right]\) by the unknown \({x_1}\), and obtain the scalar multiplication of the vector \(\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right]\)by the unknown \({x_2}\).

\(\left[ {\begin{array}{*{20}{c}}{6{x_1}}\\{ - {x_1}}\\{5{x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3{x_2}}\\{4{x_2}}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\)

Now, add the vectors \(\left[ {\begin{array}{*{20}{c}}{6{x_1}}\\{ - {x_1}}\\{5{x_1}}\end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}}{ - 3{x_2}}\\{4{x_2}}\\0\end{array}} \right]\) on the left-hand side of the equation.

\(\left[ {\begin{array}{*{20}{c}}{6{x_1} - 3{x_2}}\\{ - {x_1} + 4{x_2}}\\{5{x_1}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\)

The unknowns \({x_1}\) and \({x_2}\) must satisfy the system of equations in order to make the equation \(\left[ {\begin{array}{*{20}{c}}{6{x_1} - 3{x_2}}\\{ - {x_1} + 4{x_2}}\\{5{x_1}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\)true. So, equate the vectors as shown below:

\(\begin{array}{*{20}{c}}{6{x_1} - 3{x_2} = 1}\\{ - {x_1} + 4{x_2} = - 7}\\{5{x_1} + 0\left( {{x_2}} \right) = - 5}\end{array}\)

Thus, the system of equations is:

\(\begin{aligned}{*{20}{c}}{6{x_1} - 3{x_2} = 1}\\{ - {x_1} + 4{x_2} = - 7}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,5{x_1} = - 5}\end{aligned}\)