Question

The vectors \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right],{{\mathop{\rm v}\nolimits} _2} = \left[ {\begin{array}{*{20}{c}}2\\{ - 8}\end{array}} \right],{{\mathop{\rm v}\nolimits} _3} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\7\end{array}} \right]\) span \({\mathbb{R}^2}\) but do not form a basis. Find two different ways to express \(\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\) as a linear combination of \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}\).

Short answer

The two different ways to express \(\left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)\) as a linear combination are \(5{{\mathop{\rm v}\nolimits} _1} - 2{{\mathop{\rm v}\nolimits} _2}\) and \(10{{\mathop{\rm v}\nolimits} _1} - 3{{\mathop{\rm v}\nolimits} _2} + {{\mathop{\rm v}\nolimits} _3}\).

Step-by-step solution

Write the vector equation as an augmented matrix

To solve the vector equation \({{\mathop{\rm x}\nolimits} _1}\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right] + {{\mathop{\rm x}\nolimits} _2}\left[ {\begin{array}{*{20}{c}}2\\{ - 8}\end{array}} \right] + {{\mathop{\rm x}\nolimits} _3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\), convert it into the augmented matrix shown below:

\(\left[ {\begin{array}{*{20}{c}}1&2&{ - 3}&1\\{ - 3}&{ - 8}&7&1\end{array}} \right]\)

Apply the row operation

Perform an elementary row operation to produce the row-reduced echelon form of the matrix.

At row 2, multiply row 1 by 3 and add it to row 2.

\( \sim \left( {\begin{array}{*{20}{c}}1&2&{ - 3}&1\\0&{ - 2}&{ - 2}&4\end{array}} \right)\)

At row 2, multiply row 2 by \( - \frac{1}{2}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&2&{ - 3}&1\\0&1&1&{ - 2}\end{array}} \right)\)

At row 1, multiply row 2 by 2 and subtract it from row 1.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{ - 5}&5\\0&1&1&{ - 2}\end{array}} \right)\)

Hence, you can consider \({{\mathop{\rm x}\nolimits} _1} = 5 + 5{{\mathop{\rm x}\nolimits} _3}\) and \({{\mathop{\rm x}\nolimits} _2} = - 2 - {{\mathop{\rm x}\nolimits} _3}\), where \({{\mathop{\rm x}\nolimits} _3}\) is any real number.

Determine two different ways to express \(\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\) as a linear combination

Let \({{\mathop{\rm x}\nolimits} _3} = 0\) and \({{\mathop{\rm x}\nolimits} _3} = 1\) give two different ways to express \(\left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)\) as a linear combination of the vectors \(5{{\mathop{\rm v}\nolimits} _1} - 2{{\mathop{\rm v}\nolimits} _2}\) and \(10{{\mathop{\rm v}\nolimits} _1} - 3{{\mathop{\rm v}\nolimits} _2} + {{\mathop{\rm v}\nolimits} _3}\).

Thus, the two ways to express \(\left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)\) as a linear combination are \(5{{\mathop{\rm v}\nolimits} _1} - 2{{\mathop{\rm v}\nolimits} _2}\) and \(10{{\mathop{\rm v}\nolimits} _1} - 3{{\mathop{\rm v}\nolimits} _2} + {{\mathop{\rm v}\nolimits} _3}\).