Question

Let \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .

Short answer

\({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right]\)

Step-by-step solution

Determine vector \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\)

It is noted that the second row of the matrix is twice the first row. Therefore, when \({\mathop{\rm v}\nolimits} = \left( {1, - 3,4} \right)\), which is the first row of the matrix.

\(\begin{array}{c}{{\mathop{\rm uv}\nolimits} ^T} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right]\end{array}\)

Thus, \({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right]\).