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Q32E
Expert-verifiedLet \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .
\({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right]\)
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, \({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right]\).
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