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Q32E

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Found in: Page 191

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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

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

See the step by step solution

## Step 1: 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, $${\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right]$$.