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Q40E

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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

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# (M) Let $$H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}$$ and $$K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}$$, where$${{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)$$Then $$H$$ and $$K$$ are subspaces of $${\mathbb{R}^3}$$. In fact, $$H$$ and $$K$$ are planes in $${\mathbb{R}^3}$$ through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as $${c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}$$ and also as $${c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}$$. To build w, solve the equation $${c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}$$ for the unknown $${c_j}'{\mathop{\rm s}\nolimits}$$.)

The nonzero vector w that generates that line is $$\left( {1, - 2, - 1} \right)$$.

See the step by step solution

## Step 1: Write the augmented matrix

$${\mathop{\rm w}\nolimits}$$ can be written as $${c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}$$ and $${c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}$$ because the line lies in both $$H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}$$ and $$K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}$$. To determine $${\mathop{\rm w}\nolimits}$$, find scalars (c) which solve $${c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} + {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4} = 0$$.

Consider the augmented matrix shown below:

$$\left( {\begin{array}{*{20}{c}}{{{\mathop{\rm v}\nolimits} _1}}&{{{\mathop{\rm v}\nolimits} _2}}&{ - {{\mathop{\rm v}\nolimits} _3}}&{ - {{\mathop{\rm v}\nolimits} _4}}&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&1&{ - 2}&0&0\\3&3&1&{12}&0\\8&4&{ - 5}&{28}&0\end{array}} \right)$$

## Step 2: Convert the matrix into the row-reduced echelon form

Consider the matrix $$A = \left( {\begin{array}{*{20}{c}}5&1&{ - 2}&0&0\\3&3&1&{12}&0\\8&4&{ - 5}&{28}&0\end{array}} \right)$$.

Use the code in MATLAB to obtain the row-reduced echelon form as shown below:

$$\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left( {5\,\,\,1\,\,\, - 2\,\,\,0\,\,\,0;\,3\,\,\,3\,\,\,1\,\,\,\,12\,\,\,0;\,\,\,\,8\,\,\,4\,\,\, - 5\,\,\,28\,\,\,0} \right)\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm rref}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{array}$$

$$\left( {\begin{array}{*{20}{c}}5&1&{ - 2}&0&0\\3&3&1&{12}&0\\8&4&{ - 5}&{28}&0\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&{\frac{{ - 10}}{3}}&0\\0&1&0&{\frac{{26}}{3}}&0\\0&0&1&{ - 4}&0\end{array}} \right)$$

## Step 3: Determine the nonzero vector w that generates the line

The vector of $${c_j}'{\mathop{\rm s}\nolimits}$$ is a multiple of $$\left( {\frac{{10}}{3},\frac{{ - 26}}{3},4,1} \right)$$.

For example, $$\left( {10, - 26,12,3} \right)$$ yields $${\mathop{\rm w}\nolimits} = 10{{\mathop{\rm v}\nolimits} _1} - 26{{\mathop{\rm v}\nolimits} _2} = 12{{\mathop{\rm v}\nolimits} _3} + 3{{\mathop{\rm v}\nolimits} _4} = \left( {24, - 48, - 24} \right)$$. The other option for $${\mathop{\rm w}\nolimits}$$ is $$\left( {1, - 2, - 1} \right)$$.

Thus, the nonzero vector w that generates the line is $$\left( {1, - 2, - 1} \right)$$.

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