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Q37E

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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

# (M) Determine whether w is in the column space of $$A$$, the null space of $$A$$, or both, where$${\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)$$

$${\mathop{\rm w}\nolimits}$$ is in Col A, and w is not in Nul A.

See the step by step solution

## Step 1: Write an augmented matrix

Consider the augmented matrix $$\left( {\begin{array}{*{20}{c}}A&{\mathop{\rm w}\nolimits} \end{array}} \right)$$ shown below:

$$\left( {\begin{array}{*{20}{c}}A&w\end{array}} \right) = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1&1\\{ - 5}&{ - 1}&0&{ - 2}&1\\9&{ - 11}&7&{ - 3}&{ - 1}\\{19}&{ - 9}&7&1&{ - 3}\end{array}} \right)$$

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

Consider the matrix $$A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1&1\\{ - 5}&{ - 1}&0&{ - 2}&1\\9&{ - 11}&7&{ - 3}&{ - 1}\\{19}&{ - 9}&7&1&{ - 3}\end{array}} \right)$$.

Use the following code in MATLAB to obtain the row-reduced echelon form.

$$\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left( {7\,\,\,6\,\,\, - 4\,\,\,1\,\,\,1;\,\, - 5\,\,\, - 1\,\,\,0\,\,\, - 2\,\,\,1;\,\,9\,\,\, - 11\,\,\,7\,\,\, - 3\,\,\, - 1;\,19\,\,\, - 9\,\,\,7\,\,\,1\,\,\, - 3} \right)\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm rref}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{array}$$

$$\left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1&1\\{ - 5}&{ - 1}&0&{ - 2}&1\\9&{ - 11}&7&{ - 3}&{ - 1}\\{19}&{ - 9}&7&1&{ - 3}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&{\frac{{ - 1}}{{95}}}&{\frac{1}{{95}}}\\0&1&0&{\frac{{39}}{{19}}}&{\frac{{ - 20}}{{19}}}\\0&0&1&{\frac{{267}}{{95}}}&{\frac{{ - 172}}{{95}}}\\0&0&0&0&0\end{array}} \right)$$

## Step 3: Determine whether w is in the column space of A

A typical vector v in Col A has the property that the equation $$A{\mathop{\rm x}\nolimits} = {\mathop{\rm v}\nolimits}$$ is consistent.

The system of equations of matrix A is consistent.

Thus, $${\mathop{\rm w}\nolimits}$$ is in Col A.

## Step 4: Determine whether w is in the null space of A

A typical vector v in Nul A has the property that $$A{\mathop{\rm v}\nolimits} = 0$$.

Use the code in MATLAB to compute the matrix $${\mathop{\rm Aw}\nolimits}$$ as shown below:

$$\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left( {7\,\,\,6\,\,\, - 4\,\,\,1;\,\, - 5\,\,\, - 1\,\,\,0\,\,\, - 2;\,\,9\,\,\, - 11\,\,\,7\,\,\, - 3;\,19\,\,\, - 9\,\,\,7\,\,\,1} \right)\\ > > {\mathop{\rm w}\nolimits} = \left( {1;\,\,1;\,\, - 1;\,\, - 3} \right)\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm A}\nolimits} * w\end{array}$$

$$\begin{array}{c}{\mathop{\rm A}\nolimits} {\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{14}\\0\\0\\0\end{array}} \right)\end{array}$$

Thus, w is not in Nul A.