Q29E Construct a $3 \times 3$ matri... [FREE SOLUTION]

91影视

Linear Algebra and its Applications

David C. Lay, Steven R. Lay and Judi J. McDonald

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 483 pages

ISBN: 978-03219822384

Chapter 1: Q29E (page 1)

Construct a $3 \times 3$ matrix, not in echelon form, whose columns span ${\mathbb{R}^3}$. Show that the matrix you construct has the desired property.

Short Answer

Expert verified

The required matrix is $A = \left[ {\begin{array}{*{20}{c}}0&0&3\\0&4&0\\5&0&0\end{array}} \right]$.

Step by step solution

Writing the conditions for the echelon form

The matrix is in the echelon form if it satisfies the following conditions:

Non-zero rows should be positioned above zero rows.
Each row's leading entry should bein the column to the right of the row above its leading item.
In each column, all items below the leading entry must be zero.

Constructing a matrix in the non-echelon matrix

Assume that the matrix is $A = \left[ {\begin{array}{*{20}{c}}0&0&3\\0&4&0\\5&0&0\end{array}} \right]$.

Here, each row's leading entry is not in the column to the right of the row above its leading item. So, the matrix is in non-echelon form.

Constructing an arbitrary vector in ${\mathbb{R}^3}$

Suppose ${\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\\{{b_3}}\end{array}} \right]$ is the required vector.

Re-arrange the assumed vector as shown below:

$\begin{array}{c}{\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\\{{b_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{0 + 0 + {b_1}}\\{0 + {b_2} + 0}\\{{b_3} + 0 + 0}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0\\0\\{{b_3}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\{{b_2}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{b_1}}\\0\\0\end{array}} \right]\end{array}$

Checking if the column matrix span is ${\mathbb{R}^3}$

Simplify vector ${\bf{b}} = \left[ {\begin{array}{*{20}{c}}0\\0\\{{b_3}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\{{b_2}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{b_1}}\\0\\0\end{array}} \right]$ further.

$\begin{array}{c}{\bf{b}} = \left[ {\begin{array}{*{20}{c}}0\\0\\{\frac{{5{b_3}}}{5}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\{\frac{{4{b_2}}}{4}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{\frac{{3{b_1}}}{3}}\\0\\0\end{array}} \right]\\ = \frac{{{b_3}}}{5}\left[ {\begin{array}{*{20}{c}}0\\0\\5\end{array}} \right] + \frac{{{b_2}}}{4}\left[ {\begin{array}{*{20}{c}}0\\4\\0\end{array}} \right] + \frac{{{b_1}}}{3}\left[ {\begin{array}{*{20}{c}}3\\0\\0\end{array}} \right]\end{array}$

Thus, it shows that the columns of the assumed matrix span are ${\mathbb{R}^3}$.

Therefore, the required matrix is $A = \left[ {\begin{array}{*{20}{c}}0&0&3\\0&4&0\\5&0&0\end{array}} \right]$.