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Q12E

Expert-verifiedFound in: Page 191

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Find a basis for the set of vectors in **\({\mathbb{R}^{\bf{2}}}\)** on the line **\(y = {\bf{5}}x\)**.**

A basis for \(Nul A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}\).

The set of all **homogeneous** equation solutions, \(A{\bf{x}} = 0\), is \(Nul A\). It is called **null space** of *A*.

The line equation \(y = 5x\) can also be written as \(y - 5x = 0\).

In the matrix form, it is \(A = \left( {\begin{array}{*{20}{c}}5&{ - 1}\end{array}} \right)\).

** **

From the above equation, \(y\) corresponds to the **pivot** position. So, \(y\) is the basic variable, and \(x\) is the free variable.

Let \(x = a\).

Substitute the value \(x = a\) in the equation \(y = 5x\) to obtain the general solution

\(y = 5a\).

** **

Obtain the vector in the parametric form using \(x = a\) and \(y = 5a\).

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right) = \left( {\begin{array}{*{20}{c}}a\\{5a}\end{array}} \right)\\ = a\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\end{array}\)

It can also be written as shown below:

\(\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right) = x\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\)

The basis of vector is shown below:

\( \Rightarrow \left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}\)

Thus, \({\bf{x}} = x\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\), and a basis for \(Nul A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}\).

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