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Q25E

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Linear Algebra and its Applications
Found in: Page 1
Linear Algebra and its Applications

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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Short Answer

Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).

  1. Is \(b\) in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)? How many vectors are in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)?
  2. Is \(b\) in \(W\)? How many vectors are in W.
  3. Show that \({a_1}\) is in W. [Hint: Row operations are unnecessary.]

  1. The set \(\left\{ {{a_1},{a_2},{a_3}} \right\}\) contains three vectors and \(b\) is not one of them.
  2. The system of the augmented matrix is consistent. There are infinitely many vectors in \(W = \left\{ {{a_1},{a_2},{a_3}} \right\}\). So, \(b\) is in \(W\).
  3. It is proven that \({a_1}\) is in \(W\) .
See the step by step solution

Step by Step Solution

Step 1: Identify whether b  is in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\) 

a.

A vector equation \({x_1}{a_1} + {x_2}{a_2} + ... + {x_n}{a_n} = b\) has the same solution set as the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{....}&{{a_n}}&b\end{array}} \right]\).

Write the coefficient matrix into a set of vectors.

\(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\3\\6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 2}\\3\end{array}} \right]} \right\}\)

The set \(\left\{ {{a_1},{a_2},{a_3}} \right\}\) contains three vectors and \(b\) is not one of them.

Step 2: Apply row operation to identify that \(b\) is in \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\)

b.

A vector equation \({x_1}{a_1} + {x_2}{a_2} + ... + {x_n}{a_n} = b\) has the same solution set as the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{....}&{{a_n}}&b\end{array}} \right]\). In particular, \(b\) can be generated by a linear combination of \({{\mathop{\rm a}\nolimits} _1},...{a_n}\), if and only if, there exists a solution to the linear system corresponding to the augmented matrix.

Write the given matrix in an augmented matrix.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&3&{ - 2}&1\\{ - 2}&6&3&{ - 4}\end{array}} \right]\)

Perform an elementary row operation to produce the first augmented matrix.

Apply sum of row 3 and \(2\) times of row 1 at row 3.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&3&{ - 2}&1\\0&6&{ - 5}&4\end{array}} \right]\)

Step 3: Apply row operation

Perform an elementary row operation to produce the second augmented matrix.

Apply sum of row 3 and \( - 2\) times of row 2 at row 3.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&3&{ - 2}&1\\0&0&{ - 1}&2\end{array}} \right]\)

The system of the augmented matrix is consistent. Thus, \(b\) is in \(W\).

Perform an elementary row operation to produce the third augmented matrix.

Multiply row 2 by \(\frac{1}{3}\) and row 3 by \( - 1\) .

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&1&{ - \frac{2}{3}}&{\frac{1}{3}}\\0&0&1&{ - 2}\end{array}} \right]\)

Step 4: Convert the matrix into the equation

To obtain the solution of the vector equations, it is required to convert the augmented matrix into the system of equations.

Write the obtained matrix \(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&1&{ - \frac{2}{3}}&{\frac{1}{3}}\\0&0&1&{ - 2}\end{array}} \right]\) into the equation notation.

\(\begin{array}{l}{x_1} - 4{x_3} = 4\\{x_2} - \frac{2}{3}{x_3} = \frac{1}{3}\\{x_3} = - 2\end{array}\)

The set \(W\) contains infinitely many vectors.

Step 5: Show that vector \({a_1}\) is in \(W\)

Span \(\left\{ {{v_1},{v_2},...,{v_p}} \right\}\) is a collection of all vectors that can be written in the form \({c_1}{v_1} + {c_2}{v_2} + ... + {c_p}{v_p}\).

Write vector \({a_1}\) in a linear combination.

\({a_1} = 1{a_1} + 0{a_2} + 0{a_3}\)

Hence, it has been shown that \({a_1}\) is in \(W\).

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