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Q25E

Expert-verifiedFound in: Page 1

Book edition
5th

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

Pages
483 pages

ISBN
978-03219822384

**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 \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).**

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

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

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.

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.

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

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 \( - 1\) .

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

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.

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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