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

In Exercises 13 and 14, determine if \({\mathop{\rm b}\nolimits} \) is a linear combination of the vectors formed from the columns of the matrix \(A\).
**14. \(A = \left[ {\begin{array}{{20}{c}}1&{ - 2}&{ - 6}\\0&3&7\\1&{ - 2}&5\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{{20}{c}}{11}\\{ - 5}\\9\end{array}} \right]\)**

\({\mathop{\rm b}\nolimits} \) is a linear combination \({{\mathop{\rm a}\nolimits} _1},{{\mathop{\rm a}\nolimits} _2}\), and \({{\mathop{\rm a}\nolimits} _3}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Rewrite the matrix into a vector equation

In \({\mathbb{R}^2}\), the sum of two vectors \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) is the vector addition \({\mathop{\rm u}\nolimits} + v\), which is obtained by adding the corresponding entries of \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \).

The scalar multiple of a vector \({\mathop{\rm u}\nolimits} \) by real number \(c\) is the vector \(c{\mathop{\rm u}\nolimits} \) obtained by multiplying each entry in \({\mathop{\rm u}\nolimits} \) by \(c\).

Use scalar multiplication and vector addition to rewrite the matrix as a vector equation by

\(\left[ {\begin{array}{*{20}{c}}{{x_1} - 2{x_2} - 6{x_3}}\\{3{x_2} + 7{x_3}}\\{{x_1} - 2{x_2} + 5{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{11}\\{ - 5}\\9\end{array}} \right]\)

Step 2: Write the matrix into a vector equation

The vectors on the left and right sides are equal if and only if their corresponding entries are both equal. Thus, \({x_1}\) and \({x_2}\) make the vector equation \({x_1}{a_1} + {x_2}{a_2} = b\) if and only if \({x_1}\) and \({x_2}\) satisfy the system.

Write the matrix into a vector equation.

\(\begin{aligned}{c}{x_1} - 2{x_2} - 6{x_3} &= 11\\3{x_2} + 7{x_3} &= - 5\\{x_1} - 2{x_2} + 5{x_3} &= 9\end{aligned}\)

Step 3: Convert the vector equation into an augmented matrix

A vector equation \({{\mathop{\rm x}\nolimits} _1}{a_1} + {x_2}{a_2} + ... + {x_n}{a_n} = b\) has the same solution set as the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{...}&{{a_n}}&b\end{array}} \right]\).

Thus, the augmented matrix for the vector equations \({x_1} - 2{x_2} - 6{x_3} = 11,\,3{x_2} + 7{x_3} = - 5\) and \({x_1} - 2{x_2} + 5{x_3} = 9\) is represented as:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 6}&{11}\\0&3&7&{ - 5}\\1&{ - 2}&5&9\end{array}} \right]\)

Step 4: Apply row operation

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

Replace row 3 by adding -1 times row 1 to row 3.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 6}&{11}\\0&3&7&{ - 5}\\0&0&{11}&{ - 2}\end{array}} \right]\)

Step 5: Apply row operation

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

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

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 6}&{11}\\0&1&{\frac{7}{3}}&{ - \frac{5}{3}}\\0&0&{11}&{ - 2}\end{array}} \right]\)

Step 6: Apply row operation

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

Multiply row 3 by \(\frac{1}{{11}}\).

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 6}&{11}\\0&1&{\frac{7}{3}}&{ - \frac{5}{3}}\\0&0&1&{\frac{{ - 2}}{{11}}}\end{array}} \right]\)

Step 7: Convert the matrix into the equation

As it is known that the vector \({\mathop{\rm y}\nolimits} \) defined by \(y = {c_1}{v_1} + .... + {c_p}{v_p}\) is called a linear combination of \({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\) with weights \({c_1},{c_2},...,{c_p}\).

To obtain the solution of the vector equations, you have to convert the augmented matrix into vector equations.

Write the obtained matrix \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 6}&{11}\\0&1&{\frac{7}{3}}&{ - \frac{5}{3}}\\0&0&1&{\frac{{ - 2}}{{11}}}\end{array}} \right]\) into the equation notation.

\(\begin{aligned}{c}{x_1} - 2{x_2} - 6{x_3} &= 11\\{x_2} + \frac{7}{3}{x_3} &= - \frac{{ - 5}}{3}\\{x_3} &= \frac{{ - 2}}{{11}}\end{aligned}\)

So, if \({{\mathop{\rm a}\nolimits} _1},{a_2}\), and \({{\mathop{\rm a}\nolimits} _3}\) are the three columns of \(A\), the system of equations corresponding to the vector equation \({x_1}{{\mathop{\rm a}\nolimits} _1} + {x_2}{{\mathop{\rm a}\nolimits} _2} + {x_3}{{\mathop{\rm a}\nolimits} _3} = {\mathop{\rm b}\nolimits} \) is consistent.

Hence, \({\mathop{\rm b}\nolimits} \) is a linear combination \({{\mathop{\rm a}\nolimits} _1},{{\mathop{\rm a}\nolimits} _2}\), and \({{\mathop{\rm a}\nolimits} _3}\).

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Linear Algebra and its Applications

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Short Answer

Step by Step Solution

Step 1: Rewrite the matrix into a vector equation

Step 2: Write the matrix into a vector equation

Step 3: Convert the vector equation into an augmented matrix

Step 4: Apply row operation

Step 5: Apply row operation

Step 6: Apply row operation

Step 7: Convert the matrix into the equation

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Linear Algebra and its Applications

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Short Answer

Step by Step Solution

Step 1: Rewrite the matrix into a vector equation

Step 2: Write the matrix into a vector equation

Step 3: Convert the vector equation into an augmented matrix

Step 4: Apply row operation

Step 5: Apply row operation

Step 6: Apply row operation

Step 7: Convert the matrix into the equation

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