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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

In Exercises 13 and 14, determine if \(b\) is a linear combination of the vectors formed from the columns of the matrix \(A\).

13. \(A = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&2\\0&3&5\\{ - 2}&8&{ - 4}\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}3\\{ - 7}\\{ - 3}\end{array}} \right]\)

\({\mathop{\rm b}\nolimits} \) is not a linear combination of columns \({{\mathop{\rm a}\nolimits} _1},{{\mathop{\rm a}\nolimits} _2}\), and \({{\mathop{\rm a}\nolimits} _3}\).

See the step by step solution

Step by Step Solution

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

\(\left[ {\begin{array}{*{20}{c}}{{x_1} - 4{x_2} + 2{x_3}}\\{3{x_2} + 5{x_3}}\\{ - 2{x_1} + 8{x_2} - 4{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\{ - 7}\\{ - 3}\end{array}} \right]\)

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.

\(\begin{aligned}{c}{x_1} - 4{x_2} + 2{x_3} &= 3\\3{x_2} + 5{x_3} &= - 7\\ - 2{x_1} + 8{x_2} - 4{x_3} =& - 3\end{aligned}\)

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 \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{...}&{{a_n}}&b\end{array}} \right]\).

The augmented matrix for the vector equations \({x_1} - 4{x_2} + 2{x_3} = 3,3{x_2} + 5{x_3} = - 7\) and \( - 2{x_1} + 8{x_2} - 4{x_3} = - 3\) is represented as:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&2&3\\0&3&5&{ - 7}\\{ - 2}&8&{ - 4}&{ - 3}\end{array}} \right]\)

Step 4: Apply row operation

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

Replace row 3 by adding 2 times row 1 to row 3

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&2&3\\0&3&5&{ - 7}\\0&0&0&3\end{array}} \right]\)

Step 5: Determine whether \(b\) is a linear combination of the vector

The vector \({\mathop{\rm y}\nolimits} \) is 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 \(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&2&3\\0&3&5&{ - 7}\\0&0&0&3\end{array}} \right]\) into the equation notation.

\(\begin{aligned}{c}{x_1} - 4{x_2} + 2{x_3} &= 3\\3{x_2} + 5{x_3} &= - 7\\0{x_3} &= 3\end{aligned}\)

So, \({{\mathop{\rm a}\nolimits} _1},{a_2}\), and \({{\mathop{\rm a}\nolimits} _3}\) are denoted as 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 inconsistent.

Hence, \({\mathop{\rm b}\nolimits} \) does not represent a linear combination of the columns of \(A\).


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