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Expert-verified Found in: Page 1 ### 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 # In Exercises 11 and 12, determine if $${\rm{b}}$$ is a linear combination of $${{\mathop{\rm a}\nolimits} _1},{a_2}$$ and $${a_3}$$.12.

$${\mathop{\rm b}\nolimits}$$ 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 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 to rewrite the matrix into a vector equation

$${x_1}\left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\2\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\5\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}2\\0\\8\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 5}\\{11}\\{ - 7}\end{array}} \right]$$

## Step 2: 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 $${x_1}\left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\2\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\5\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}2\\0\\8\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 5}\\{11}\\{ - 7}\end{array}} \right]$$ is represented as:

$$\left[ {\begin{array}{*{20}{c}}1&0&2&{ - 5}\\{ - 2}&5&0&{11}\\2&5&8&{ - 7}\end{array}} \right]$$

## Step 3: Apply row operation

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

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

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

## Step 4: Apply row operation

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

Sum of $$- 2$$ times row 1 and row 3 at row 3

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

## Step 5: Apply row operation

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

Sum of $$- 1$$ times row 2 and row 3 at row 3

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

## Step 6: Determine whether $$b$$ is a linear combination of the column $$A$$

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 the system of equations.

Write the obtained matrix $$\left[ {\begin{array}{*{20}{c}}1&0&2&{ - 5}\\0&5&4&1\\0&0&0&2\end{array}} \right]$$ into the equation notation.

$$\begin{array}{c}{x_1} + 2{x_3} = - 5\\5{x_2} + 4{x_3} = 1\\0 = 2\end{array}$$

The system of equations corresponding to the vector equation $${x_1}{{\mathop{\rm a}\nolimits} _1} + {x_2}{{\mathop{\rm a}\nolimits} _2} + {x_3}{{\mathop{\rm a}\nolimits} _3} = {\mathop{\rm b}\nolimits}$$ has no solution. Hence, $${\mathop{\rm b}\nolimits}$$ is not a linear combination of columns $${{\mathop{\rm a}\nolimits} _1},{{\mathop{\rm a}\nolimits} _2}$$, and $${{\mathop{\rm a}\nolimits} _3}$$.

Thus, $${\mathop{\rm b}\nolimits}$$ is not a linear combination of columns $${{\mathop{\rm a}\nolimits} _1},{{\mathop{\rm a}\nolimits} _2}$$, and $${{\mathop{\rm a}\nolimits} _3}$$. ### Want to see more solutions like these? 