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### 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 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 $${{\mathop{\rm a}\nolimits} _3}$$.

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 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 $${{\mathop{\rm a}\nolimits} _3}$$.