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Construct three different augmented matrices for linear systems whose solution set is \({x_1} = - 2,{x_2} = 1,{x_3} = 0\).

Short Answer

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The three different augmented matrices are \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\0&1&1&1\\0&0&1&0\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\0&1&1&1\\1&0&1&{ - 2}\end{array}} \right]\), and \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\2&1&1&{ - 3}\\1&0&1&{ - 2}\end{array}} \right]\).

Step by step solution

01

Write the given solution set in an augmented matrix

To express a system in theaugmented matrix form, extract the coefficients of the variables and the constants, and place these entries in the column of the matrix.

Now, begin with a simple augmented matrix for which the solution is \({x_1} = - 2\), \({x_2} = 1\), \({x_3} = 0\).

For the solution set \({x_1} = - 2\), \({x_2} = 1\), \({x_3} = 0\), the augmented matrix is as follows:

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

02

Apply the row operation

A basic principle states that row operations do not affect the solution set of a linear system.

Perform an elementary row operation on the matrix \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\0&1&0&1\\0&0&1&0\end{array}} \right]\) to produce the first augmented matrix.

Replace row two with the sum of the second and the third rows; i.e., \({R_2} \to {R_2} + {R_3}\).

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

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

03

Apply the row operation

Perform an elementary row operation on the matrix \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\0&1&1&1\\0&0&1&0\end{array}} \right]\) to produce the second augmented matrix.

Replace row three with the sum of rows one and three; i.e., \({R_3} \to {R_3} + {R_1}\).

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

After the row operation, the matrix becomes the following:

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

04

Apply the row operation

Perform an elementaryrow operation on the matrix \(\left( {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\0&1&1&1\\1&0&1&{ - 2}\end{array}} \right)\) to produce the third augmented matrix.

Replace row two with the sum of rows one and two; i.e., \({R_2} \to {R_2} + 2{R_1}\).

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

After the row operation, the matrix becomes the following:

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

Thus, the three different augmented matrices obtained are \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\0&1&1&1\\0&0&1&0\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\0&1&1&1\\1&0&1&{ - 2}\end{array}} \right]\), and \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\2&1&1&{ - 3}\\1&0&1&{ - 2}\end{array}} \right]\).

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Give separate answers for each part.

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a. Every elementary row operation is reversible.

b. A \(5 \times 6\)matrix has six rows.

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24.

a. Any list of five real numbers is a vector in \({\mathbb{R}^5}\).

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c. The weights \({{\mathop{\rm c}\nolimits} _1},...,{c_p}\) in a linear combination \({c_1}{v_1} + \cdot \cdot \cdot + {c_p}{v_p}\) cannot all be zero.

d. When are \({\mathop{\rm u}\nolimits} \) nonzero vectors, Span \(\left\{ {u,v} \right\}\) contains the line through \({\mathop{\rm u}\nolimits} \) and the origin.

e. Asking whether the linear system corresponding to an augmented matrix \(\left[ {\begin{array}{*{20}{c}}{{{\rm{a}}_{\rm{1}}}}&{{{\rm{a}}_{\rm{2}}}}&{{{\rm{a}}_{\rm{3}}}}&{\rm{b}}\end{array}} \right]\) has a solution amounts to asking whether \({\mathop{\rm b}\nolimits} \) is in Span\(\left\{ {{a_1},{a_2},{a_3}} \right\}\).

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