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Q25E
Expert-verifiedLet \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).
a.
A vector equation \({x_1}{a_1} + {x_2}{a_2} + ... + {x_n}{a_n} = b\) has the same solution set as the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{....}&{{a_n}}&b\end{array}} \right]\).
Write the coefficient matrix into a set of vectors.
\(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\3\\6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 2}\\3\end{array}} \right]} \right\}\)
The set \(\left\{ {{a_1},{a_2},{a_3}} \right\}\) contains three vectors and \(b\) is not one of them.
b.
A vector equation \({x_1}{a_1} + {x_2}{a_2} + ... + {x_n}{a_n} = b\) has the same solution set as the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{....}&{{a_n}}&b\end{array}} \right]\). In particular, \(b\) can be generated by a linear combination of \({{\mathop{\rm a}\nolimits} _1},...{a_n}\), if and only if, there exists a solution to the linear system corresponding to the augmented matrix.
Write the given matrix in an augmented matrix.
\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&3&{ - 2}&1\\{ - 2}&6&3&{ - 4}\end{array}} \right]\)
Perform an elementary row operation to produce the first augmented matrix.
Apply sum of row 3 and \(2\) times of row 1 at row 3.
\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&3&{ - 2}&1\\0&6&{ - 5}&4\end{array}} \right]\)
Perform an elementary row operation to produce the second augmented matrix.
Apply sum of row 3 and \( - 2\) times of row 2 at row 3.
\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&3&{ - 2}&1\\0&0&{ - 1}&2\end{array}} \right]\)
The system of the augmented matrix is consistent. Thus, \(b\) is in \(W\).
Perform an elementary row operation to produce the third augmented matrix.
Multiply row 2 by \(\frac{1}{3}\) and row 3 by \( - 1\) .
\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&1&{ - \frac{2}{3}}&{\frac{1}{3}}\\0&0&1&{ - 2}\end{array}} \right]\)
To obtain the solution of the vector equations, it is required to convert the augmented matrix into the system of equations.
Write the obtained matrix \(\left[ {\begin{array}{*{20}{c}}1&0&{ - 4}&4\\0&1&{ - \frac{2}{3}}&{\frac{1}{3}}\\0&0&1&{ - 2}\end{array}} \right]\) into the equation notation.
\(\begin{array}{l}{x_1} - 4{x_3} = 4\\{x_2} - \frac{2}{3}{x_3} = \frac{1}{3}\\{x_3} = - 2\end{array}\)
The set \(W\) contains infinitely many vectors.
Span \(\left\{ {{v_1},{v_2},...,{v_p}} \right\}\) is a collection of all vectors that can be written in the form \({c_1}{v_1} + {c_2}{v_2} + ... + {c_p}{v_p}\).
Write vector \({a_1}\) in a linear combination.
\({a_1} = 1{a_1} + 0{a_2} + 0{a_3}\)
Hence, it has been shown that \({a_1}\) is in \(W\).
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