Question

Let \(A = \left[ {\begin{array}{*{20}{c}}2&0&6\\{ - 1}&8&5\\1&{ - 2}&1\end{array}} \right]\), let \(b = \left[ {\begin{array}{*{20}{c}}{10}\\3\\3\end{array}} \right]\) , and let \(W\) be the set of all linear combinations of the columns of \(A\).

1. Is \(b\) in \(W\)?
2. Show that the third column of \(A\) is in \(W\).

Short answer

1. \(b\)is a linear combination of \(A\), which means \(b\) is in \(W\).
2. The third column \({a_3}\) is in \(W\).

Step-by-step solution

Write the matrix into an augmented matrix

Avector 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 into an augmented matrix.

\(\left[ {\begin{array}{*{20}{c}}2&0&6&{10}\\{ - 1}&8&5&3\\1&{ - 2}&1&3\end{array}} \right]\)

Apply row operation to identify that \(b\) is in \(W\)

a.

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

Multiply row 1 by \(\frac{1}{2}\).

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

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

Add 1 time row 1 to row 2 and add \( - 1\) times row 1 to row 3.

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

Apply row operation to identify that \(b\) is in \(W\)

Perform an elementaryrow operation to produce the third augmented matrix.

Multiply row 2 by \(\frac{1}{8}\).

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

Apply sum of 2 times of row 2 and row 3 at row 3

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

When \({a_1},{a_2}\), and \({a_3}\) are the three columns of \(A\), the system of equations corresponding to the vector equation \({x_1}{a_1} + {x_2}{a_2} + {x_3}{a_3} = b\) is consistent.

Thus, \(b\) is a linear combination of \(A\), which means \(b\) is in \(W\).

Show that the third column of A is in W

b.

If vector\({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\)in\({\mathbb{R}^n}\)are given with scalars\({c_1},{c_2},...,{c_p}\), then the vector\({\mathop{\rm y}\nolimits} \)defined by\(y = {c_1}{v_1} + .... + {c_p}{v_p}\)is called alinear combination of\({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\)with weights\({c_1},{c_2},...,{c_p}\).

Let\({a_1},{a_2}\), and \({a_3}\) represent the three columns of \(A\). Since \({{\mathop{\rm a}\nolimits} _3} = 0{a_1} + 0{a_2} + 1{a_3}\) is a linear combination of the columns of \(A\), so \({a_3}\) is in \(W\).

Hence, the third column \({a_3}\) is in \(W\).