91Ó°ÊÓ

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.

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

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.

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

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 \(W\).

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 \(W\).