Write each vector as a linear combination of the vectors in...

Write each vector as a linear combination of the vectors in S. (If not possible, enter IMPOSSIBLE.) S = {(6, -7, 8, 6), (4, 6, -4, 1)} (a) u = (2, 19, -16, -4) u = \begin{pmatrix} -1 \\ \\ \end{pmatrix} s_1 + \begin{pmatrix} 3 \\ \\ \end{pmatrix} s_2 (b) v = \begin{pmatrix} \frac{57}{2}, \frac{123}{4}, -18, \frac{21}{2} \end{pmatrix} v = \begin{pmatrix} \frac{3}{2} \\ \\ \end{pmatrix} s_1 + \begin{pmatrix} \frac{9}{4} \\ \\ \end{pmatrix} s_2 (c) w = (-4, -14, \frac{21}{8}, \frac{59}{8}) w = \begin{pmatrix} -1 \\ \\ \end{pmatrix} s_1 + \begin{pmatrix} \frac{1}{8} \\ \\ \end{pmatrix} s_2 (d) z = (4, -2, 3, \frac{13}{4}) z = \begin{pmatrix} \frac{1}{2} \\ \\ \end{pmatrix} s_1 + \begin{pmatrix} \frac{1}{4} \\ \\ \end{pmatrix} s_2

Transcript

00:01 Hi, in this question we have to write each vectors in the linear combination of vectors.

00:07 Here it is given that s is equal to 1, 2, minus 2, 2, minus 1, 1.

00:16 Here we can say that s1 is equal to 1, 2, minus 2 and s2 is equal to 2, minus 1, 1.

00:24 Now, first one it is given the vector z is equal to minus 11, 3 and minus 3.

00:32 So that here the linear combination will be alpha into s1 plus beta into s2.

00:39 So that here alpha and beta are the scalars.

00:45 So that here we can write it as minus 11, 3, minus 3 equal to alpha into 1, 2, minus 2 plus beta into 2, minus 1, 1.

00:57 Now we have alpha plus 2 beta equal to minus 11, 2 alpha minus beta equal to 3 and minus 2 alpha plus beta equal to minus 3.

01:10 Now we can consider the equations as 1, 2 and 3.

01:14 Now by solving 1 and 2, that is here first equation multiplied by 1 which gives alpha plus 2 beta equal to minus 11 and the second equation can be multiplied by 2 so that it will be 4 alpha minus 2 beta equal to 6 such that here we will get 5 alpha equals to minus 5 so that alpha will be minus 1.

01:49 Now by substituting alpha equal to minus 1 in first equation, we will get minus 1 plus 2 beta equal to minus 11 then beta will be minus 5 so that here we can say that minus 11, 3, minus 3 which is z equal to minus 1 into 1, 2, minus 2 plus minus 5 into 2, minus 1, 1.

02:19 Hence we can say that z is a linear combination of the vectors in s and then we are having that v equal to minus 1, minus 5 and 5 so that here also we have to write v as a linear combination of vectors in s such that it will be 1, minus 5, 5 equal to alpha into 1, 2, minus 2 plus beta of 2, minus 1, 1...

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