Exercise 7 < b m a t r i x > , v =

Exercise 7 \begin{bmatrix} 2 \\ 4 \\ 3 \\ 2 \\ 4 \\ 5 \end{bmatrix}, v = \begin{bmatrix} 2 \\ 4 \\ 6 \\ 1 \\ 7 \\ 3 \\ 5 \end{bmatrix}, w = \begin{bmatrix} 2 \\ 4 \\ 0 \\ 5 \\ 2 \\ 3 \\ 5 \end{bmatrix}, and z = \begin{bmatrix} 2 \\ 4 \\ 3 \\ 7 \\ 5 \\ 3 \\ 5 \end{bmatrix}. Let u = a. Are the sets \{u, v\}, \{u, w\}, \{u, z\}, \{v, w\}, \{v, z\}, \{w, z\} each linearly independent? Why or why not? b. Does the answer to Part (a) imply that \{u, v, w, z\} is linearly indepen- dent? c. To determine if \{u, v, w, z\} is linearly dependent, is it wise to check if, say, w is a linear combination of u, v, and z? d. Is \{u, v, w, z\} linearly dependent?

Transcript

00:01 Given that the set s equal to a, b, c, and the set t equal to d, e, comma, f are two linearly independent sets of n vector.

00:31 The solution of part a is the set a, b, b, b, c, c, comma, c, and a set a, b, c, d comma e comma f could be linearly independent or linearly linearly dependent depending on the values of a dot dot comma f the explanation is suppose the function f equal to a plus p plus c and d equal to this is the null vector then s and t are linearly independent but set a comma b comma c comma b comma e comma f is not as one multiplied by a plus b plus c plus d plus zero multiplied by e equal to f here f can be expressed as linearly combination of a comma b comma c comma d comma e vectors therefore the set a comma b comma c comma d e comma f is not linearly independent again if s intersection t equal to null set then given set is linearly independent in part p the set a plus t b b plus e comma c plus f could be independent or dependent on the value of a comma dot dot dot f here let us suppose that a plus d equal to alpha b plus c equal to beta and c plus f equal to gamma and consider the relation c1 multiplied by alpha plus c2 multiplied by beta plus c3 multiplied by gamma equal to zero as c i belongs to r the value of i is 0 1 2 3 and so on that implies that c1 a plus p plus c2 multiplied by b plus e plus c3 multiplied by c plus f equal to 0 simplify it we will get c1 multiplied by a c2 multiplied by b cp multiplied by cp multiplied by c is the one set plus the other set is c1 multiplied by d plus c2 multiplied by e plus c3 multiplied by f is equal to 0.

05:54 That implies that either c1a plus c2b plus c3c equal to 0 and c1d plus c2e plus c3f is equal to 0 or c1a plus c2b plus c2b plus c3 b plus c1, c3 c is equal to c1d plus c2 e plus c2 e plus c3 f is equal to 0 when c1 multiplied by a c2 multiplied by b plus c3 multiplied by c equal to c1 multiplied by d plus c2 multiplied by e plus c3 multiplied by f that implies that c1 multiplied by a minus b plus c2 multiplied by b minus e c3 multiplied by c minus f equal to zero that implies that now it is a equal to if a equal to b and b equal to e and c equal to f this for then s intersection t is not equal to 5.

07:37 Hence, the set a plus d, comma b plus e, comma c plus f is not be linearly independent.

07:59 In part c, the set a, a plus b, comma, a plus b, and a plus c is of always linearly independent.

08:22 Suppose alpha 1 equal to a plus b plus c and alpha 2 is equal to a plus b and alpha 3 is equal to a.

08:38 Let us consider the relation c1 multiplied by alpha 1 plus 2 multiplied by alpha 2 plus c3 multiplied by alpha 3 is equal to 0...

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