Question: == = (3, 1,-1) are three 5. (S.15). Let u = (-1, 1, 2), u = (1,-1,3), us vectors. Find expansion of the vector

== = (3, 1,-1) are three 5. (S.15). Let u = (-1, 1, 2), u = (1,-1,3), us vectors. Find expansion of the vector v= (3, 2, 4) via vectors u, u2, u3; 6.(S.10). Solve the following system, by determinant methods (Cramer's Rule) x1+x2 x31, 2x1 + x2 x3 = 2, 1-2x2 x3 = -1;

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