The set {(x, y, z) | x > 0, y > 0, z < 0} is a domain in R³? justify
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manas444
1. Determine if each list of vectors is linearly dependent or independent. Justify your answers. (a) (1, 2),(2, 1) Solution. We write c1(1, 2) +c2(2, 1) = (0, 0) and solve for c1 and c2. This yields two equations c1 + 2c2 = 0 and 2c1 + c2 = 0, and it follows that c1 = c2 = 0 is the only solution. Hence the two vectors are linearly independent. (b) (2, −2),(−2, 2) Solution. Notice that (−2, 2) = −(2, −2). Since one vector is a scalar multiple of the other, they are linearly dependent. (c) (1, 2, 1),(1, 3, 1),(0, −1, 0) Solution. We write c1(1, 2, 1) + c2(1, 3, 1) + c3(0, −1, 0) = (0, 0, 0) which yields only 2 equations c1 + c2 = 0 and 2c1 + 3c2 − c3 = 0 in 3 unknowns. Hence there must be nonzero solutions: for instance, c1 = 1, c2 = −1, c3 = −1 is such a solution. This means the vectors are linearly dependent. (d) (1, 0, 0),(1, 1, 0),(1, 1, 1) Solution. We write c1(1, 0, 0) + c2(1, 1, 0) + c3(1, 1, 1) = (0, 0, 0), which yields 3 equations c1 +c2 +c3 = 0, c1 +c2 = 0 and c3 = 0. It follows that c1 = c2 = c3 = 0 is the only solution, and thus the vectors are linearly independent. (e) x + 1, x2 + 2x, x2 − 2 in the vector space P2 of polynomials of degree less than or equal to 2. Solution. We write a(x + 1) + b(x 2 + 2x) + c(x 2 − 2) = 0, and compare the coefficients of each different power of x on each side of the equation. This yields 3 equations (the first comes from looking at the constant terms, the second from the x-terms and the third from the x 2 -terms): a−2c = 0, a+2b = 0 and b+c = 0. Solving for a, b, c, we see that there is a free variable, and one nonzero solution is given by a = 2, b = −1, c = 1. Hence the vectors are linearly dependent.
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