14.1

1. requires 8 = 2x - y and 9 = + 3 y. Solving these equations gives x = 3 and >• = —2.

3. The determinant of the matrix with the three vectors as columns is equal to 3, so the vectors are linearly independent.

5. Suppose a (a + b) + £(b + c) + y (a + c) = 0. Then (a + y) a + (a + £)b + (P + y)c = 0. Because a, b, and c are linearly independent, a + y = 0, a + = 0, and £ + y = 0. It follows that a = £ = y = 0, which means that a + b, b + c, and a + c are linearly independent. The vectors a - b, b + c, and a + c are linearly dependent because (a - b) -r (b + c) - (a + c) = 0.

7. Both these two statements follow imrtiediately from the definitions.

14.2

1. (a) 1. (The determinant of the matrix is 0, so the rank is less than 2. Because not all entries are 0, the rank is 1.) (b) 2 (c) 2 (d) 3 (e) 2 (f) 3

3"A=(6 2)andB=(_3 Here r(AB) = 0 and r(BA) = 1.

14.3

1. (a) No solutions, (b) x, = 1 + (2/3)b, x2 = 1 +a- (5/3)b, x3 = a, and = b, with a, b arbitrary. Two degrees of freedom, (c) = (-1/3K

x2 = (5/3)a, X2 = a, and x4 = l, with a arbitrary. One degree of freedom, (d) No solutions, (e) x\ = x2 = x3 = 0 is the only solution. There are 0 degrees of freedom, (f) X] = a, x2 = —a, X3 = —a, and x4 = a, with a arbitrary. One degree of freedom.

3. For a 0 and a # 7, the system has a unique solution. For a = 0 and b = 9/2, or for a = 7 and b = 10/3, the system has an infinite number of solutions. For other values of the parameters, there are no solutions.

5. (a) Unique solution for p ^ 3. For p = 3 and q — 0, there are infinitely many solutions (1 degree of freedom). For p = 3 and q ^ 0, there are no solutions, (b) For p # 3, only 2 = 0 is orthogonal to the three vectors. For p = 3, the vector z = (—a, 0, a) is orthogonal to the three vectors, for all values of a. (C) Let the n vectors be a; = (an,..., ain), i = 1, n. If b = (¿>i,..., b„) is orthogonal to each of these n vectors, then the scalar product of b with each a; is 0, anbi+----\- ainbn = 0 (i = !,...,«)

Because aj, ..., a„ are linearly independent, this homogeneous system of equations only has the solution b\ — ■ • ■ = b„ = 0, so b = 0.

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