MATH 51 MIDTERM 1 SOLUTIONS

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1 MATH 5 MIDTERM SOLUTIONS. Compute the following: (a). 2 [ = ] (b). [ ] = [ ] (c) = (d). 2 = (e). [ ] [ ] 2 [ ] = 7 7 2

2 2. Find the inverse of the matrix A = We write the augmented matrix [A I 3 ] and perform row operations until we get the identity on the left. In the first step we do row2-row and row3-3row to get then we do row3-2row2 and we get finally we do row-3row and row2-2row3 and we are left with From here we deduce. A =

3 3(a). Consider the following matrix A and its row reduced echelon form rref(a): A = , rref(a) = (You do not need to check this.) Find a basis for the column space C(B) of B. If we take the columns of A correspondig to the position of the pivots in rref(a) we get the basis , 3,

4 3(b). As in part (a), A = , rref(a) = Find some vectors that span the nullspace N(A) of A. We want to find the x R 5 that satisfy Ax = 0. For this we use rref(a) and get the equations x = 3x 3 x 4 x 2 = x 3 + x 4 x 5 = 0 This implies that we can write x as x x 2 x 3 x 4 = x x and therefore a basis for N(A) is the set 0, x 5 4

5 4(a). Consider the equation Ax = b where A = Find the 2 0 condition(s) on the vector b for this equation to have a solution. (Your answer should be one or more equations involving the components b i of b = b.) b 2 b 3 We write the augmented matrix 3 0 b b b 3 and proceed to row reduce it. First we do row2-2row and row3-row and get 3 0 b 0 0 b 2 2b 0 0 b 3 b bext we do row3+row2 to get 0 0 b 3(b 2 2b ) 0 0 b 2 2b b 3 b + b 2 2b and for the system to have a solution we must have b 3 b + b 2 2b = b 2 + b 3 3b = 0 5

6 4(b). Find a matrix B such that N(B) = C(A), where A = (This is the matrix from part (a).) First we note that by part a) we have C(A) = b b 2 R 3 / 3b + b 2 + b 3 = 0 We might rewrite the equation 3b + b 2 + b 3 as [ ] 3 b = 0 b 3 and from this we can tell that the matrix B = [ 3 ] will satisfy N(B) = C(A). b 2 b 3 6

7 [ 3 5(a). Find all solutions of the equation Ax =, where 6] A = [ ] We write the augmented matrix [ 2 2 ] and proceed to row reduce it. First we do row2-5row to get [ 2 2 ] and then row-2row2 [ ] The equivalent system of equations is now x = + 2x 2 7x 4 x 3 = + 3x 4 From here we deduce that the set of solutions is x x

8 5(b). Find a basis for the nullspace of A, where A is the matrix in part (a). From part a) we get the basis 2 7 0,

9 6(a). Suppose S is the sphere in R 3 of radius centered at the origin. Suppose that A, B, and C are three points on S and that side AC of the triangle ABC is a diameter of the sphere. Prove using vectors that the triangle has a right angle at B. If we let A = (a, a 2, a 3 ), then since AC is a diameter we must have C = A = ( a, a 2, a 3 ). Let B = (b, b 2, b 3 ). To prove that the triangle ABC has a right angle at B is the same as proving that the vectors AB = B A and CB = B C are perpendicular. Now we compute AB CB = (B A) (B C) = b a b 2 a 2 b + a b 2 + a 2 b 2 a 3 b 3 + a 3 = (b a )(b + a ) + (b 2 a 2 )(b 2 + a 2 ) + (b 3 a 3 )(b 3 + a 3 ) = b 2 a 2 + b 2 2 a b 2 3 a 2 3 = (b 2 + b b 2 3) (a 2 + a a 3 3) And since A and B are points in the sphere we have b 2 + b b 2 3 = a 2 + a a 2 3 =, and that implies AB CB = 0. 9

10 6(b). Suppose A is a matrix with 4 columns. Suppose u and v are linearly independent vectors in R 4 such that Au = Av = 0. Prove that if x, y, and z are any vectors in R 4, then the vectors Ax, Ay, and Az must be linearly dependent. We can rephrase teh fact that Au = Av = 0 by saying that u, v N(A). By condition of the problem u and v are linearly independent and that implies dimn(a) 2. A has 4 columns, therefore by proposition 2.4 (page 77 in the book) we must have rank(a) + nullity(a) = 4 and this implies nullity(a) 2. Since the dimension of the column space is at most 2 any three vectors must be linearly dependent. 0

11 7(a). Find a parametric equation for the line L through the points A = (, 0, 3) and B = (2,, 5). First we find the direction vector v = B A = (3,, 2). parametric equation of the line is x x 2 = 0 + t x 3 Now the

12 7(b). Find a point C on L such the line OC and the line L are perpendicular to each other. (Here L is the line from part (a), and the line OC is the line through the origin O = (0, 0, 0) and C.) If we look at the equation we found in part a), any point of the line is given by a value of t, let us assume that the point C is given by + 3t C = t 3 + 2t The line OC has direction C and the line L has direction (3,, 2). In order for two lines to be perpendicular this two vectors must be perpendicular, ie. we must have + 3t t 3 = t t + t t = 0 4t = 3 And from here we find C = ( 23 4, 3 4, 36 4 ). t = 3 4 2

13 8. Find a matrix A such that: A [ 3 =, A ] 3 [ 7 =, A 3] 3 2 = [ 7 8]. Let us name the vectors v =, v 2 = 3 and v 3 = 3 2. We can find e = (v 2 2 v ) and e 2 = v 3 v 2. This implies Ae = 2 (Av Av 2 ) = [ [ 4 2 = 2 2] 6] [ 0 Ae 2 = Av 3 Av 2 = 5] We notice that e 3 = v e e 2 and using the above results we get [ ] Ae 3 = Av Ae Ae 2 = 0 Finally, Ae, Ae 2 and Ae 3 are the columns of A, therefore we obtain [ ] 2 0 A =

14 9(a). Suppose is an equilateral triangle in R 3 each of whose sides has length 7. Let A, B, and C be the corners of. Find ( 2 ) ( AB 3 ) AC. [Notation: AB is the vector beginning at A and ending at B.] We do the computation ( 2 ) ( AB 3 ) AC ( ) ( ) = 6 AB AC = 6 AB AC cos θ Where AB = AC = 7, and θ = π/3 is the angle that AB and AC make (visually, to find the angle between the vectors we need to draw them starting from the same point). Thus, we find that the answer is (6)(7)(7)( ) =

15 9(b). Suppose u, v, and w are unit vectors in R 4 such that each one is orthogonal to the other two. Find a number c so that u + 2v + 3w and 5u + v + cw are orthogonal to each other. (Recall that a unit vector is a vector whose norm is.) The problem is equivalent to find c such that (u + 2v + 3w) (5u + v + cw) = 0. Now we have (u + 2v + 3w) (5u + v + cw) = 5u u + u v + cu w +0v u + 2v v + 2cv w +5w u + 3w v + 3cw w = 5 u v 2 + 3c w 2 Here we used the fact that the vectors u, v and w are orthogonal, ie. u v = u w = v w = 0. Now we use the fact that they have norm and the equation reduces to c = 0, therefore c =

16 0. Short answer questions. (No explanations required.) (a). Suppose that a linear subspace V is spanned by vectors v, v 2,..., v k. What, if anything, can you conclude about the dimension of V? Answer: dimv k. Since we have k vectors spanning V a set with more than k vectors must be linearly dependent (proposition 2. in the book). Since a basis consists of linearly independent vectors we must have dimv k (b). Suppose that a linear subspace W contains a set {w, w 2,..., w k } of k linearly independent vectors. What, if anything, can you conclude about the dimension of W? Answer: dimw k. If we had a basis with less than k elements, again by proposition 2. a set of k vectors could not be linearly independent. (c). Suppose u v > 0. What, if anything, can you conclude about the angle θ between u and v? [Note: by definition, the angle θ between two nonzero vectors is in the interval 0 θ π.] Answer: 0 θ < π. Since u v = u v cosθ and u v > 0 we must 2 have cosθ > 0. (d). Suppose that A is a k m matrix (i.e., a matrix with k rows and m columns.) If k < m, what, if anything, can you conclude about the number of solutions of Ax = b? Answer: Either there is no solution or if a solution exist there are infinitely many. This is because according to proposition 2.4 we have rank(a)+nullity(a) = m, and since the columns of A are vectors in R k we have rank(a) k, therefore nullity(a) > 0. If b is in the column space of A we will have infinitely many solutions, otherwise we will have no solution. (e). Suppose A is a matrix with 5 rows and 4 columns. Suppose that the equation Ax = 0 has only one solution. What, if anything, can you conclude about the dimension of the column space of A? Answer: dimc(a) = 4. The only solution for Ax = 0 must be x = 0 and therefore N(A) = 0. Again we obtain the answer using proposition