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Practice Exam I Linear Algebra September 21, 2004 1. Find a vector orthogonal to both 1,2,1 and 1,0,1 . Let [ x, y , z ] be an arbitrary vector that is orthogonal to both of the given vectors. So [ x, y, z ] [1,2,1] x 2 y z 0 and [ x, y, z ] [1,0,1] x z 0 . Now solving the second equation we have that z = x and substituting this into the first equation we have 2y = 0 or y = 0. Letting x = s where s is an arbitrary number then z = s too. So [s, 0 ,s] is the general solution for a vector that is orthogonal to the given two vectors and in particular [1,0,1] or [-1, 0 ,-1] are two of the infinite vectors that would answer the problem. 2. Assume the 3x3 matrix B reduces to A by the following steps. 3 R3 3 R2 1 R2 2 4 R2 B R B1 R B2 R A Give the elementary matrices whose product (in the proper order) with B gives A. The elementary matrix that would work for the first row operation is 1 0 0 E1 0 1 0 . The elementary matrix that would work for the second row 0 3 1 1 0 0 operation is E 2 0 0 1 . The elementary matrix that would work for the last 0 1 0 1 0 0 row operation is E1 0 4 0 . So E3 E 2 E1 is the product of elementary 0 0 1 matrices whose product with B gives A. 3. Let u [2,1,5] and v [1,2,2]. Find the following. (a) u (b) u v (c) u v (d) 3u 2v (e) The angle between u and v (f) The distance between u and v (a) u (2) 2 1 5 2 30 (b) 6 2 (e) 68.58 or 1.20 radians (f) 27 (d) [-8, 7, 11] 1 1 1 2 4. Find the inverse of the matrix 0 1 0 1 0 0 0 0 1 1 1 0 2 1 1 1 Use algorithm we discussed in class to get the inverse to be 1 1 2 2 0 0 . 1 1 1 0 0 0 2 0 0 3 1 1 2 1 5. Let A 1 3 , B , and C . Find the following if 2 1 1 3 5 1 0 3 possible, if not say why not. (a) AB (b) BA 0 6 2 (a) 6 0 4 6 3 3 (c) C B 3 6 (b) 3 6 (d) C A 1 5 0 (c) 1 6 2 (d) can’t find this because the matrices C and A are of different sizes. 6. Solve the following system of equations using a matrix method. 3x 6 y 3z 0 y 2z 2 3x 7 y 6 z 2 3 6 3 0 Set up the augmented matrix 0 1 2 2 and solve it to get the solution 3 7 6 2 x [4,2,0] . 1 2 1 7. Let A 2 4 2 . Give a basis for the null space, column space and row 3 6 3 space of the matrix A. 1 2 1 Row reduce the matrix down to 0 0 0 . The basis for the column space of 0 0 0 A is {[1,2,3]} , the basis for the row space of A is {[1,-2,-1]}, and the basis for the null space of A is {[2,1,0], [1,0,1]}. 8. For vectors u, v in n prove that if u v and u v are perpendicular then u v . Proof: if u v and u v are perpendicular then the dot product between them must be 0. That is (u v) (u v) u (u v) v(u v) u u u v v u v v u v 0 u v . The reason why we can cancel the middle dot products is because u v v u . 1 1 2 3 1 2 9. Let A 1 and B 1 . Find A and ( AB) . 1 2 1 1 2 3 1 4 and ( AB) 1 B 1 A 1 A 2 A 1 A 1 . 3 5 0 1 10. Prove ( A B) C A ( B C ) for all m x n matrices A, B, and C . Proof: Since A, B, and C have the same size then (A+B)+C and A+(B+C) will have the same size. We only need to prove that the (i,j)th entry of (A+B)+C is the same as the (i,j)th entry of A+(B+C). So the (i,j)th entry of (A+B)+C is (ai , j bi , j ) ci , j ai , j (bi , j ci , j ) (because of associativity property of real numbers). The right side of this expression is the (i,j) th entry of A+(B+C) and hence we are done. 11. Determine if {[1,3,2], [2,5,3], [4,0,1]} are linearly dependent or independent. To determine if the vectors are linearly independent or dependent we must show that r[1,3,2] s[2,5,3] t[4,0,1] 0 has a solution of [0,0,0] or a nonzero solution. This is equivalent to solving the homogeneous system in matrix form 2 4 1 3 5 0 . This is equivalent to showing the matrix row reduces to the 2 3 1 identity or not. Thus if we row reduce the matrix, we will find out that it does row reduced to the identity which says that the original vectors are independent. 2 3 1 12. Given the matrix D 4 1 2 . 1 0 1 (a) Row reduce the matrix D to row echelon form. (b) What is the rank of D? (c) Solve Dx 0 . (d) Determine the nullity of D. (a) (b) (c) (d) 1 1 1 4 2 We get row echelon form is 0 1 0 . 0 0 1 The rank of D is 3. That is rank(D)=3. Using (a) we can solve the homogenous system. The solution is [ 0, 0, 0]. The nullity of D is the number of free variables in the solution in (c). Thus the nullity(D) = 0.