Test1_sol

Test1_sol - MOOREV MAT 342 TEST 1 2/23/2009 90 MINUTES NO...

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Unformatted text preview: MOOREV MAT 342 TEST 1 2/23/2009 90 MINUTES NO CALCULATOR NO NOTES NAME gg(>l’L)rIeiZ 2h) %(Zg\;/ 1. Let A be a 3x5 matrix. What must be the dimension of a matrix B if of the products AB and BA are defined? a) 3x3 b) 3x5 c) 5x3 d) 5x5 Answe<€j\) E:ZE§ _‘L d 1 2. aijg] Wm} The second row of. A + 2B is L""&,[XW The (3,2) entry of the product AB is '"Q 3. If A is the matrix on the right and A2 = I , A = x 0 then what is the value of x ? "‘ 1 1 x: —| Scratch Work: Ag; Cl3«|]\©l ¢E’S;—§>:~V I 5 vw/ Hf NB .”]:X;:f 1”] :1 7K 71—4 :3; xzti X3", )“WZO :? xz~l “:3:i L 5/ L 10 3. 10 L1, \0 5, 15’ W3 72% ‘7160 ‘8, w q, 11;” [6‘ [5 n: W MOORE MAT 342 TEST 1 2/23/2009 90 MINUTES NO CALCULATOR NO NOTES 4. In computing the inverse of a 3x3 matrix A 1 0 2 0 2 2 the augmented matrix [All] is formed and 0 1 —3 —1 1 l A is reduced to I by row operations. The O 0 l 2 0 1 - matrix on the right was obtained from a particular A after several operations. Complete the row reduction operations to find A'l. Eng» iozgflél 7v; )0 ammo < 0173.411 mama o x 05: L1 69 <7 I :2~C> l yulészj) c; I: 3, C) l /q*] N -—Li 1 C9 " 5‘ pr K0] 5. A = 1 0 0 Compute A-1 by performing row operations on th 1 i 2 augmented matrix [AEI]. Write A'1 as a product of three elementary matrices. / E3 ' ’ [A/IE;1002100 [001100 M("/Z}) )00 100 . “32690 "'5'? @{OHIO 1.1.1, mo 41o up: my Rabbis w- 1 H01 DO) 0—H 1 53- . /q_‘ . /b\ :: &:3.£::Z ‘ 1.1— .4 A 1‘, C) C) \ O I fix ' 010]»[015' 391% “H “101 0&/ MOORE MAT 342 TEST 1 2/23/2009 QQ’MINUTES NO CALCULATOR NO NOTES- 6. The matrix on the right is the reduced 11 —1 0 row echelon form of the augmented 0 0 1 matrix for a system of 3 equations in the 0 0 0 varibles x1, x2, x3, X4 and x5. The number of free variables in the solution is 2L 7. Let AX = B be the matrix form of a system of equations, x —1. _1 112 ‘X = y , B = 2 and A = 2 1 0 z -3 1 2 2 The value of y is C) Scratch Work: MOORE MAT 342 TEST 1 2/23/2009 90 MINUTES NO CALCULATOR NO NOTES 8. Each of the matrices below is the reduced row echelon form of the augmented matrix for a system of linear equations. (1)10000 (2)100 (3)100 (4)1ooor(5)1101 '0 1 1 l O O 0 1 0 1 1 0 1 0 O O 0 1 O 0 0 O 0 O G 0 0 0 0 O O l 1 Which systems above have no solution ? answer Which systems above have more than one solution ? answer I Which systems have exactly one solution ? answer (z) (4) 9. Augmented matrices of the form [AEB] are given below. By reducing A to reduced row echelon form (if A is not already of that form), find the solution of AX = B in each case. (a) 1 0 1:2 (b) 1 1 132 (c) 1 032 01133 0 0 0:0 01:3 000:0 ' 0003.0 00:0 (0? >61: 14%; (g) x’izgqqcrp'x} (c) >921 X$;3”['X3 ){Z’e ('rX’lz MOORE MAT 342 TEST 1 2/23/2009 90 MINUTES NO CALCULATOR NO NOTES 10. For what value of k is the system below consistent? What is the' solution for that value of k? xl- x2+ 2x3 = 0 -x1+ 2x2— 3x3 = l = k X- X 2 3 MOORE MAT 342 TEST 1 2/23/2009 90 MINUTES NO CALCULATOR NO NOTES' 11. Find the determinant of the four matrices below. This can be done by inspection. A=[100] B=0001 c: 1113 D: 40791 0 1 0 1 O 0 O 0 3 1 1 O 1 6 3 2 1 0 1 0 1 0 0 0 0 2 2 O O 3 3 3 '0 O 1 O -1 -l -1 —3 O 0 O 0 l 0 O 0 1 6 det(A) = I det(B) = -[ det(C) = O det(D) = “’12! 12.The first and third rows of A = _g_ _g_ cof(A) are given. Finish by F filling in the second row. —L—- —J— 0 1 Scratch Work: Mg_(fino‘ 7 I l 1 ‘3 fiaafljfl’ l 1 I '3 ‘1 C? r C}5 l I ‘——"49 Ag'} 1 ] 6X59(: -: LQEIZ'IL) '(:D 00 '23” 007.3 *4 -l -1 —'5 aye) CD ' naval .i. WfiM/ufifte'D—a OW 57, MV<C~I)H’51 arc-3 35 OOOLE 430001 MOORE MAT 342 TEST 1 2/23/2009 90 MINUTES NO CALCULATOR NO NOTES 13.Let A and B be square matrices with det(A2+AB) = 4 and det(A"1) = 2. Compute det(A+B) . , ’ V ATVs :- MW“) 7' M(A”1»(/fi AB>>:M(A"D.MA§AB>:x—Lt:§ ...
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Test1_sol - MOOREV MAT 342 TEST 1 2/23/2009 90 MINUTES NO...

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