practiceExam1b

# practiceExam1b - Math 310 Hour Exam 1(Solutions Prof S...

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Unformatted text preview: Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: 26 Sept 1994 Problem 1: Use row—reduced echelon form (Gauss—Jordan) to ﬁnd all solutions of 2 3 1 \$1 1 1 1 1 \$2 = 3 3 4 2 \$3 4 E12 1 1 1 3 A2 2X1A3 3X1 3 AI1X2yAglx2 1 O 2 8 —>2311 —i—5 —> 01—1—5 3 4 2 4 —1 —5 0 0 0 0 Thus \$3 is free variable; With :03— — 011 in A17: 0, homogenous solutions are a(—2 1 1). With 5133 : 0 in Ar = b, particular solution is (8 — 5 0). 80 general: (8 — 201, —5 + or, oz). Problem 2: 1 0 0 (a) ISA: 1 1 0 invertible? If so, use elementary row operations to ﬁnd A‘l. 0 1 1 100100 _1X1100100 _1X2100100 A A 1100102—>010—1103—>010—1 10 011001 011001 0011—11 NOWA 1 is on the right. 1 (b) Use row operations to ﬁnd the LU decomposition of A = ( 1 1 0 0 1 A2—1;<1( and from inverse of the row operation we get L = A%X1(Ig) = ( Is A singular? How many solutions does A1: = 0 have? (toprovv)det(A)=1(3-1—1-1)—2(1-1—1-3)=2+4=6. Thus A is non-singular; and Ax = 0 has unique solution x = O. O b—‘O \_/ OOH OI—‘O l—‘l—‘O ) is echelon-form so gives U; OI—‘H OI—‘O I—IDO V Problem 3: ODD—lI—l I—‘WN ,_.._.o (a) Find the determinant of A = ( (b) Use Cramer’s rule (determinants) to solve ( 2 3 > < x1 ) = ( 2 > First, det(A):2~2—3-3=4—9=—5. 2 3 2 2 Nextdet(5 2)—4—15——11anddet<3 5)—10—6—4 Sow1=15—1andx2=—§. Problem 4: Show that the set S consisting of 2X2 symmetric matrices (AT 2 A) forms a subspace of R2”. IfA,B E S then AT :A, BT : B. Then (A + B)T 2 AT + BT : A + B; so A + B is symmetric, and hence is also in S. Similarly ifA E S and c is scalar, then (cA)T : C(AT) : C(A) 2 CA, so that CA is symmetric, and hence is also in S. ...
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