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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(31—11)—2(11—13)=2+4=6.
Thus A is nonsingular; and Ax = 0 has unique solution x = O. O
b—‘O
\_/ OOH
OI—‘O
l—‘l—‘O ) is echelonform 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—33=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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 Spring '08
 Smith

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