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Unformatted text preview: e. Then there will be inﬁnitely many solutions (two parallel lines become one
single line).
The equation y D 1 comes from elimination (subtract x C y D 5 from x C 2y D 6).
Then x D 4 and 5x 4y D c D 16.
1
(a) Another solution is 2 .x C X; y C Y; z C Z/. (b) If 25 planes meet at two points,
they meet along the whole line through those two points.
Elimination leads to an upper triangular system; then comes back substitution.
2x C 3y C z D 8
xD2
y C 3z D 4 gives y D 1 If a zero is at the start of row 2 or 3,
8z D 8
z D 1 that avoids a row operation.
2x 3y D 3
2x 3y
D3
2x 3y D 3
xD3
y C z D 1 and
4x 5y C z D 7 gives
y C z D 1 and y D 1
2x
y 3z D 5
5z D 0
zD0
2y C 3z D 2
Subtract 2 row 1 from row 2, subtract 1 row 1 from row 3, subtract 2 row 2 from
row 3
Subtract 2 times row 1 from row 2 to reach .d 10/y z D 2. Equation (3) is y z D 3.
If d D 10 exchange rows 2 and 3. If d D 11 the system becomes singular.
The second pivot position will contain 2 b . If b D 2 we exchange with row 3. If
b D 1 (singular case) the second equation is y z D 0. A solution is .1; 1; 1/.
0x C 0y C 2z D 4
Exchange
0x C 3y C 4z D 4
Example of
x C 2y C 2z D 5
but then
x C 2y C 2z D 5
(a) 2 exchanges
(b)
0x C 3y C 4z D 6
break down 0x C 3y C 4z D 6
(exchange 1 and 2, then 2 and 3)
(rows 1 and 3 are not consistent)
If row 1 D row 2, then row 2 is zero after the ﬁrst step; exchange the zero row with row
3 and there is no third pivot. If column 2 D column 1, then column 2 has no pivot.
Example x C 2y C 3z D 0, 4x C 8y C 12z D 0, 5x C 10y C 15z D 0 has 9 different
coefﬁcients but rows 2 and 3 become 0 D 0: inﬁnitely many solutions.
Row 2 becomes 3y 4z D 5, then row 3 becomes .q C 4/z D t 5. If q D 4 the
system is singular—no third pivot. Then if t D 5 the third equation is 0 D 0. Choosing
z D 1 the equation 3y 4z D 5 gives y D 3 and equation 1 gives x D 9.
Singular if row 3 is a combination of rows 1 and 2. From the end view, the three planes
form a triangle. This happens if rows 1 C 2 D row 3 on the left side but not the right
side: x C y C z D 0, x 2y z D 1, 2x y D 4. No parallel planes but still no solution.
345 3 4 5 21 (a) Pivots 2; 2 ; 3 ; 4 in the equations 2x C y D 0; 2 y C z D 0; 3 z C t D 0; 4 t D 5 after elimination. Back substitution gives t D 4; z D 3; y D 2; x D 1. (b) If
the offdiagonal entries change from C1 to 1, the pivots are the same. The solution is
.1; 2; 3; 4/ instead of . 1; 2; 3; 4/.
6 22 The ﬁfth pivot is 5 for both matrices (1’s or
nC1
.
n 1’s off the diagonal). The nth pivot is Solutions to Exercises 11 23 If ordinary elimination leads to x C y D 1 and 2y D 3, the original second equation 24
25
26 27
28
29 30
31
32 could be 2y C `.x C y/ D 3 C ` for any `. Then ` will be the multiplier to reach
2y D 3.
a2
Elimination fails on
if a D 2 or a D 0.
aa
a D 2 (equal columns), a D 4 (equal rows), a D 0 (zero column).
Solvable for s D 10 (add the two pairs of equations to get a C b C c C d on the left sides,
12 and 2 C s on the right sides). The four equations for a; b; c; d are singular! Two
2
3
2
3
1100
1
100
13
04
1 1 07
61 0 1 07
60
solutio...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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