Unformatted text preview: mn picture) are changed.
If z D 2 then x C y D 0 and x y D z give the point .1; 1; 2/. If z D 0 then
x C y D 6 and x y D 4 produce .5; 1; 0/. Halfway between those is .3; 0; 1/.
If x; y; z satisfy the ﬁrst two equations they also satisfy the third equation. The line
1
L of solutions contains v D .1; 1; 0/ and w D . 1 ; 1; 1 / and u D 2 v C 1 w and all
2
2
2
combinations c v C d w with c C d D 1.
Equation 1 C equation 2 equation 3 is now 0 D 4. Line misses plane; no solution.
Column 3 D Column 1 makes the matrix singular. Solutions .x; y; z/ D .1; 1; 0/ or
.0; 1; 1/ and you can add any multiple of . 1; 0; 1/; b D .4; 6; c/ needs c D 10 for
solvability (then b lies in the plane of the columns).
Four planes in 4dimensional space normally meet at a point. The solution to Ax D
.3; 3; 3; 2/ is x D .0; 0; 1; 2/ if A has columns .1; 0; 0; 0/; .1; 1; 0; 0/; .1; 1; 1; 0/,
.1; 1; 1; 1/. The equations are x C y C z C t D 3; y C z C t D 3; z C t D 3; t D 2.
(a) Ax D .18; 5; 0/ and (b) Ax D .3; 4; 5; 5/. Solutions to Exercises 8 10 Multiplying as linear combinations of the columns gives the same Ax . By rows or by
11
12
13
14 15
16 17 18 19 20 21 22 columns: 9 separate multiplications for 3 by 3.
Ax equals .14; 22/ and .0; 0/ and (9; 7/.
Ax equals .z; y; x/ and .0; 0; 0/ and (3; 3; 6/.
(a) x has n components and Ax has m components (b) Planes from each equation
in Ax D b are in ndimensional space, but the columns are in mdimensional space.
2x C 3y C z C 5t D 8 is Ax D b with the 1 by 4 matrix A D Œ 2 3 1 5 . The
solutions x ﬁll a 3D “plane” in 4 dimensions. It could be called a hyperplane.
10
01
(a) I D
(b) P D
01
10
01
1
0
ı
ı
2
90 rotation from R D
, 180 rotation from R D
D I.
10
0
1
"
#
"
#
010
001
P D 0 0 1 produces .y; z; x/ and Q D 1 0 0 recovers .x; y; z/. Q is the
100
010
inverse of P .
"
#
100
10
1 1 0 subtract the ﬁrst component from the second.
ED
and E D
11
001
"
#
#
"
100
100
0 1 0 , E v D .3; 4; 8/ and E 1 E v recovers
E D 0 1 0 and E 1 D
101
101
.3; 4; 5/.
00
10
projects onto the y axis.
projects onto the x axis and P2 D
P1 D
01
00
0
5
5
.
and P2 P1 v D
has P1 v D
vD
0
0
7
p
p
1
p2
p2 rotates all vectors by 45ı . The columns of R are the results from
RD
2
2
2
rotating .1; 0/ and .0; 1/!
"#
x
The dot product Ax D Œ 1 4 5 y D .1 by 3/.3 by 1/ is zero for points .x; y; z/
z
on a plane in three dimensions. The columns of A are onedimensional vectors. 23 A D Œ 1 2 I 3 4 and x D Œ 5 2 0 and b D Œ 1 7 0 . r D b A x prints as zero. 24 A v D Œ 3 4 5 0 and v 0 v D 50. But v A gives an error message from 3 by 1 times 3 by 3. 25 ones.4; 4/ ones.4; 1/ D Œ 4 4 4 4 0 ; B w D Œ 10 10 10 10 0 . 26 The row picture has two lines meeting at the solution (4; 2). The column picture will have 4.1; 1/ C 2. 2 ; 1/ D 4(column 1) C 2(column 2) D right side .0; 6/.
27 The row picture shows 2 planes in 3dimensional space. The column picture is in
2dimensional space. The solut...
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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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