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Solving Linear Equations and Matrices

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Added on  2023/06/05

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This article explains how to solve linear equations and matrices using Gaussian elimination. It includes step-by-step solutions and examples for finding the rank of a matrix and solving systems of equations. Desklib provides study material, solved assignments, essays, and dissertations on this topic.

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Solution
Q1)
L1
X -1 = y-2 = z-3
L2
X + 1 = y2
2 = z1
2
P containing L1
X – 1 = y – 2 = z = 3
X =1, y = 2, z = 3
(1, 2, 3) lies on plane P
Taking R = (1, 2, 3)
x+1
1 = y2
2 = z1
2
(-1, 2, 1) lies on the L2
Write S = (-1, 2,1) vector <1, 2, 2>
RS=¿(-1, 2, 1) – (-1, 2, 3)
RS=¿(21, 0, -2)
RSV =2, 0 ,2>¿<1 , 2, 2>¿
| i j k
2 0 2
1 2 2 |
normal = i(0 + 40 – j(-4 + 2) + k(-4-0)
normal vector = 4i + 2j – 4k

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= (-2 + u-w, 2u –w, -2+ 2u – w)
Normal vector=0
| i j k
2+uw 2 uw 2+2uw
4 2 4 | = 0
I(-80 + 4w +4-4u + 2w)-j(+8-4u+ 4w+ 8-8u+4w) + k(-4 + 2u-2w-8u + 4w)
It is known that
a(x – x1) + b(y –y1) + z(z –z1) = 0
(a, b, c) are normal vectors
*x1, y1, z1) are on the plane P
(a, b, c) = (4, 2, -4)
(x, y, z) = (1, 2, 3)
4(x-1) + 2(y-2) – 4(z-3)= 0
4x + 2y – 4z = -4
a) Plane P
4x + 2y -4z + 4 = 0
b) x1
1 = y 2
1 = z3
1
X = 1 +w, y = 2 + w. z= 3 + w
A = (1 + w, z +w, 3 = W)
L2 = x1
1 = y 2
2 = z3
2
X = -1 + U, Y = 2 + U, Z = 1 +2U
B = (-1 + u, 2 + 2u, 1 + 2u)
AB= (1+ u ,2+2 u , 1+ 2u )(1+ w , 2+ w , 3+2)
i(-12u + 6w + 4) – j(-12u +8w +16) + k(-6u + 2w -4) =0
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-12u + 6w + 4 = 0
-12u +8w +16 = 0
-6u + 2w -4 = 0
W = -6
Substituting to, -12u + 6w + 4 = 0
U = -8/3
A = (1 + w, 2 + w, 3 + w)
= (1-6, 2-6, 3-6)
A = (-5, -4, -3)
B = (-1-8/3, 2-16/3, 1-16/3)
B = (-11/3, -10.3, -13, 3)
Q2) Use Gaussian elimination
a) Writing in matrix form
( a b
0 1
1 2
c
2
1
d
1
0
1 2 3
2 3 1
4
1 | k
3
2
6
0 )
R1 ↔ R2 (interchange the 1 and 2 rows)
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( a b
1 2
0 1
c
1
2
d
0
1
1 2 3
2 3 1
4
1
| k
2
3
6
0 )
R3 - 1 R1 → R3 (multiply 1 row by 1 and subtract it from 3 row); R4 - 2 R1 → R4 (multiply 1
row by 2 and subtract it from 4 row)
( a b
1 2
0 1
c
1
2
d
0
0
0 0 4
0 1 1
4
1
| k
2
3
8
4 )
R1 - 2 R2 → R1 (multiply 2 row by 2 and subtract it from 1 row); R4 + 1 R2 → R4 (multiply 2
row by 1 and add it to 4 row)
(a b
1 0
0 1
c
3
2
d
2
1
0 0 4
0 0 1
4
2
| k
8
3
8
7 )
R3 / 4 → R3 (divide the 3 row by 4)
( a b
1 0
0 1
c
3
2
d
2
1
0 0 1
0 0 4
1
2 | k
8
3
2
7 )

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R1 - 3 R3 → R1 (multiply 3 row by 3 and subtract it from 1 row); R2 + 2 R3 → R2 (multiply 3
row by 2 and add it to 2 row); R4 + 1 R3 → R4 (multiply 3 row by 1 and add it to 4 row)
( a b
1 0
0 1
c
0
0
d
5
3
0 0 1
0 0 0
1
3
| k
14
7
2
9 )
R4 / 3 → R4 (divide the 4 row by 3)
( a b
1 0
0 1
c
0
0
d
5
3
0 0 1
0 0 0
1
1
| k
14
7
2
3 )
R1 + 5 R4 → R1 (multiply 4 row by 5 and add it to 1 row); R2 - 3 R4 → R2 (multiply 4 row by
3 and subtract it from 2 row); R3 - 1 R4 → R3 (multiply 4 row by 1 and subtract it from 3 row)
( a b
1 0
0 1
c
0
0
d
0
0
0 0 1
0 0 0
0
1
| k
1
2
1
3 )
a = -1
b = 2
c = 1
d = -3
b)
(2 3 4
1 2 3
3 5 7|5
4
9 )
R1 / 2 → R1 (divide the 1 row by 2)
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(1 1.5 2
1 2 3
3 5 7|2.5
4
9 )
R2 - 1 R1 → R2 (multiply 1 row by 1 and subtract it from 2 row); R3 - 3 R1 → R3 (multiply 1
row by 3 and subtract it from 3 row)
(1 1.5 2
0 0.5 1
0 0.5 1|2.5
1.5
1.5 )
R2 / 0.5 → R2 (divide the 2 row by 0.5)
(1 1.5 2
0 1 2
0 0.5 1|2.5
3
1.5 )
R1 - 1.5 R2 → R1 (multiply 2 row by 1.5 and subtract it from 1 row); R3 - 0.5 R2 → R3
(multiply 2 row by 0.5 and subtract it from 3 row)
(1 0 1
0 1 2
0 0 0 |2
3
0 )
X1 – X3 = -2
X2 + X3 = 3
Q3)
3x + 2y + z = 1
2x + y + 6z = 3
4x + 4y + z = 6
Write the equation in the form of Ax = B
[3 2 1
2 1 6
4 4 1 ][ x
y
z ]=
[1
3
6 ]
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Matrix [A|B]
(3 2 1
2 1 6
4 4 1|1
3
6 )
Write the matrix in row form
R1 R3
[ 4 4 1
2 1 6
3 2 1
6
3
1 ]
R2 R2 -1/2 *R1
[ 4 4 1
0 1 11
2
2 2 1
6
0
1 ]
R3 R3 -3/4 *R1
[ 4 4 1
0 1 11
2
0 1 1
4
6
0
7
2 ]R3 R3 – 1*R2
[4 4 1
0 1 11
2
0 0 21
4
6
0
7
2 ]So,
Rank of A = Rank of (AB) = number of variables

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[4 4 1
0 1 11
2
0 0 21
4 ] [ x
y
z ]=
[ 6
0
7
2 ]
4x + 4y + z = 6
-y + 11/2y = 0
-21/4z = -7/2
Z = 74
221
= 2/3
Y = 11/2*z
Y = 11/2 * 2/3
= 11/3
4x + 4*11/3 + 2/3 = 6
X = -7/3
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