Algebra Assignment: Gaussian Elimination and Linear Equations Solution

This assignment provides detailed solutions to problems involving linear equations and Gaussian elimination. The first question focuses on finding the equation of a plane containing a line and a point, utilizing vector operations and normal vectors. The second question demonstrates the application of Gaussian elimination to solve systems of linear equations, including row operations and matrix transformations. The third question involves solving a system of three linear equations using matrix representation and row reduction techniques, determining the rank of the coefficient matrix and finding the values of the variables. Desklib offers a wealth of similar solved assignments and past papers to aid students in their studies.

Solution
Q1)
L1
X -1 = y-2 = z-3
L2
X + 1 = y−2
2 = z−1
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 = y−2
2 = z−1
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)⃗
RS∗⃗V =←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

= (-2 + u-w, 2u –w, -2+ 2u – w)
Normal vector=0
| i j k
−2+u−w 2 u−w −2+2u−w
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) x−1
1 = y −2
1 = z−3
1
X = 1 +w, y = 2 + w. z= 3 + w
A = (1 + w, z +w, 3 = W)
L2 = x−1
1 = y −2
2 = z−3
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

-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)

( 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 )

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)

(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 ]

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

[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 = 7∗4
2∗21
= 2/3
Y = 11/2*z
Y = 11/2 * 2/3
= 11/3
4x + 4*11/3 + 2/3 = 6
X = -7/3