Fsc Part 1 Mathematics (Complete Solution)

Q3
Solve the following systems by reducing their augumented matriced to the echelon form and reduce the echelon forms.

(i)  
x1 - 2x2 - 2x3 = -1
2x1 + 3x2 + x3 = 1
5x1 - 4x2 - 3x3 = 1
}
   
(ii)  
x + 2y + z = 2
2x + y + 2z = -1
2x + 3y - z = 9
}
(iii)  
x1 + 4x2 + 2x3 = 2
2x1 + x2 - 2x3 = 9
3x1 + 2x2 - 2x3 = 12
}

Solution

(i)  
x1 - 2x2 - 2x3 = -1
2x1 + 3x2 + x3 = 1
5x1 - 4x2 - 3x3 = 1
}
The augmented matrix of the system is
 
1 -2 -2 : -1
2 3 1 : 1
5 -4 -3 : -1
 
By adding (-2)R1 to R2 and (-5)R1 to R3, we get
 
1 -2 -2 : -1
2 3 1 : 1
5 -4 -3 : -1
 
R
~
 
1 -2 -2 : -1
(-2)1 + 2 (-2)(-2) + 3 (-2)(-2) + 1 : (-2)(-1) + 1
(-5)1 + 5 (-5)(-2) - 4 (-5)(-2) - 3 : (-5)(-1) + 1
 
By (-2)R1 + R2
and
(-5)R1 + R3
R
~
 
1 -2 -2 : -1
0 7 5 : 3
0 6 7 : 6
 
R
~
 
1 -2 -2 : -1
0 6 7 : 6
0 7 5 : 3
 
By R2 ↔ R3
R
~
 
1 -2 -2 : -1
0 1
7
6
: 1
0 7 5 : 3
 
BY
1
6
R2
R
~
 
1 -2 -2 : -1
0 1
7
6
: 1
0 0
- 19
6
: -4
 
By R3 + (-7)R2
R
~
 
1 -2 -2 : -1
0 1
7
6
: 1
0 0 1 :
24
19
 
By (-
6
19
) R3
................ (A)

The equivalent System in the echelon form is

x1 - 2x2 - 2x3 = -1
         (1)
x2 +
7
6
x3 = 1
         (2)
x3 =
24
19
         (3)

Put x3 =
24
19
in (2)
x2 +
7
6
(
24
19
) = 1
x2 +
28
19
= 1
x2 = 1 -
28
19
x2 =
19 - 28
19
x2 = -
9
19
Substituting     x2 = -
9
19
and x3 =
24
19
    in (1)
x1 - 2(-
9
19
) - 2(
24
19
) = -1
x1 +
18
19
-
48
19
= -1
x1 = -1 -
18
19
+
48
19
x1 =
-19 - 18 + 48
19
x1 =
11
19

Reduce the matrix in (A) to reduced echelon form
R
~
 
1 -2 -2 : -1
0 1
7
6
: 1
0 0 1 :
24
19
 
 
R
~
 
1 2(1) - 2
2(
7
6
)
- 2
: 2(1) - 1
0 1
7
6
: 1
0 0 1 :
24
19
 
By R1 + 2R2
 
R
~
 
1 0
1
3
: 1
0 1
7
6
: 1
0 0 1 :
24
19
 
 
R
~
 
1 0 0 :
11
19
0 1
7
6
: 1
0 0 1 :
24
19
 
By R1 + (-
1
3
) R3
R
~
 
1 0 0 :
11
19
0 1 0 : -
9
19
0 0 1 :
24
19
 
By R2 + (-
7
6
) R3

The Reduced echelon form is
x1 =
11
19
x2 = -
9
19
x3 =
24
19

(ii)  
x + 2y + z = 2
2x + y + 2z = -1
2x + 3y - z = 9
}
The augmented matrix of the system is
 
1 2 1 : 2
2 1 2 : -1
2 3 -1 : 9
 
By adding (-2)R1 to R2 and (-2)R1 to R3, we get
 
1 2 1 : 2
2 1 2 : -1
2 3 -1 : 9
 
R
~
 
1 2 1 : 2
2 + (-2)1 1 + (-2)2 2 + (-2)1 : -1 + (-2)2
2 + (-2)1 3 + (-2)2 -1 + (-2)1 : 9 + (-2)2
 
By R2 + (-2)R1
and
R3 + (-2)R1
R
~
 
1 2 1 : 2
0 -3 0 : -5
0 -1 -3 : 5
 

Now replace R2 with R3
R
~
 
1 2 1 : 2
0 -1 -3 : 5
0 -3 0 : -5
 
By R2 ↔ R3

Multiply R2 by (-1)
R
~
 
1 2 1 : 2
0 1 3 : -5
0 -3 0 : -5
 
BY (-1)R2

By adding (3)R2 to R3
R
~
 
1 2 1 : 2
0 1 3 : -5
0 0 9 : -20
 
By R3 + 3R2
R
~
 
1 2 1 : 2
0 1 3 : -5
0 0 1 :
-20
9
 
By (
1
9
) R3
...................... (A)

The equivalent System in the echelon form is

x + 2y + z = 2
         (1)
y + 3z = -5
         (2)
z = -
20
9
         (3)

Put z = -
20
9
in (2)
y + 3(-
20
9
) = -5
y -
20
3
= -5
y = -5 +
20
3
y =
-15 + 20
3
y =
5
3
Substituting     y =
5
3
and z = -
20
9
    in (1)
x + 2(
5
3
) + (-
20
9
) = 2
x +
10
3
-
20
9
= 2
x = 2 -
10
3
+
20
9
x =
18 - 30 + 20
9
x =
8
9
Hence       x =
8
9
, y =
5
3
, z = -
20
9

Reduce the matrix in (A) to reduced echelon form
R
~
 
1 2 1 : 2
0 1 3 : -5
0 0 1 :
-20
9
 

Adding (-2)R2 to RR1
R
~
 
1 0 -5 : 12
0 1 3 : -5
0 0 1 :
-20
9
 
By R1 - 2R2
 
R
~
 
1 0 0 :
8
9
0 1 3 : -5
0 0 1 :
-20
9
 
By R1 + 5 R3
 
R
~
 
1 0 0 :
8
9
0 1 0 :
5
3
0 0 1 :
-20
9
 
By R2 - 3R3

The Reduced echelon form is
x =
8
9
, y =
5
3
, z = -
20
9

(iii)  
x1 + 4x2 + 2x3 = 2
2x1 + x2 - 2x3 = 9
3x1 + 2x2 - 2x3 = 12
}
The augmented matrix of the system is
 
1 4 2 : 2
2 1 -2 : 9
3 2 -2 : 12
 
By adding (-2)R1 to R2 and (-3)R1 to R3, we get
 
1 4 2 : 2
2 1 -2 : 9
3 2 -2 : 12
 
R
~
 
1 4 2 : 2
2 + (-2)1 1 + (-2)4 -2 + (-2)2 : 9 + (-2)2
3 + (-3)1 2 + (-3)4 -2 + (-3)2 : 12 + (-3)2
 
By R2 + (-2)R1
and
R3 + (-5)R1
R
~
 
1 4 2 : 2
0 -7 -6 : 5
0 -10 -8 : 6
 
R
~
 
1 4 2 : 2
0 -10 -8 : 6
0 -7 -6 : 5
 
By R2 ↔ R3

Multiply R2 by (-
1
10
)
R
~
 
1 4 2 : 2
0 1
4
5
: -
3
5
0 -7 -6 : 5
 
BY (-
1
10
)R2

Adding 7R2 to R3
R
~
 
1 4 2 : 2
0 1
4
5
: -
3
5
0 0 -
2
5
:
4
5
 
By R3 + 7R2
R
~
 
1 4 2 : 2
0 1
4
5
: -
3
5
0 0 1 : -2
 
By (-
5
2
) R3
...................... (A)

The equivalent System in the echelon form is

x1 + 4x2 + 2x3 = 2
         (1)
x2 +
4
5
x3 = -
3
5
         (2)
x3 = -2
         (3)

Put x3 = -2 in (2)
x2 +
4
5
( -2 ) = -
3
5
x2 -
8
5
= -
3
5
x2 = -
3
5
+
8
5
x2 =
-3 + 8
5
x2 = 1
Substituting     x2 = 1 and x3 = -2     in (1)
x1 + 4(1) + 2(-2) = 2
x1 + 4 - 4 = 2
x1 = 2
Hence       x1 = 2,     x2 = 1,     x3 = -2

Reduce the matrix in (A) to reduced echelon form
R
~
 
1 4 2 : 2
0 1
4
5
: -
3
5
0 0 1 : -2
 
 
R
~
 
1 0 -
6
5
:
22
5
0 1
4
5
: -
3
5
0 0 1 : -2
 
By R1 + (-4)R2
 
R
~
 
1 0 0 : 2
0 1
4
5
: -
3
5
0 0 1 : -2
 
By R1 + (
6
5
) R3
R
~
 
1 0 0 : 2
0 1 0 : 1
0 0 1 : -2
 
By R2 + (-
4
5
) R3

The Reduced echelon form is
x1 = 2
x2 = 1
x3 = -2

Other Topics

;