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

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

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

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