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Faculty of Engineering and Applied Science
2000 Fall
[Note: some browsers do not display symbols correctly. Read
“-” as “-” (minus) and
“ú” as “ú” (set of real numbers).]A set of instructions on how to use Dr. George’s row reduction computer program is available at this link.
|
For each of the following systems of linear equations |
| (i) | rewrite as a matrix equation |
| (ii) | identify the coefficient matrix, the matrix of constants, and the augmented matrix |
| (iii) | use elementary operations to reduce the augmented matrix to either row echelon form, or reduced row echelon form |
| (iv) | write the solution, if any, of the system of linear equations |
| x | + | 3y | - | 2z | = | 8 |
| 3y | - | 5z | = | 11 | ||
| 2x | + | 5y | - | 8z | = | 19 |
The matrix equation is Ax = b, where
[ 1 3 -2 ] [ x ] [ 8 ]
A = [ 0 3 -5 ] , x = [ y ] , b = [ 11 ]
[ 2 5 -8 ] [ z ] [ 19 ]
A is the coefficient matrix, b is the matrix of [right side] constants, and the augmented matrix is
[ 1 3 -2 | 8 ]
[A|b] = [ 0 3 -5 | 11 ]
[ 2 5 -8 | 19 ]
Using the row reduction program on this linear system, we obtain
Program to handle the Arithmetic in the Row Reduction of a Linear System VisualBASIC version (c)1999, Glyn George. Linear System: Row 1: 1/1 3/1 -2/1 8/1 Row 2: 0/1 3/1 -5/1 11/1 Row 3: 2/1 5/1 -8/1 19/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 1 from row 3 . Linear System: Row 1: 1/1 3/1 -2/1 8/1 Row 2: 0/1 3/1 -5/1 11/1 Row 3: 0/1 -1/1 -4/1 3/1 ----------------------------------------------------------------------------- Selected row operation: Swap rows 3 and 2 . Linear System: Row 1: 1/1 3/1 -2/1 8/1 Row 2: 0/1 -1/1 -4/1 3/1 Row 3: 0/1 3/1 -5/1 11/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 2 by -1 / 1 . Linear System: Row 1: 1/1 3/1 -2/1 8/1 Row 2: 0/1 1/1 4/1 -3/1 Row 3: 0/1 3/1 -5/1 11/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 3 / 1 times row 2 from row 3 . Linear System: Row 1: 1/1 3/1 -2/1 8/1 Row 2: 0/1 1/1 4/1 -3/1 Row 3: 0/1 0/1 -17/1 20/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 3 by -17 / 1 . Linear System: Row 1: 1/1 3/1 -2/1 8/1 Row 2: 0/1 1/1 4/1 -3/1 Row 3: 0/1 0/1 1/1 -20/17 -----------------------------------------------------------------------------[Note: One can stop the row reduction at this row echelon form and use back substitution to obtain the solution. Here we continue to reduced row echelon form.]
----------------------------------------------------------------------------- Selected row operation: Subtract 3 / 1 times row 2 from row 1 . Linear System: Row 1: 1/1 0/1 -14/1 17/1 Row 2: 0/1 1/1 4/1 -3/1 Row 3: 0/1 0/1 1/1 -20/17 ----------------------------------------------------------------------------- Selected row operation: Subtract -14 / 1 times row 3 from row 1 . Linear System: Row 1: 1/1 0/1 0/1 9/17 Row 2: 0/1 1/1 4/1 -3/1 Row 3: 0/1 0/1 1/1 -20/17 ----------------------------------------------------------------------------- Selected row operation: Subtract 4 / 1 times row 3 from row 2 . Linear System: Row 1: 1/1 0/1 0/1 9/17 Row 2: 0/1 1/1 0/1 29/17 Row 3: 0/1 0/1 1/1 -20/17 ----------------------------------------------------------------------------- Program execution terminated.
The unique solution is therefore
(x, y, z) =
(9/17, 29/17, -20/17) =
(1/17) × (9, 29, -20).
| -3x | + | 6y | + | 16z | = | 36 |
| x | - | 2y | - | 5z | = | -11 |
| 2x | - | 3y | - | 8z | = | 15 |
The matrix equation is Ax = b, where
[-3 6 16 ] [ x ] [ 36 ]
A = [ 1 -2 -5 ] , x = [ y ] , b = [-11 ]
[ 2 -3 -8 ] [ z ] [ 15 ]
A is the coefficient matrix, b is the matrix of [right side] constants, and the augmented matrix is
[-3 6 16 | 36 ]
[A|b] = [ 1 -2 -5 |-11 ]
[ 2 -3 -8 | 15 ]
Using the row reduction program on this linear system, we obtain
Program to handle the Arithmetic in the Row Reduction of a Linear System VisualBASIC version (c)1999, Glyn George. Linear System: Row 1: -3/1 6/1 16/1 36/1 Row 2: 1/1 -2/1 -5/1 -11/1 Row 3: 2/1 -3/1 -8/1 15/1 ----------------------------------------------------------------------------- Selected row operation: Swap rows 1 and 2 . Linear System: Row 1: 1/1 -2/1 -5/1 -11/1 Row 2: -3/1 6/1 16/1 36/1 Row 3: 2/1 -3/1 -8/1 15/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -3 / 1 times row 1 from row 2 . Linear System: Row 1: 1/1 -2/1 -5/1 -11/1 Row 2: 0/1 0/1 1/1 3/1 Row 3: 2/1 -3/1 -8/1 15/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 1 from row 3 . Linear System: Row 1: 1/1 -2/1 -5/1 -11/1 Row 2: 0/1 0/1 1/1 3/1 Row 3: 0/1 1/1 2/1 37/1 ----------------------------------------------------------------------------- Selected row operation: Swap rows 3 and 2 . Linear System: Row 1: 1/1 -2/1 -5/1 -11/1 Row 2: 0/1 1/1 2/1 37/1 Row 3: 0/1 0/1 1/1 3/1 -----------------------------------------------------------------------------[Note: One can stop the row reduction at this row echelon form and use back substitution to obtain the solution. Here we continue to reduced row echelon form.]
----------------------------------------------------------------------------- Selected row operation: Subtract -2 / 1 times row 2 from row 1 . Linear System: Row 1: 1/1 0/1 -1/1 63/1 Row 2: 0/1 1/1 2/1 37/1 Row 3: 0/1 0/1 1/1 3/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -1 / 1 times row 3 from row 1 . Linear System: Row 1: 1/1 0/1 0/1 66/1 Row 2: 0/1 1/1 2/1 37/1 Row 3: 0/1 0/1 1/1 3/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 3 from row 2 . Linear System: Row 1: 1/1 0/1 0/1 66/1 Row 2: 0/1 1/1 0/1 31/1 Row 3: 0/1 0/1 1/1 3/1 ----------------------------------------------------------------------------- Program execution terminated.
The unique solution is therefore
(x, y, z) =
(66, 31, 3).
| 3x | + | 2y | - | z | = | 4 |
| 2x | + | 5y | + | z | = | 10 |
| 4x | + | 9y | - | z | = | -21 |
The matrix equation is Ax = b, where
[ 3 2 -1 ] [ x ] [ 4 ]
A = [ 2 5 1 ] , x = [ y ] , b = [ 10 ]
[ 4 9 -1 ] [ z ] [-21 ]
A is the coefficient matrix, b is the matrix of [right side] constants, and the augmented matrix is
[ 3 2 -1 | 4 ]
[A|b] = [ 2 5 1 | 10 ]
[ 4 9 -1 |-21 ]
Using the row reduction program on this linear system, we obtain
Program to handle the Arithmetic in the Row Reduction of a Linear System VisualBASIC version (c)1999, Glyn George. Linear System: Row 1: 3/1 2/1 -1/1 4/1 Row 2: 2/1 5/1 1/1 10/1 Row 3: 4/1 9/1 -1/1 -21/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 1 by 3 / 1 . Linear System: Row 1: 1/1 2/3 -1/3 4/3 Row 2: 2/1 5/1 1/1 10/1 Row 3: 4/1 9/1 -1/1 -21/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 1 from row 2 . Linear System: Row 1: 1/1 2/3 -1/3 4/3 Row 2: 0/1 11/3 5/3 22/3 Row 3: 4/1 9/1 -1/1 -21/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 4 / 1 times row 1 from row 3 . Linear System: Row 1: 1/1 2/3 -1/3 4/3 Row 2: 0/1 11/3 5/3 22/3 Row 3: 0/1 19/3 1/3 -79/3 ----------------------------------------------------------------------------- Selected row operation: Divide row 2 by 11 / 3 . Linear System: Row 1: 1/1 2/3 -1/3 4/3 Row 2: 0/1 1/1 5/11 2/1 Row 3: 0/1 19/3 1/3 -79/3 ----------------------------------------------------------------------------- Selected row operation: Subtract 19 / 3 times row 2 from row 3 . Linear System: Row 1: 1/1 2/3 -1/3 4/3 Row 2: 0/1 1/1 5/11 2/1 Row 3: 0/1 0/1 -28/11 -39/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 3 by -28 / 11 . Linear System: Row 1: 1/1 2/3 -1/3 4/3 Row 2: 0/1 1/1 5/11 2/1 Row 3: 0/1 0/1 1/1 429/28 -----------------------------------------------------------------------------[Note: One can stop the row reduction at this row echelon form and use back substitution to obtain the solution. Here we continue to reduced row echelon form.]
----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 3 times row 2 from row 1 . Linear System: Row 1: 1/1 0/1 -7/11 0/1 Row 2: 0/1 1/1 5/11 2/1 Row 3: 0/1 0/1 1/1 429/28 ----------------------------------------------------------------------------- Selected row operation: Subtract -7 / 11 times row 3 from row 1 . Linear System: Row 1: 1/1 0/1 0/1 39/4 Row 2: 0/1 1/1 5/11 2/1 Row 3: 0/1 0/1 1/1 429/28 ----------------------------------------------------------------------------- Selected row operation: Subtract 5 / 11 times row 3 from row 2 . Linear System: Row 1: 1/1 0/1 0/1 39/4 Row 2: 0/1 1/1 0/1 -139/28 Row 3: 0/1 0/1 1/1 429/28 ----------------------------------------------------------------------------- Program execution terminated.
The unique solution is therefore
(x, y, z) =
(39/4, -139/28, 429/28) =
(1/28) × (273, -139, 429).
| x | + | 2y | - | 3z | = | -5 | ||
| 2x | + | 4y | - | 6z | + | w | = | -8 |
| 6x | + | 13y | - | 17z | + | 4w | = | -21 |
The matrix equation is Ax = b, where
[ 1 2 -3 0 ] [ x ] [ -5 ]
A = [ 2 4 -6 1 ] , x = [ y ] , b = [ -8 ]
[ 6 13 -17 4 ] [ z ] [-21 ]
[ w ]
A is the coefficient matrix, b is the matrix of [right side] constants, and the augmented matrix is
[ 1 2 -3 0 | -5 ]
[A|b] = [ 2 4 -6 1 | -8 ]
[ 6 13 -17 4 | -21 ]
Using the row reduction program on this linear system, we obtain
Program to handle the Arithmetic in the Row Reduction of a Linear System VisualBASIC version (c)1999, Glyn George. Linear System: Row 1: 1/1 2/1 -3/1 0/1 -5/1 Row 2: 2/1 4/1 -6/1 1/1 -8/1 Row 3: 6/1 13/1 -17/1 4/1 -21/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 1 from row 2 . Linear System: Row 1: 1/1 2/1 -3/1 0/1 -5/1 Row 2: 0/1 0/1 0/1 1/1 2/1 Row 3: 6/1 13/1 -17/1 4/1 -21/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 6 / 1 times row 1 from row 3 . Linear System: Row 1: 1/1 2/1 -3/1 0/1 -5/1 Row 2: 0/1 0/1 0/1 1/1 2/1 Row 3: 0/1 1/1 1/1 4/1 9/1 ----------------------------------------------------------------------------- Selected row operation: Swap rows 3 and 2 . Linear System: Row 1: 1/1 2/1 -3/1 0/1 -5/1 Row 2: 0/1 1/1 1/1 4/1 9/1 Row 3: 0/1 0/1 0/1 1/1 2/1 -----------------------------------------------------------------------------[Note: One can stop the row reduction at this row echelon form and use back substitution to obtain the solution. Here we continue to reduced row echelon form.]
----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 2 from row 1 . Linear System: Row 1: 1/1 0/1 -5/1 -8/1 -23/1 Row 2: 0/1 1/1 1/1 4/1 9/1 Row 3: 0/1 0/1 0/1 1/1 2/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -8 / 1 times row 3 from row 1 . Linear System: Row 1: 1/1 0/1 -5/1 0/1 -7/1 Row 2: 0/1 1/1 1/1 4/1 9/1 Row 3: 0/1 0/1 0/1 1/1 2/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 4 / 1 times row 3 from row 2 . Linear System: Row 1: 1/1 0/1 -5/1 0/1 -7/1 Row 2: 0/1 1/1 1/1 0/1 1/1 Row 3: 0/1 0/1 0/1 1/1 2/1 ----------------------------------------------------------------------------- Program execution terminated.
The one parameter family of solutions is
therefore
(x, y, z, w)
= (5t - 7,
1 - t,
t, 2) =
(-7, 1, 0, 2)
+ t (5, -1, 1, 0).
Geometrically, this is a line in
ú4.
| x | + | 2y | - | 3w | = | 4 | ||
| 2x | + | 4y | + | z | - | 6w | = | -8 |
| 3x | + | 6y | - | 2z | - | 2w | = | 12 |
The matrix equation is Ax = b, where
[ 1 2 0 -3 ] [ x ] [ 4 ]
A = [ 2 4 1 -6 ] , x = [ y ] , b = [ -8 ]
[ 3 6 -2 -2 ] [ z ] [ 12 ]
[ w ]
A is the coefficient matrix, b is the matrix of [right side] constants, and the augmented matrix is
[ 1 2 0 -3 | 4 ]
[A|b] = [ 2 4 1 -6 | -8 ]
[ 3 6 -2 -2 | 12 ]
Using the row reduction program on this linear system, we obtain
Program to handle the Arithmetic in the Row Reduction of a Linear System VisualBASIC version (c)1999, Glyn George. Linear System: Row 1: 1/1 2/1 0/1 -3/1 4/1 Row 2: 2/1 4/1 1/1 -6/1 -8/1 Row 3: 3/1 6/1 -2/1 -2/1 12/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 1 from row 2 . Linear System: Row 1: 1/1 2/1 0/1 -3/1 4/1 Row 2: 0/1 0/1 1/1 0/1 -16/1 Row 3: 3/1 6/1 -2/1 -2/1 12/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 3 / 1 times row 1 from row 3 . Linear System: Row 1: 1/1 2/1 0/1 -3/1 4/1 Row 2: 0/1 0/1 1/1 0/1 -16/1 Row 3: 0/1 0/1 -2/1 7/1 0/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -2 / 1 times row 2 from row 3 . Linear System: Row 1: 1/1 2/1 0/1 -3/1 4/1 Row 2: 0/1 0/1 1/1 0/1 -16/1 Row 3: 0/1 0/1 0/1 7/1 -32/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 3 by 7 / 1 . Linear System: Row 1: 1/1 2/1 0/1 -3/1 4/1 Row 2: 0/1 0/1 1/1 0/1 -16/1 Row 3: 0/1 0/1 0/1 1/1 -32/7 -----------------------------------------------------------------------------[Note: One can stop the row reduction at this row echelon form and use back substitution to obtain the solution. Here we continue to reduced row echelon form.]
----------------------------------------------------------------------------- Selected row operation: Subtract -3 / 1 times row 3 from row 1 . Linear System: Row 1: 1/1 2/1 0/1 0/1 -68/7 Row 2: 0/1 0/1 1/1 0/1 -16/1 Row 3: 0/1 0/1 0/1 1/1 -32/7 ----------------------------------------------------------------------------- Program execution terminated.
The one parameter family of solutions is
therefore
(x, y, z, w)
= (-68/7 - 2t,
t, -16,
-32/7)
= (-68/7, 0, -16,
-32/7)
+ t (-2, 1, 0, 0)
= (1/7) × (-68, 0,
-112, -32)
+ t (-2, 1, 0, 0) .
Geometrically, this is a line in
ú4.
| x1 | - | 3x3 | - | x5 | = | 1 | ||||
| x2 | - | 2x3 | - | x4 | = | -2 | ||||
| x1 | + | x2 | - | 5x3 | - | x4 | - | x5 | = | -1 |
| 2x1 | - | 3x2 | + | 2x4 | = | 4 |
The matrix equation is Ax = b, where
[ 1 0 -3 0 -1 ] [ x_1 ] [ 1 ]
A = [ 0 1 -2 -1 0 ] [ x_2 ] , b = [ -2 ]
[ 1 1 -5 -1 -1 ] , x = [ x_3 ] [ -1 ]
[ 2 -3 0 2 0 ] [ x_4 ] [ 4 ]
[ x_5 ]
A is the coefficient matrix, b is the matrix of [right side] constants, and the augmented matrix is
[ 1 0 -3 0 -1 | 1 ]
[A|b] = [ 0 1 -2 -1 0 | -2 ]
[ 1 1 -5 -1 -1 | -1 ]
[ 2 -3 0 2 0 | 4 ]
Using the row reduction program on this linear system, we obtain
Program to handle the Arithmetic in the Row Reduction of a Linear System VisualBASIC version (c)1999, Glyn George. Linear System: Row 1: 1/1 0/1 -3/1 0/1 -1/1 1/1 Row 2: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 3: 1/1 1/1 -5/1 -1/1 -1/1 -1/1 Row 4: 2/1 -3/1 0/1 2/1 0/1 4/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 1 / 1 times row 1 from row 3 . Linear System: Row 1: 1/1 0/1 -3/1 0/1 -1/1 1/1 Row 2: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 3: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 4: 2/1 -3/1 0/1 2/1 0/1 4/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 1 from row 4 . Linear System: Row 1: 1/1 0/1 -3/1 0/1 -1/1 1/1 Row 2: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 3: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 4: 0/1 -3/1 6/1 2/1 2/1 2/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 1 / 1 times row 2 from row 3 . Linear System: Row 1: 1/1 0/1 -3/1 0/1 -1/1 1/1 Row 2: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 3: 0/1 0/1 0/1 0/1 0/1 0/1 Row 4: 0/1 -3/1 6/1 2/1 2/1 2/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -3 / 1 times row 2 from row 4 . Linear System: Row 1: 1/1 0/1 -3/1 0/1 -1/1 1/1 Row 2: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 3: 0/1 0/1 0/1 0/1 0/1 0/1 Row 4: 0/1 0/1 0/1 -1/1 2/1 -4/1 ----------------------------------------------------------------------------- Selected row operation: Swap rows 4 and 3 . Linear System: Row 1: 1/1 0/1 -3/1 0/1 -1/1 1/1 Row 2: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 3: 0/1 0/1 0/1 -1/1 2/1 -4/1 Row 4: 0/1 0/1 0/1 0/1 0/1 0/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 3 by -1 / 1 . Linear System: Row 1: 1/1 0/1 -3/1 0/1 -1/1 1/1 Row 2: 0/1 1/1 -2/1 -1/1 0/1 -2/1 Row 3: 0/1 0/1 0/1 1/1 -2/1 4/1 Row 4: 0/1 0/1 0/1 0/1 0/1 0/1 -----------------------------------------------------------------------------[Note: One can stop the row reduction at this row echelon form and use back substitution to obtain the solution. Here we continue to reduced row echelon form.]
----------------------------------------------------------------------------- Selected row operation: Subtract -1 / 1 times row 3 from row 2 . Linear System: Row 1: 1/1 0/1 -3/1 0/1 -1/1 1/1 Row 2: 0/1 1/1 -2/1 0/1 -2/1 2/1 Row 3: 0/1 0/1 0/1 1/1 -2/1 4/1 Row 4: 0/1 0/1 0/1 0/1 0/1 0/1 ----------------------------------------------------------------------------- Program execution terminated.
The two parameter family of solutions is
therefore
(x1, x2, x3,
x4, x5) =
(1 + 3s + t,
2 + 2s + 2t,
s, 4 + 2t, t)
= (1, 2, 0, 4, 0) + s (3, 2, 1, 0, 0)
+ t (1, 2, 0, 2, 1).
Geometrically, this is a plane in
ú5.
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