1.1: 5,9,17,23,27,35
1.2: 3,9,15,31,33
DUE: Wed., Jan 31, Beginning of Class
Please note that alternative row operations are possible that will result in
same solution set.
1.1: 5) The point that lies on the two lines (the intersection point) is the
solution of the system:
x_1 + 4 x_2 = 7
x_1 - x_2 = -1
The augmented matrix of the system is
[ 1 4 7 ]
[ 1 -1 -1 ]
To eliminate the x_1 in the second equation, do the row operation:
-1 row1 + row2 --> row2 (Replace row2 by the result of this row operation):
[ 1 4 7 ]
[ 0 -5 -8 ]
Multiply row2 by -1/5:
[ 1 4 7 ]
[ 0 1 8/5]
Now the matrix is in upper triangular form. The equation notation is (it is
not necessary to rewrite the equivalent linear system; we do it here just
for clarity):
x_1 + 4 x_2 = 7 Back substitution gives x_1 = -4 (8/5) + 7,
x_2 = 8/5 thus x_1 = 7 - 32/5 = 3/5
The back substitution above is equivalent to doing one more row operation
-4 row2 + row1:
[ 1 0 3/5 ]
[ 0 1 8/5 ]
This matrix is row equivalent to the original matrix, so the systems have
the same solution sets (x_1 = 3/5, x_2 = 8/5).
1.1:9) The system has a unique solution: x_3 = 0, x_2 = 0, x_1 = 0.
To see this, use back substitution, or use row operations to get in reduced
row echelon form:
[ 1 -5 7 0 ]
[ 0 1 3 0 ]
[ 0 0 1 0 ]
-3 row3 + row2; and -7 row3 + row1:
[ 1 -5 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]
5 row2 + row1:
[ 1 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]
1.1:17) The augmented matrix of the system is
[-2 -3 4 5 ]
[ 0 1 -2 4 ]
[ 1 3 -1 2 ]
Interchange rows 1 and 3:
[ 1 3 -1 2 ]
[ 0 1 -2 4 ]
[-2 -3 4 5 ]
2 row1 + row3:
[ 1 3 -1 2 ]
[ 0 1 -2 4 ]
[ 0 3 2 9 ]
-3 row2 + row3:
[ 1 3 -1 2 ]
[ 0 1 -2 4 ]
[ 0 0 8 -3 ]
so the system is consistent (in fact there is a unique solution).
1.1:23) Determine the value(s) of h such that the augmented matrix is that
of a consistent linear system.
[ 1 h -2 ]
[ -4 2 10 ]
4 row1 + row2:
[ 1 h -2 ]
[ 0 4h+2 2 ]
The only way that the system would be inconsistent (i.e. no solution)
is if 4 h + 2 = 0, that is h = -1/2, for then the 2nd row gives 0 = 2, a
contradiction.
The system is consistent for all h not equal to -1/2.
1.1:27) Do the following lines have a common point of intersection?
2 x_1 + 3 x_2 = -1
6 x_1 + 5 x_2 = 0
2 x_1 - 5 x_2 = 7
Determine if the system is consistent by row reduction:
[ 2 3 -1 ]
[ 6 5 0 ]
[ 2 -5 7 ]
-1 row1 + row3; and -3 row1 + row2:
[ 2 3 -1 ]
[ 0 -4 3 ]
[ 0 -8 8 ]
Multiply row3 by -1/8:
[ 2 3 -1 ]
[ 0 -4 3 ]
[ 0 1 -1 ]
4 row3 + row2:
[ 2 3 -1 ]
[ 0 0 1 ]
[ 0 1 -1 ]
Now the second row says 0 x_1 + 0 x_2 = 1; i.e. 0 = 1, a contradiction.
Thus there is no solution; the system is inconsistent.
An accurate graph of the three lines would also show that they don't
intersect in a common point, but row reduction proves it.
1.1.35) Write an appropriate system of equations for T_1,...,T_6.
Node1: T_1 = ( 10 + 20 + T_2 + T_4)/4
Node2: T_2 = (T_1 + 20 + T_3 + T_5)/4
Node3: T_3 = (T_2 + 20 + 40 + T_6)/4
Node4: T_4 = ( 10 + T_1 + T_5 + 20)/4
Node5: T_5 = (T_4 + T_2 + T_6 + 20)/4
Node6: T_6 = (T_5 + T_3 + 40 + 20)/4
Rewrite these equations in the standard form for a linear system, collecting
all variables on the left side:
4 T_1 - T_2 - T_4 = 30
- T_1 + 4 T_2 - T_3 - T_5 = 20
- T_2 + 4 T_3 - T_6 = 60
- T_1 + 4 T_4 - T_5 = 30
- T_2 - T_4 + 4T_5 - T_6 = 20
- T_3 - T_5 + 4 T_6 = 60
1.2.3) a) Reduced row echelon. Note each of the three leading 1's (in pivot
columns 1,2, and 4) is the only nonzero entry in its column.
b) Echelon form
1.2.9) Find the general solution for the system represented by the
augmented matrix
[ 0 3 6 9 ]
[ -1 1 -2 -1 ]
Multiply row2 by -1; then interchange row1 and row2:
[ 1 -1 2 1 ]
[ 0 3 6 9 ]
Multiply row2 by 1/3:
[ 1 -1 2 1 ]
[ 0 1 2 3 ]
row2 + row 1 (replace result in row1):
[ 1 0 4 4 ]
[ 0 1 2 3 ]
Hence x_3 is free variable; and x_1, x_2 are basic variables. Write the
general solution as
x_1 = 4 - 4 x_3
x_2 = 3 - 2 x_3
x_3 free
1.2.15) Note the system is in echelon form and is consistent (since rightmost
column is not a pivot column). To put in reduced echelon form, we need one
row operation: 2 row2 + row1
[ 1 0 0 0 1 -1 ]
[ 0 1 0 0 -3 1 ]
[ 0 0 0 1 5 -4 ]
[ 0 0 0 0 0 0 ]
The variables x_1, x_2, and x_4 corresponding to respective pivot columns 1, 2,
and 4 are basic variables. x_3 and x_5 are free variables.
x_1 = -1 - x_5
x_2 = 1 + 3 x_5
x_3 free
x_4 = -4 - 5 x_5
x_5 free
1.2.31) (Undetermined system)
The system is given to be consistent so there is either a unique
solution or an infinite # of solutions. Let n = # of variables (unknowns);
let m = # equations = # rows in the matrix representing the system. We are
given that n > m.
In reduced echelon form of the matrix, there are at most m leading 1 entries
(at most m rows with a leading entry are possible), so at most m basic
variables. The remaining variables (n-m of them, where n-m >= 1) are free
variables (can be assigned any value). Thus there are an infinite # of
solutions, one for each of the infinite # of real numbers we can assign to
the free variable.
1.2.33) Yes, the system can be consistent. For your particular 3 x 2 system,
the reduced echelon form must have at least one row of all zeros. For example
pick 3 lines that intersect in a common point:
- x_1 + x_2 = 1
x_1 + x_2 = 1
2 x_1 + x_2 = 1
The reduced row echelon form for the augmented matrix here is
[ 1 0 0 ]
[ 0 1 1 ]
[ 0 0 0 ]
Thus the system is consistent.
The lines intersect at the unique solution x_1 = 0, x_2 = 1.