Problem:Let A be the 3 by 4 matrix given by, |
> A := matrix(3,4,[a,0,b,2, a,a,4,4, 0,a,2,b]);
[a 0 b 2]
[ ]
A := [a a 4 4]
[ ]
[0 a 2 b]
suppose A is the augmented matrix for a linear system.
Find the values of a and b for which the system have:
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Solution:
| Let us perform elementary row operations to get the rref of A. To save typing we alias the maple commands that we need. |
> alias(P=swaprow,Q=mulrow,R=addrow):
| First permute rows 1 and 2 and then (assuming a is not zero) divide the new first row by a. |
> A2 := Q( P(A,1,2), 1,1/a );
[1 1 4/a 4/a]
[ ]
A2 := [a 0 b 2 ]
[ ]
[0 a 2 b ]
| to get zeroes below the leading 1 we replace the second row with the result of adding (-a) times the first row to the second row. |
> A3 := R(A2,1,2,-a);
[1 1 4/a 4/a]
[ ]
A3 := [0 -a -4 + b -2 ]
[ ]
[0 a 2 b ]
| now replace the third row by the result of adding the second to the third row. |
> A4 := R(A3,2,3,1);
[1 1 4/a 4/a ]
[ ]
A4 := [0 -a -4 + b -2 ]
[ ]
[0 0 -2 + b -2 + b]
| To get rid of the condition that a must be different than zero we can multiply the first row by a. |
> A5 := Q(A4,1,a);
[a a 4 4 ]
[ ]
A5 := [0 -a -4 + b -2 ]
[ ]
[0 0 -2 + b -2 + b]
> sol := backsub(A5);
[ -2 + b -2 + b ]
sol := [- ------, ------, 1]
[ a a ]
| Therefore there is a unique solution when a is not zero AND b is not 2. Notice that when b =2 the last row of A5 is a row of zeroes. When b=2 have, |
> A6 := subs({b=2},evalm(A5));
[a a 4 4]
[ ]
A6 := [0 -a -2 -2]
[ ]
[0 0 0 0]
> sol1 := backsub(A6);
[ -1 + _t[1] -1 + _t[1] ]
sol1 := [-2 ----------, -2 ----------, _t[1]]
[ a a ]
| Thus, when b=2 but a is not 0 we have a one-parameter solution. Maple uses _t[1] as the parameter. Now when both a=0 and b=2 the system becomes, |
> A7 := subs({a=0,b=2},evalm(A5));
[0 0 4 4]
[ ]
A7 := [0 0 -2 -2]
[ ]
[0 0 0 0]
| By adding (1/2)row1 to row 2 we get |
> A8 := R(A7,1,2,1/2);
[0 0 4 4]
[ ]
A8 := [0 0 0 0]
[ ]
[0 0 0 0]
> sol2 := backsub(A8);
sol2 := [_t[2], _t[1], 1]
| so when b=2 and a=0 the system has a two-parameter set of solutions. In this case all we know is that z=1 but x and y are free to take any values. The only other possibility to consider is when a=0 but b is not equal to 2, |
> A9 := subs({a=0},evalm(A5));
[0 0 4 4 ]
[ ]
A9 := [0 0 -4 + b -2 ]
[ ]
[0 0 -2 + b -2 + b]
| since b is not 2, we can divide the third row by b-2 to get, |
> A10 := Q(A9,3,1/(b-2));
[0 0 4 4]
[ ]
A10 := [0 0 -4 + b -2]
[ ]
[0 0 1 1]
| and performing 2 more R operations we get A10 in rref. |
> A11 := R( R(A10,3,1,-4), 3,2,4-b );
[0 0 0 0 ]
[ ]
A11 := [0 0 0 2 - b]
[ ]
[0 0 1 1 ]
| well actually A11 is not in rref... we need to reorder the rows and divide row2 by (2-b). |
> A12 := P(Q(A11,2,1/(2-b)),1,3);
[0 0 1 1]
[ ]
A12 := [0 0 0 1]
[ ]
[0 0 0 0]
| Now it is clear that the system is inconsistent. Look at the second row it says that 0=1 ! |