# No, one or Infinitely many solutions

## 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: a unique solution a one-parameter solution a two-parameter solution no solution

### Solution:

 Let us perform elementary row operations to get the rref of A. To save typing we alias the maple commands that we need.

 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 !

Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>