# Problem9:

Use the result of the previous problem to find the inverse of the matrix,

> ;

```                             [ 1    1     0    0]
[                  ]
[-1    1     0    0]
[                  ]
[ 1    1     1    1]
[                  ]
[ 1    1    -1    1]```

### SOLUTION:

 We partition the matrix into,

> ;

```                                   [A    0]
[      ]
[B    A]```
 where,

> A := matrix(2,2,[1,1,-1,1]); B := matrix(2,2,[1,1,1,1]);

```                                     [ 1    1]
A := [       ]
[-1    1]

[1    1]
B := [      ]
[1    1]```
 as it was seen in Problem8 the inverse is also partitioned into 4 blocks with the inverse of A in the main diagonal, 0 on the upper right hand side and C in the lower left hand side. The matrix C is given by,

> C := - inverse(A) &* B &* inverse(A);

```                       //[1/2    -1/2]     \    [1/2    -1/2]\
C := -||[           ] &* B| &* [           ]|
\\[1/2    1/2 ]     /    [1/2    1/2 ]/```
> evalm(C);
```                                   [ 0    0]
[       ]
[-1    0]```
 Thus, the inverse of the original 4x4 matrix is,

> ;

```                         [1/2    -1/2     0      0  ]
[                          ]
[1/2    1/2      0      0  ]
[                          ]
[ 0      0      1/2    -1/2]
[                          ]
[-1      0      1/2    1/2 ]```

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