# Problem12:

Find the inverse of each of the following matrices, where a,b,c,d, and e are all nonzero,

> ;

```                                 [a    0    0    0]
[                ]
[0    b    0    0]
A := [                ]
[0    0    c    0]
[                ]
[0    0    0    d]

[0    0    0    a]
[                ]
[0    0    b    0]
B := [                ]
[0    c    0    0]
[                ]
[d    0    0    0]

[e    0    0    0]
[                ]
[1    e    0    0]
C := [                ]
[0    1    e    0]
[                ]
[0    0    1    e]```

### SOLUTION:

 Well if you have maple you can just do it! First enter the matrices to maple.

> A := diag(a,b,c,d): B:=matrix(4,4,[0,0,0,a,0,0,b,0,0,c,0,0,d,0,0,0]):
> D1 := matrix(2,2,[e,0,1,e]); Z := matrix(2,2,0); G :=matrix(2,2,[0,1,0,0]);

```                                      [e    0]
D1 := [      ]
[1    e]

[0    0]
Z := [      ]
[0    0]

[0    1]
G := [      ]
[0    0]```
> C := blockmatrix(2,2,[D1,Z,G,D1]);
```                                 [e    0    0    0]
[                ]
[1    e    0    0]
C := [                ]
[0    1    e    0]
[                ]
[0    0    1    e]```
 Notice the cool way to enter A, Z, and C into maple! Now compute the inverses. Here are three different ways:

> Ainv := inverse(A);

```                              [1/a     0      0      0 ]
[                        ]
[ 0     1/b     0      0 ]
Ainv := [                        ]
[ 0      0     1/c     0 ]
[                        ]
[ 0      0      0     1/d]```
> Binv := evalm(1/B);
```                              [ 0      0      0     1/d]
[                        ]
[ 0      0     1/c     0 ]
Binv := [                        ]
[ 0     1/b     0      0 ]
[                        ]
[1/a     0      0      0 ]```
> Cinv := evalm(C^(-1));
```                          [ 1/e        0         0        0 ]
[                                 ]
[   1                             ]
[- ----     1/e        0        0 ]
[    2                            ]
[   e                             ]
[                                 ]
Cinv := [  1          1                   ]
[ ----     - ----     1/e       0 ]
[   3          2                  ]
[  e          e                   ]
[                                 ]
[   1        1          1         ]
[- ----     ----     - ----    1/e]
[    4        3          2        ]
[   e        e          e         ]```

#### But that's not the point!!

Let's get these inverses by using what we know about matrix multiplication.

#### To Compute the Inverse of A:

 Notice that A can be obtained from the 4x4 identity by multiplying the first row by a, the second row by b, the third by c and the last by d. Let's denote by E1,E2,E3,E4 the elemtary matrices of those elementary row operations. We can then write:

> ;

`                               A = E4 E3 E2 E1 I`
 now to compute the inverse of A we can use the theorem that says that the inverse of a product of matrices is the product of the inverses in reverse order. Thus,

> ;

```                                -1
A   = F1 F2 F3 F4```
 where the F's are the inverses of the E's. The operation that reverses multiplication of a row by a number is division by the same number i.e. multiplication of the same row by one over the number. Thus, the last formula says that the inverse of A is obtained by dividing the last row of the 4x4 identity by d, dividing the third row by c etc... Exactly what we got as Ainv above. Notice that there is really nothing special about 4x4 matrices. By the same reasoning we can see that the inverse of any nxn diagonal matrix will exist provided that none of the diagonal entries are zero. The inverse is then the diagonal matrix of the inverses.

#### To Compute the Inverse of B:

 Notice that the matrix B is obtained from the 4x4 identity by first getting A (as before) and then swapping column 1 with colum 4 then swapping column 2 with column 3. It is easy to check that elementary column operations are done by POST-multiplying by elementary matrices. We can then write:

> ;

`                           B = E4 E3 E2 E1 I G1 G2`
 where G1 and G2 are the elementary matrices that perform the swapping of columns. If we recall the the inverses of the G's are the same G's, we have,

> ;

```                  -1
B   = G2 G1 F1 F2 F3 F4```
 and this last equation says that the inverse of B can be obtained by swapping rows 1 and 4 and rows 2 and 3 of the inverse of A. Notice, that we need to swap rows and not columns since the G's are now PRE multiplying the F's.

#### To Compute the Inverse of C:

 It is not obvious how to write C as a product of elementary matrices. We know, however, that if C has an inverse then (and only then) it can be written as a product of elem. matrices. The way to find these elem. matrices is to perform elem. row operations on C until we get the identity. Let us suppose that it takes k elem. row operations to take C into the 4x4 identity matrix I4. Let us write the corresponding elementary matrices by E1,E3,..., Ek (don't get confused with the E's above....). We can then write:

> ;

`                               Ek ... E2 E1 C = I4`
 this shows that the inverse of C is the product (Ek...E2 E1). A standard algorithm for keeping track of this product (and therefore get the inverse of C at the end) is to perform the elementary row operations not just on C but on C augmented by I4, i.e. on

> C_I4 := augment(C,diag(1,1,1,1));

```                        [e    0    0    0    1    0    0    0]
[                                    ]
[1    e    0    0    0    1    0    0]
C_I4 := [                                    ]
[0    1    e    0    0    0    1    0]
[                                    ]
[0    0    1    e    0    0    0    1]```
 bring this augmented matrix into rref (reduced row echelon form) to obtain I4 augmented with the inverse of C. Watch,

> rref(C_I4);

```            [1    0    0    0     1/e        0         0        0 ]
[                                                     ]
[                       1                             ]
[0    1    0    0    - ----     1/e        0        0 ]
[                        2                            ]
[                       e                             ]
[                                                     ]
[                      1          1                   ]
[0    0    1    0     ----     - ----     1/e       0 ]
[                       3          2                  ]
[                      e          e                   ]
[                                                     ]
[                       1        1          1         ]
[0    0    0    1    - ----     ----     - ----    1/e]
[                        4        3          2        ]
[                       e        e          e         ]```
 the right hand 4x4 block matrix is just the product (Ek...E2 E1) which is the inverse of C. Extra: can you find the E's explicitly?

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