> with(linalg):
Problem 1.

We have a 4 by 4 matrix of coefficients, and we can find its inverse by augmenting it with the 4 by 4 identity matrix and then doing a Gauss-Jordan elimination on the new matrix.

> A:=matrix([[1,2,4,8],[1,3,9,27],[1,4,16,64],[1,5,25,125]]);
 A := [ 1 2 4 8 ] [ 1 3 9 27 ] [ 1 4 16 64 ] [ 1 5 25 125 ]

> Id:=matrix([[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]);
 Id := [ 1 0 0 0 ] [ 0 1 0 0 ] [ 0 0 1 0 ] [ 0 0 0 1 ]

> B:=augment(A,Id);
 B := [ 1 2 4 8 1 0 0 0 ] [ 1 3 9 27 0 1 0 0 ] [ 1 4 16 64 0 0 1 0 ] [ 1 5 25 125 0 0 0 1 ]

> B:=gaussjord(B);
 B := [ 1 0 0 0 10 -20 15 -4 ] [ 0 1 0 0 -47/6 19 -31/2 13/3 ] [ 0 0 1 0 2 -11/2 5 -3/2 ] [ 0 0 0 1 -1/6 1/2 -1/2 1/6 ]

This gives us the inverted matrix of coefficients as the right half of this matrix (B).

Now, the math we're going to do is as follows.

The original system of linear equations can be rewritten. The left sides of the equations as the product of two matrices, A * the column matrix C, where
 C := [ x ] [ y ] [ z ] [ t ]

The right side of the system we are to solve can also be written as a column matrix D:
 D := [ 1 ] [ 2 ] [ 3 ] [ 4 ]

So we have that A * C = D. But since we've found that A is invertible by adjoining the corresponding identity matrix to the right of A and eliminating it, we can multiply the inverse of A to the left side of both equations and get that

A-1 * A * C = A-1 * D.

And since A-1 * A = I by the definition of the inverse of a matrix, And I*C=C by the definition of the identity matrix, then we have that

C = A-1 * D.

So I'll have Maple solve this and I'll have the solution for #1.

For convenience, I'll call A-1 (G).
> G:=delcols(B,1..4);
 G := [ 10 -20 15 -4 ] [ -47/6 19 -31/2 13/3 ] [ 2 -11/2 5 -3/2 ] [ -1/6 1/2 -1/2 1/6 ]

> D:=[1,2,3,4];

D := [1, 2, 3, 4]

> evalm(G&*D);
 [ -1 1 0 0 ]

Since this matrix equals the matrix (C), we have the solution that:
```
x = -1
y = 1
z = 0
t = 0.

```

Problem 2. Let A be an upper triangular matrix of order n with zeroes on the main diagonal. Prove that An=0.

For the matrix A specified here, we know that all entries A[r,s] for which r 3 s are zeroes. That is, A looks something like
 [ 0 A[1,2] A[1,3] ... A[1,n-1] A[1,n] ] [ 0 0 A[2,3] ... A[2,n-1] A[2,n] ] [ 0 0 0 ... A[3,n-1] A[2,n] ] .......................... [ 0 0 0 ... 0 A[n-1,n] ] [ 0 0 0 ... 0 0 ]

Let us represent A both as a column of the rows r1, r2, . . . , rn and as a row of the columns c1, c2, . . . , cn:
```
r1
r2
A = r3 = [c1   c2   c3   . . .   cn].
. . .
rn

```

Consider what happens when we multiply A by itself to get A2. To obtain the (i,j)-entry of A2, we take ri times cj. If j = 1, then c1 = 0, and so all entries of the first column in A2 are zero. If j = 2, then c1 = [ A[1, 2] 0 0 . . . 0 ], and A2[i, 2] will be zero if the first member of ri is zero. This is true for all iP every row starts with a zero. So the entire second column of A2 consists of zeroes.

This can be extended over all j up to n. In any column cj, the first j11 terms are non-zero, and in any row rj11, the first j11 terms are zeroes. So every entry (j11, j) in A2 is zero. To further generalize this, note that in any row rj1k, where n > j > k > 0, the first j1k terms are zeroes. Therefore, every entry A2[i,j] for which i is less than or equal to j11 is zero. So A2 looks like
 [ 0 0 A2[1,3] ... A2[1,n-1] A2[1,n] ] [ 0 0 0 ... A2[2,n-1] A2[2,n] ] [ 0 0 0 ... A2[3,n-1] A2[2,n] ] .......................... [ 0 0 0 ... 0 0 ] [ 0 0 0 ... 0 0 ]
We see that the first diagonal above the main diagonal of A2 consists of zeroes.

Now take any matrix B whose main diagonal, first upper diagonal,..., m-th upper diagonal consist of zeroes. This means that B[i,j]=0 whenever i is greater than or equal to j-m. Let us look what happens when we multiply B by A. Take BA[i,j] where i is greater than or equal to j-m-1 It is equal to the product of the i-th row of B and the j-th column of A. The i-th row of B looks like this:

[0,0,....,0, B[i,i+m], B[i, i+m+1],....]

That it the first i+m entries of this row are zeroes. Only the first j-1 entries of the j-th column of A may be non-zeroes. Since j-1 is less than or equal to i+m (remember that i is greater than or equal to j-m-1), the product of these row and column is zero. Thus the (i,j)-th entry of BA is zero whenever i is greater than or equal to j-m-1. This means that the main diagonal and m+1 upper diagonals of BA consist of zeroes.

Now let B=A2. The fact proved in the last paragraph shows that the firt two upper diagonals in A3 consist of zeroes. Then take B=A3. We can conclude that the first three upper diagonals in A4 consist of zeroes. Repeating this process we can conclude that the first n-1 upper diagonals in An consist of zeroes, that is An is the zero n by n matrix.

Problem 3. Prove that the symmetric matrix
 [ 1 9 ] [ 9 1 ]

cannot be represented in the form A*AT where A is a square matrix which has only real (not arbitrary complex) entries.

Let A be a general 2 by 2 matrix, and find the product of A and its transpose:

> A:=matrix([[a,b],[c,d]]);
 A := [ a b ] [ c d ]

> evalm(A&*transpose(A));
 A := [ a2 + b2 ac + bd ] [ ac + bd c2 + d2 ]

If this is to represent the matrix given above, then we may equate the corresponding entries of the matrices:
```
a2 + b2 = 1
ac + bd = 9                                               (1)
c2 + d2 = 1.

```

Since a and b are reals, the first equation gives that |b| does not exceed 1 and |a| does not exceed 1 (indeed A2 is greater than or equal to 0, so b2 does not exceed 1, so |b| does not exceed 1, similar for a). This same argument applies to the last equation, so |c|,|d| do not exceed 1.

Since |a|,|b| do not exceed 1, we can say that |ab| does not exceed 1, and since |c|,|d| do not exceed 1, we can also say that |cd| does not exceed 1. We may now use the given fact that |x + y| does not exceed |x| + |y| for all x, y to say that |ab + cd| does not exceed |ab| + |cd| which does not exceed 1 + 1 = 2. So the second equation in set (1) above can never be satisfied, and thus the matrix given in the problem can never be represented as A*AT.