Here's a fun way to test your matrix multiplication program.

Choose a natural number $N, the size of your matrices. Here, I'll use

our $N = 6;
[download]

for brevity, but I recommend $N = 10; for your tests.

Create the triangular matrices C and I with the following recursive constructions.

our $C;
for my $k (0 .. $N - 1) {
    for my $l (0 .. $N - 1) {
        my $e;
        if (0 == $k) {
            if (0 == $l) {
                $e = 1;
            } else {
                $e = 0;
            }
        } else {
            $e = 0;
            if (0 < $l) {
                $e += $$C[$k-1][$l-1];
            }
            if ($l < $N - 1) {
                $e += $$C[$k-1][$l+1];
            }
        }
        $$C[$k][$l] = $e;
    }
}

our $U;
for my $k (0 .. $N - 1) {
    for my $l (0 .. $N - 1) {
        my $e;
        if (0 == $k) {
            if (0 == $l) {
                $e = 1;
            } else {
                $e = 0;
            }
        } elsif (1 == $k) {
            if (1 == $l) {
                $e = 2;
            } else {
                $e = 0;
            }
        } else {
            $e = -$$U[$k-2][$l];
            if (0 < $l) {
                $e += 2 * $$U[$k-1][$l-1];
            }
        }
        $$U[$k][$l] = $e;
    }
}
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This will look something like this.

=begin output

C = (
   1   0   0   0   0   0
   0   1   0   0   0   0
   1   0   1   0   0   0
   0   2   0   1   0   0
   2   0   3   0   1   0
   0   5   0   4   0   1
)

U = (
   1   0   0   0   0   0
   0   2   0   0   0   0
  -1   0   4   0   0   0
   0  -4   0   8   0   0
   1   0 -12   0  16   0
   0   6   0 -32   0  32
)

=end output
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Now compute the matrix product C·C^T, where C^T is the transpose of C. For this, you'll of course need to write a matrix transposition function, but that should be easier than the matrix multiplication function.

You should get a matrix like this:

C*C^T = (
   1   0   1   0   2   0
   0   1   0   2   0   5
   1   0   2   0   5   0
   0   2   0   5   0  14
   2   0   5   0  14   0
   0   5   0  14   0  42
)
[download]

Note that any antidiagonal of this matrix contains the same number in all positions. This should be true for larger $N as well. If it's not, then either you have made a mistake in one of your functions, or I have made a mistake in this description.

Further, compute the matrix C·U. You should get a diagonal matrix.