If $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \cdots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}$$, then the inverse of $$\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$$ is:

We have to evaluate the product

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \cdots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix}$$

and we are told that the result equals

$$\begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}.$$

First, recall the general multiplication rule for 2 × 2 matrices. If we have two matrices of the special upper-triangular form $$A=\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}, \qquad B=\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix},$$ then their product is obtained as follows:

$$AB=\begin{bmatrix} 1\cdot 1+ a\cdot 0 & 1\cdot b + a\cdot 1\\ 0\cdot 1+1\cdot 0 & 0\cdot b + 1\cdot 1 \end{bmatrix} =\begin{bmatrix} 1 & a+b \\ 0 & 1 \end{bmatrix}.$$

So, multiplying two such matrices simply adds their upper-right entries. Because of associativity of matrix multiplication, we can apply this fact successively. Hence, the product of all the given matrices is

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \cdots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 1+2+3+\dots+(n-1) \\ 0 & 1 \end{bmatrix}.$$

The sum in the upper-right corner is the sum of the first $$n-1$$ natural numbers. The well-known formula for this sum is

$$1+2+3+\dots+(n-1)=\frac{(n-1)n}{2}.$$

Therefore the product becomes

$$\begin{bmatrix} 1 & \dfrac{(n-1)n}{2} \\ 0 & 1 \end{bmatrix}.$$

We are told that this equals $$\begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}.$$ So the upper-right entries must be equal, giving the equation

$$\frac{(n-1)n}{2}=78.$$

Multiplying both sides by 2, we get

$$(n-1)n=156.$$

Expanding and arranging in standard quadratic form,

$$n^2-n-156=0.$$

To solve this quadratic, compute the discriminant: $$\Delta = (-1)^2 - 4(1)(-156)=1+624=625.$$ Taking the square root, $$\sqrt{625}=25$$.

Using the quadratic formula $$n=\dfrac{-b\pm\sqrt{\Delta}}{2a}$$ with $$a=1$$ and $$b=-1$$, we get

$$n=\frac{\,1\pm25\,}{2}.$$

This gives two roots: $$n=\frac{26}{2}=13 \quad\text{or}\quad n=\frac{-24}{2}=-12.$$

The problem concerns positive integers, so we accept

$$n=13.$$

Now we must find the inverse of the matrix $$\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 13 \\ 0 & 1 \end{bmatrix}.$$

For any matrix of the form $$\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix},$$ the inverse is obtained by changing the sign of $$a$$, because

$$\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -a \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & a-a \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$$

which is the identity matrix. Therefore, the inverse of $$\begin{bmatrix} 1 & 13 \\ 0 & 1 \end{bmatrix}$$ is

$$\begin{bmatrix} 1 & -13 \\ 0 & 1 \end{bmatrix}.$$

Hence, the correct answer is Option D.