If `[{:(1, 1), (0,1):}]*[{:(1, 2), (0,1):}]*[{:(1, 3), (0,1):}]cdotcdotcdot[{:(1, n-1), (0,1):}] = [{:(1, 78), (0,1):}]`, then the inverse of `[{:(1, n), (0,1):}]` is

Options

A. `[(1,0),(12,1)]`
B. `[(1,-13),(0,1)]`
C. `[(1,0),(13,1)]`
D. `[(1,-12),(0,1)]`

Answer

To solve the problem, we need to find the inverse of the matrix given the product of a series of matrices. Let's break down the solution step by step.

### Step 1: Understanding the Product of Matrices
We are given a product of matrices of the form:

\[ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \cdots \begin{pmatrix} 1 & n-1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 78 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Finding the General Form of the Product
The product of these matrices can be represented as:

\[ \begin{pmatrix} 1 & S \\ 0 & 1 \end{pmatrix} \]

where \( S \) is the sum of the first \( n-1 \) integers. The sum of the first \( n-1 \) integers is given by:

\[ S = \frac{(n-1)n}{2} \]

### Step 3: Setting Up the Equation
From the problem, we know:

\[ \frac{(n-1)n}{2} = 78 \]

Multiplying both sides by 2 gives:

\[ (n-1)n = 156 \]

### Step 4: Rearranging the Equation
Rearranging the equation, we get:

\[ n^2 - n - 156 = 0 \]

### Step 5: Solving the Quadratic Equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = -1, c = -156 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-156) = 1 + 624 = 625 \]

Now substituting into the quadratic formula:

\[ n = \frac{1 \pm \sqrt{625}}{2} = \frac{1 \pm 25}{2} \]

This gives us two possible values for \( n \):

\[ n = \frac{26}{2} = 13 \quad \text{and} \quad n = \frac{-24}{2} = -12 \]

Since \( n \) must be positive, we take \( n = 13 \).

### Step 6: Forming the Matrix
Now, we can form the matrix for \( n = 13 \):

\[ A = \begin{pmatrix} 1 & 13 \\ 0 & 1 \end{pmatrix} \]

### Step 7: Finding the Inverse of the Matrix
The inverse of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:

\[ \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix \( A \):

- \( a = 1, b = 13, c = 0, d = 1 \)
- The determinant \( ad - bc = 1 \cdot 1 - 0 \cdot 13 = 1 \)

Thus, the inverse is:

\[ A^{-1} = \begin{pmatrix} 1 & -13 \\ 0 & 1 \end{pmatrix} \]

### Final Answer
The inverse of the matrix \( A \) is:

\[ \begin{pmatrix} 1 & -13 \\ 0 & 1 \end{pmatrix} \]