if `[{:(1,2,a),(0,1,4),(0,0,1):}]^n=[{:(1,18,2007),(0,1,36),(0,0,1):}]` then find the value of n

Explanation

To solve the problem, we need to find the value of \( n \) such that:

\[ \left[\begin{array}{ccc} 1 & 2 & a \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{array}\right]^n = \left[\begin{array}{ccc} 1 & 18 & 2007 \\ 0 & 1 & 36 \\ 0 & 0 & 1 \end{array}\right] \]

### Step 1: Analyze the Matrix Exponentiation

We start by recognizing that we are dealing with a matrix raised to the power \( n \). The matrix we have is a triangular matrix, which simplifies the exponentiation process.

### Step 2: Calculate the First Power of the Matrix

Let's denote the matrix as \( A = \left[\begin{array}{ccc} 1 & 2 & a \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{array}\right] \).

Calculating \( A^2 \):

\[ A^2 = A \cdot A = \left[\begin{array}{ccc} 1 & 2 & a \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{array}\right] \cdot \left[\begin{array}{ccc} 1 & 2 & a \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{array}\right] \]

Calculating the entries:

- First row, first column: \( 1 \cdot 1 + 2 \cdot 0 + a \cdot 0 = 1 \)
- First row, second column: \( 1 \cdot 2 + 2 \cdot 1 + a \cdot 0 = 2 + 2 = 4 \)
- First row, third column: \( 1 \cdot a + 2 \cdot 4 + a \cdot 1 = a + 8 \)

- Second row, first column: \( 0 \cdot 1 + 1 \cdot 0 + 4 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 2 + 1 \cdot 1 + 4 \cdot 0 = 1 \)
- Second row, third column: \( 0 \cdot a + 1 \cdot 4 + 4 \cdot 1 = 4 + 4 = 8 \)

- Third row remains the same: \( 0, 0, 1 \)

Thus, we have:

\[ A^2 = \left[\begin{array}{ccc} 1 & 4 & a + 8 \\ 0 & 1 & 8 \\ 0 & 0 & 1 \end{array}\right] \]

### Step 3: Generalize the Matrix Power

From the pattern, we can see that:

- The first row, second column entry is \( 2n \).
- The second row, third column entry is \( 4n \).
- The first row, third column entry is \( a + 8(n-1) \).

### Step 4: Set Up the Equations

Now we can set up equations based on the given matrix:

1. From the first row, second column:
\[ 2n = 18 \implies n = 9 \]

2. From the second row, third column:
\[ 4n = 36 \implies n = 9 \]

3. From the first row, third column:
\[ a + 8(n-1) = 2007 \]

### Step 5: Solve for \( a \)

Substituting \( n = 9 \) into the equation for \( a \):

\[ a + 8(9-1) = 2007 \implies a + 64 = 2007 \implies a = 2007 - 64 = 1943 \]

### Conclusion

Thus, the value of \( n \) is:

\[ \boxed{9} \]