To solve the problem step by step, we need to find the trace of the matrix \( A_{10} \) and then determine the unit digit of that trace.
### Step 1: Understand the structure of the matrices
The matrices are defined as follows:
- \( A_1 = [1] \)
- \( A_2 = [(2, 3), (4, 5)] \)
- \( A_3 = [(6, 7, 8), (9, 10, 11), (12, 13, 14)] \)
From this, we can see that the \( n \)-th matrix \( A_n \) has \( n \) rows and \( n \) columns.
### Step 2: Determine the first element of \( A_{10} \)
To find the first element of \( A_{10} \), we need to find the total number of elements in all previous matrices.
The number of elements in each matrix is:
- \( A_1 \): 1 element
- \( A_2 \): 2x2 = 4 elements
- \( A_3 \): 3x3 = 9 elements
- ...
- \( A_n \): \( n^2 \) elements
The total number of elements in the first \( n \) matrices is:
\[
1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}
\]
For \( n = 9 \):
\[
\text{Total elements} = \frac{9 \times 10 \times 19}{6} = 285
\]
Thus, the first element of \( A_{10} \) is 286.
### Step 3: Find the elements of \( A_{10} \)
The matrix \( A_{10} \) will have 10 rows and 10 columns, starting from 286. The elements will be filled in a row-wise manner.
The first row will be:
- Row 1: \( 286, 287, 288, 289, 290, 291, 292, 293, 294, 295 \)
The second row will be:
- Row 2: \( 296, 297, 298, 299, 300, 301, 302, 303, 304, 305 \)
Continuing this way, the last row will be:
- Row 10: \( 385, 386, 387, 388, 389, 390, 391, 392, 393, 394 \)
### Step 4: Calculate the trace of \( A_{10} \)
The trace of a matrix is the sum of the diagonal elements. For \( A_{10} \), the diagonal elements are:
- \( 286 \) (1st row, 1st column)
- \( 297 \) (2nd row, 2nd column)
- \( 308 \) (3rd row, 3rd column)
- ...
- \( 394 \) (10th row, 10th column)
The diagonal elements can be calculated as:
- The \( k \)-th diagonal element can be expressed as:
\[
\text{Element} = 286 + (k-1) \times 11
\]
for \( k = 1, 2, \ldots, 10 \).
Calculating each diagonal element:
- For \( k = 1 \): \( 286 + 0 \times 11 = 286 \)
- For \( k = 2 \): \( 286 + 1 \times 11 = 297 \)
- For \( k = 3 \): \( 286 + 2 \times 11 = 308 \)
- For \( k = 4 \): \( 286 + 3 \times 11 = 319 \)
- For \( k = 5 \): \( 286 + 4 \times 11 = 330 \)
- For \( k = 6 \): \( 286 + 5 \times 11 = 341 \)
- For \( k = 7 \): \( 286 + 6 \times 11 = 352 \)
- For \( k = 8 \): \( 286 + 7 \times 11 = 363 \)
- For \( k = 9 \): \( 286 + 8 \times 11 = 374 \)
- For \( k = 10 \): \( 286 + 9 \times 11 = 385 \)
### Step 5: Sum the diagonal elements
Now, we sum these diagonal elements:
\[
\text{Trace} = 286 + 297 + 308 + 319 + 330 + 341 + 352 + 363 + 374 + 385
\]
Calculating this sum:
\[
\text{Trace} = 286 + 297 + 308 + 319 + 330 + 341 + 352 + 363 + 374 + 385 = 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
\]
### Step 6: Find the unit digit of the trace
To find the unit digit of the trace, we can add the unit digits of each diagonal element:
- Unit digits: \( 6, 7, 8, 9, 0, 1, 2, 3, 4, 5 \)
Now, summing these unit digits:
\[
6 + 7 + 8 + 9 + 0 + 1 + 2 + 3 + 4 + 5 = 45
\]
The unit digit of 45 is \( 5 \).
### Final Answer
The unit digit of \( \lambda \) (the trace of \( A_{10} \)) is \( \boxed{5} \).
