Let A = `((1,1,3),(5,2,6),(-2,-1,-3))` and let n be the smallest value of n `in `N such that `A^(n) = O_(3)` then det `(I_(3)+A+A^(2)+ cdots+ A^(n-1))` is equal to _____ .

Answer

To solve the problem, we need to find the smallest natural number \( n \) such that \( A^n = O_3 \) (the null matrix of order \( 3 \times 3 \)). Then, we will calculate the determinant of the matrix \( I_3 + A + A^2 + \cdots + A^{n-1} \).

### Step 1: Calculate \( A^2 \)

Given the matrix:
\[ A = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \]

We calculate \( A^2 = A \cdot A \).

\[ A^2 = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \]

Calculating each element of \( A^2 \):

- First row:
- \( (1 \cdot 1 + 1 \cdot 5 + 3 \cdot -2) = 1 + 5 - 6 = 0 \)
- \( (1 \cdot 1 + 1 \cdot 2 + 3 \cdot -1) = 1 + 2 - 3 = 0 \)
- \( (1 \cdot 3 + 1 \cdot 6 + 3 \cdot -3) = 3 + 6 - 9 = 0 \)

- Second row:
- \( (5 \cdot 1 + 2 \cdot 5 + 6 \cdot -2) = 5 + 10 - 12 = 3 \)
- \( (5 \cdot 1 + 2 \cdot 2 + 6 \cdot -1) = 5 + 4 - 6 = 3 \)
- \( (5 \cdot 3 + 2 \cdot 6 + 6 \cdot -3) = 15 + 12 - 18 = 9 \)

- Third row:
- \( (-2 \cdot 1 + -1 \cdot 5 + -3 \cdot -2) = -2 - 5 + 6 = -1 \)
- \( (-2 \cdot 1 + -1 \cdot 2 + -3 \cdot -1) = -2 - 2 + 3 = -1 \)
- \( (-2 \cdot 3 + -1 \cdot 6 + -3 \cdot -3) = -6 - 6 + 9 = -3 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & -1 & -3 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)

Next, we calculate \( A^3 = A^2 \cdot A \):
\[ A^3 = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & -1 & -3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \]

Calculating each element of \( A^3 \):

- First row:
- All elements will be \( 0 \) since the first row of \( A^2 \) is all zeros.

- Second row:
- \( (3 \cdot 1 + 3 \cdot 5 + 9 \cdot -2) = 3 + 15 - 18 = 0 \)
- \( (3 \cdot 1 + 3 \cdot 2 + 9 \cdot -1) = 3 + 6 - 9 = 0 \)
- \( (3 \cdot 3 + 3 \cdot 6 + 9 \cdot -3) = 9 + 18 - 27 = 0 \)

- Third row:
- \( (-1 \cdot 1 + -1 \cdot 5 + -3 \cdot -2) = -1 - 5 + 6 = 0 \)
- \( (-1 \cdot 1 + -1 \cdot 2 + -3 \cdot -1) = -1 - 2 + 3 = 0 \)
- \( (-1 \cdot 3 + -1 \cdot 6 + -3 \cdot -3) = -3 - 6 + 9 = 0 \)

Thus, we have:
\[ A^3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = O_3 \]

### Step 3: Determine \( n \)

Since \( A^3 = O_3 \), the smallest \( n \) such that \( A^n = O_3 \) is \( n = 3 \).

### Step 4: Calculate \( I_3 + A + A^2 \)

Now we need to calculate the determinant of:
\[ I_3 + A + A^2 \]

Where:
\[ I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Adding \( I_3 + A + A^2 \):
\[ I_3 + A + A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & -1 & -3 \end{pmatrix} \]

Calculating the sum:
\[ = \begin{pmatrix} 1 + 1 + 0 & 0 + 1 + 0 & 0 + 3 + 0 \\ 0 + 5 + 3 & 1 + 2 + 3 & 0 + 6 + 9 \\ 0 - 2 - 1 & 0 - 1 - 1 & 1 - 3 - 3 \end{pmatrix} \]
\[ = \begin{pmatrix} 2 & 1 & 3 \\ 8 & 6 & 15 \\ -3 & -2 & -5 \end{pmatrix} \]

### Step 5: Calculate the Determinant

Now we calculate the determinant of:
\[ \begin{pmatrix} 2 & 1 & 3 \\ 8 & 6 & 15 \\ -3 & -2 & -5 \end{pmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

Calculating:
\[ = 2(6 \cdot -5 - 15 \cdot -2) - 1(8 \cdot -5 - 15 \cdot -3) + 3(8 \cdot -2 - 6 \cdot -3) \]
\[ = 2(-30 + 30) - 1(-40 + 45) + 3(-16 + 18) \]
\[ = 2(0) - 1(5) + 3(2) \]
\[ = 0 - 5 + 6 = 1 \]

### Final Answer

The value of \( \text{det}(I_3 + A + A^2) \) is \( \boxed{1} \).