If A ` = [(1,1,1),(0,1,1),(0,0,1)] and M = A + A^(2) + A^(3) + . . . . + A^(20)` then the sum of all the elements of the matrix M is equal to _____

Answer

To solve the problem, we need to calculate the sum of all elements of the matrix \( M \) defined as:

\[ M = A + A^2 + A^3 + \ldots + A^{20} \]

where

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

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

To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the elements:

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

- Second row:
- \( (0 \cdot 1 + 1 \cdot 0 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 1 \cdot 0) = 1 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 1 \cdot 1) = 2 \)

- Third row:
- \( (0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 0 \cdot 1 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 0 \cdot 1 + 1 \cdot 1) = 1 \)

Thus,

\[ A^2 = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \]

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

Now we calculate \( A^3 \):

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the elements:

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

- Second row:
- \( (0 \cdot 1 + 1 \cdot 0 + 2 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 2 \cdot 0) = 1 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 2 \cdot 1) = 3 \)

- Third row:
- \( (0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 0 \cdot 1 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 0 \cdot 1 + 1 \cdot 1) = 1 \)

Thus,

\[ A^3 = \begin{pmatrix} 1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Identify the Pattern

From the calculations, we can see a pattern forming for \( A^n \):

\[ A^n = \begin{pmatrix} 1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Calculate \( M \)

Now we can express \( M \):

\[ M = A + A^2 + A^3 + \ldots + A^{20} \]

The sum of the elements in \( M \) can be calculated by summing the corresponding elements of \( A^n \):

- The first row sums to \( 1 + 1 + 1 + \ldots + 1 = 20 \)
- The second row sums to \( 0 + 1 + 2 + \ldots + 20 = 210 \)
- The third row sums to \( 0 + 0 + 0 + \ldots + 0 + \frac{20(21)}{2} = 210 \)

### Step 5: Final Calculation

The total sum of all elements in \( M \):

\[ \text{Sum} = 20 + 210 + 210 = 440 \]

Thus, the sum of all the elements of the matrix \( M \) is:

\[ \boxed{440} \]