Let `A=[(1,0,0),(1,1,0),(1,1,1)]` and `B=A^20` . Then the sum of the elements of the first column of B is

Options

A. 211
B. 251
C. 231
D. 210

Answer

To solve the problem, we need to compute \( B = A^{20} \) where \( A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \), and then find the sum of the elements of the first column of matrix \( B \).

### Step 1: Understand the structure of the matrix \( A \)

The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \]
This matrix has a specific pattern that we can exploit when raising it to a power.

### Step 2: Compute \( A^2 \)

To find \( A^{20} \), we will first compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \]

Calculating the elements:
- First row:
- \( 1 \cdot 1 + 0 \cdot 1 + 0 \cdot 1 = 1 \)
- \( 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 1 = 0 \)
- \( 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 = 0 \)

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

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

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

### Step 3: Compute \( A^3 \)

Next, we compute \( A^3 \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \]

Calculating the elements:
- First row:
- \( 1 \cdot 1 + 0 \cdot 1 + 0 \cdot 1 = 1 \)
- \( 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 1 = 0 \)
- \( 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 = 0 \)

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

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

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

### Step 4: Identify the pattern

From the calculations, we can observe a pattern. The first column of \( A^n \) appears to be:
- First element: \( 1 \)
- Second element: \( n \)
- Third element: \( \frac{n(n+1)}{2} \)

### Step 5: Compute \( A^{20} \)

Using the identified pattern, we can compute \( A^{20} \):
\[ A^{20} = \begin{pmatrix} 1 & 0 & 0 \\ 20 & 1 & 0 \\ 210 & 20 & 1 \end{pmatrix} \]

### Step 6: Sum the elements of the first column of \( B \)

Now, we need to find the sum of the elements of the first column of \( B \):
\[ 1 + 20 + 210 = 231 \]

Thus, the sum of the elements of the first column of \( B \) is \( \boxed{231} \).

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