Let A = [[1, 0, 0], [1, 1, 0], [1, 1, 1]] and B = A²⁰. Then the sum of the elements of the first col

Q: Why can't we just use the Binomial Theorem here?

You absolutely can! If you write $A = I + N$, where $N$ is the strictly lower triangular part, you'll find $N^2$ is also nilpotent and $N^3 = 0$. This makes the binomial expansion very short. It's a beautiful alternative perspective!

Q: Is there a shortcut to finding the sum of the first column without writing the whole matrix?

Yes, if you look at the column vector $v = [1, 1, 1]^T$, you can observe how $A$ transforms it. But for this specific problem, finding the general form is the most robust way to avoid errors.

Q: Why does the diagonal remain 1?

Because $A$ is a lower triangular matrix with 1s on the diagonal. The product of two lower triangular matrices is always lower triangular, and the diagonal elements of the product are simply the product of the diagonal elements of the original matrices. Since $1 \times 1 = 1$, the diagonal stays 1 forever!

Question

Let $A = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 1 & 1 \end{bmatrix}$ 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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a $3 	imes 3$ lower triangular matrix:

$$A = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 1 & 1 \end{bmatrix}$$

Our objective is to compute $B = A^{20}$ and determine the sum of the elements in the first column of $B$. Since manual multiplication is inefficient, we must identify a pattern by calculating the first few powers of $A$.

### Identifying the Pattern

Let us compute $A^2$ by multiplying $A$ with itself:

$$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix}$$

Next, we compute $A^3$ by multiplying $A^2$ with $A$:

$$A^3 = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 3 & 1 & 0 \ 6 & 3 & 1 \end{bmatrix}$$

### Generalizing the Matrix $A^n$

By observing the progression of the elements, we can deduce the general form for $A^n$. The element $a_{21}$ follows the sequence $1, 2, 3, \dots, n$, which is simply $n$.

The element $a_{31}$ follows the sequence $1, 3, 6, \dots$, which represents the sum of the first $n$ natural numbers. This is given by the formula:

$$a_{31} = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$

Thus, the general form of the matrix $A^n$ is:

$$A^n = \begin{bmatrix} 1 & 0 & 0 \ n & 1 & 0 \ \frac{n(n+1)}{2} & n & 1 \end{bmatrix}$$

### Final Calculation

To find $B = A^{20}$, we substitute $n = 20$ into our general formula. The first column of $B$ consists of the following elements:

$$b_{11} = 1$$
$$b_{21} = 20$$
$$b_{31} = \frac{20(21)}{2} = 210$$

The sum of the elements in the first column is:

$$1 + 20 + 210 = 231$$

The final result for the sum of the elements in the first column of $B$ is **231**.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 111011001$ and $B = A^{20}$ . Then the sum of the elements of the first column of B is :

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Visualized Solution

Introduction to Matrix $A$

Calculating $A^{2}$

Calculating $A^{3}$

Identifying the Pattern for $a_{21}$

Identifying the Pattern for $a_{31}$

General Form of $A^{n}$

Finding Matrix $B = A^{20}$

Sum of the First Column

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

Identifying the Pattern

Generalizing the Matrix $A^{n}$

Final Calculation

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​111​011​001​​ and B=A20. Then the sum of the elements of the first column of B is :

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Visualized Solution

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