Let A = [1 2; 3 4] and B = [a 0; 0 b], a, b ∈ N. Then (a) there cannot exist any B such that AB = BA

Q: Wait, why do we reject the negative root here?

In this specific problem, we are given that $a, b \in \mathbb{N}$, which means they must be positive integers. There are no negative roots to reject because the constraint itself restricts our domain to natural numbers.

Q: Why can't we just use energy conservation?

Energy conservation is a principle for physical systems. Here, we are dealing with linear algebra and matrix properties. The commutativity of matrices is a purely algebraic concept, not a physical one.

Q: Why are there infinitely many solutions?

We found the condition $a = b$. Since $a$ and $b$ are natural numbers, any pair $(n, n)$ where $n \in \mathbb{N}$ works. Since the set of natural numbers $\{1, 2, 3, \dots\}$ is infinite, there are infinitely many such pairs.

Question

Let $A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$, $a, b \in N$. Then

Options:

(A) there cannot exist any $B$ such that $AB = BA$

(B) there exist more than one but finite number of $B$'s such that $AB = BA$

(C) there exists exactly one $B$ such that $AB = BA$

(D) there exist infinitely many $B$'s such that $AB = BA$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two matrices, $A$ and $B$, where:

$$A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$$

The condition for the "dance" is that the matrices must commute, meaning $AB = BA$. We aim to find the relationship between $a$ and $b$ that satisfies this equality.

### The Choreography of Multiplication

First, we calculate the product $AB$:

$$AB = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} = \begin{bmatrix} (1)(a) + (2)(0) & (1)(0) + (2)(b) \ (3)(a) + (4)(0) & (3)(0) + (4)(b) \end{bmatrix} = \begin{bmatrix} a & 2b \ 3a & 4b \end{bmatrix}$$

Next, we calculate the product $BA$:

$$BA = \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} = \begin{bmatrix} (a)(1) + (0)(3) & (a)(2) + (0)(4) \ (0)(1) + (b)(3) & (0)(2) + (b)(4) \end{bmatrix} = \begin{bmatrix} a & 2a \ 3b & 4b \end{bmatrix}$$

### The Moment of Truth

For the matrices to be equal, their corresponding elements must be identical:

$$\begin{bmatrix} a & 2b \ 3a & 4b \end{bmatrix} = \begin{bmatrix} a & 2a \ 3b & 4b \end{bmatrix}$$

By comparing the elements, we observe:

1. The first row, second column gives $2b = 2a$, which simplifies to $a = b$.
2. The second row, first column gives $3a = 3b$, which also simplifies to $a = b$.

The diagonal elements $a=a$ and $4b=4b$ are satisfied for any $a$ and $b$. Thus, the necessary and sufficient condition for commutativity is **$a = b$**.

### The Infinite Possibilities

We are given that $a, b \in \mathbb{N}$, where $\mathbb{N} = \{1, 2, 3, \dots\}$. Since the condition requires $a = b$, any pair $(n, n)$ where $n \in \mathbb{N}$ will satisfy the requirement.

Examples of such matrices include:

$$B_1 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}, \quad B_2 = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}, \quad B_3 = \begin{bmatrix} 3 & 0 \ 0 & 3 \end{bmatrix}, \dots$$

Because the set of natural numbers is infinite, there are **infinitely many** such matrices $B$ that satisfy the condition $AB = BA$.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [1324]$ and $B = [a 0 0 b]$ , $a, b \in N$ . Then

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Visualized Solution

Introduction to Matrices $A$ and $B$

The Commutativity Condition

Setting up the Product $A B$

Calculating the Product $A B$

Setting up and Calculating $B A$

Equating $A B$ and $B A$

Solving for $a$ and $b$

Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Choreography of Multiplication

The Moment of Truth

The Infinite Possibilities

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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=[13​24​] and B=[a0​0b​], a,b∈N. Then

Select Answer:

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The Commutativity Condition

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Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BA

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