Problem 70 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)\) and \(B=\left(\begin{array}{cc}a & 0 \\ 0 & b\end{array}\right), a, b \in N\). Then (A) there cannot exist any \(B\) such that \(A B=B A\) (B) there exist more than one but finite number of \(B\) 's such that \(A B=B A\) (C) there exists exactly one \(B\) such that \(A B=B A\) (D) there exist infinitely many \(B\) 's such that \(A B=B A\)

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We need to determine how many matrices \( B \) exist such that the product of matrices \( A \) and \( B \) is equal to the product of \( B \) and \( A \). We will calculate \( AB \) and \( BA \) and set them equal, to solve for possible \( a \) and \( b \).

Step 2: Calculate AB

Calculate the matrix multiplication of \( A \) and \( B \):\[ AB = \left( \begin{array}{cc} 1 & 2 \ 3 & 4 \end{array} \right) \left( \begin{array}{cc} a & 0 \ 0 & b \end{array} \right) = \left( \begin{array}{cc} 1 \cdot a + 2 \cdot 0 & 1 \cdot 0 + 2 \cdot b \ 3 \cdot a + 4 \cdot 0 & 3 \cdot 0 + 4 \cdot b \end{array} \right) = \left( \begin{array}{cc} a & 2b \ 3a & 4b \end{array} \right) \]

Step 3: Calculate BA

Calculate the matrix multiplication of \( B \) and \( A \):\[ BA = \left( \begin{array}{cc} a & 0 \ 0 & b \end{array} \right) \left( \begin{array}{cc} 1 & 2 \ 3 & 4 \end{array} \right) = \left( \begin{array}{cc} a \cdot 1 + 0 \cdot 3 & a \cdot 2 + 0 \cdot 4 \ 0 \cdot 1 + b \cdot 3 & 0 \cdot 2 + b \cdot 4 \end{array} \right) = \left( \begin{array}{cc} a & 2a \ 3b & 4b \end{array} \right) \]

Step 4: Set AB equal to BA

Equate the results of \( AB \) and \( BA \):\[ \left( \begin{array}{cc} a & 2b \ 3a & 4b \end{array} \right) = \left( \begin{array}{cc} a & 2a \ 3b & 4b \end{array} \right) \]This gives the equations:1) \( a = a \) (always true)2) \( 2b = 2a \)3) \( 3a = 3b \)4) \( 4b = 4b \) (always true)

Step 5: Solve the Equations

From equations 2 and 3, both give \( a = b \). Therefore, for \( A \) and \( B \) to commute, \( a \) and \( b \) must be equal. Since \( a \) and \( b \) are natural numbers, there are infinitely many such pairs \( (a, a) \) for each natural number \( a \).

Step 6: Conclusion on Possible Solutions

Since there are infinitely many natural numbers, there are infinitely many matrices \( B \) such that \( AB = BA \) when \( a = b \). Thus, the correct answer is option (D): there exist infinitely many \( B \) 's such that \( A B = B A \).

Key Concepts

Matrix Multiplication

Matrix multiplication is a core operation in linear algebra essential for various mathematical computations. When we talk about multiplying matrices, the key process is to multiply rows from the first matrix by columns of the second matrix. This operation is not as straightforward as simple multiplication of numbers.

Let’s break down the process:
* You can only multiply two matrices if the number of columns in the first matrix (let's call it Matrix A) matches the number of rows in the second matrix (we'll call this Matrix B).
* The resulting matrix will have dimensions dictated by the number of rows in Matrix A and the number of columns in Matrix B.

In the example where matrix \( A = \begin{pmatrix}1 & 2 \ 3 & 4\end{pmatrix}\) and matrix \( B = \begin{pmatrix}a & 0 \ 0 & b\end{pmatrix}\), we perform matrix multiplication to obtain \( AB \) as follows:

\[ AB = \begin{pmatrix}1a + 2 \times 0 & 1 \times 0 + 2b \ 3a + 4 \times 0 & 3 \times 0 + 4b \end{pmatrix} = \begin{pmatrix}a & 2b \ 3a & 4b \end{pmatrix}\]

The process requires careful attention to how each element is calculated, ensuring each is the sum of multiplied terms.

Natural Numbers

Natural numbers are a fundamental concept in mathematics. They are the set of positive integers beginning from 1, including zero when specifically defined as the set of whole numbers. These numbers are represented as \( \mathbb{N} = \{1, 2, 3, 4, \ldots\}\). They are essential in counting and ordering.

In our matrix problem, the elements \( a \) and \( b \) of matrix \( B \) are natural numbers. This constraint implies that when solving for the commutation condition \( AB = BA \), \( a \) and \( b \) must be selected from the set of natural numbers.

This mathematical restriction is critical because it establishes the nature of possible solutions, indicating that for every natural number \( a \), we can find a corresponding pair where \( a = b \). Hence, the allowance for infinite solutions when considering all natural numbers as alternatives.

Step-by-Step Solution

Breaking down a problem into smaller, manageable parts is the essence of a step-by-step solution. Our goal is to unravel complex problems into easier segments which follow a logical sequence.

In the exercise provided about matrix commutation, the step-by-step solution guides us through:
* Defining the problem: Determining conditions for matrices \( A \) and \( B \) such that their multiplication order does not affect the result. This is established by \( AB = BA \).
* Performing matrix multiplications for \( AB \) and \( BA \), showing how each matrix affects the product based on its structure.
* Equating \( AB \) and \( BA \) to derive a set of equations. In this case: \( 2b = 2a \) and \( 3a = 3b \).
* Solving these equations, leading to the discovery that \( a = b \), meaning \( A \) and \( B \) will commute if \( \begin{pmatrix}a & 0 \ 0 & a\end{pmatrix} \) is used for matrix \( B \).
* Concluding with the interpretation that there's an infinite number of solutions as these conditions permit \( a \) to be any natural number.

Such structured solutions aid in developing understanding by following a path that builds upon each preceding step to arrive at the final answer.