Problem 70 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)\) and \(B=\left(\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right), a, b \in N\). 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 Bs such that AB = BA.

Step 1: Calculate AB

To determine if \(AB = BA\), we first compute \(AB\). Use matrix multiplication rules: \[AB = \left(\begin{array}{cc}1 & 2 \ 3 & 4\end{array}\right)\left(\begin{array}{cc}a & 0 \ 0 & b\end{array}\right) = \left(\begin{array}{cc}1 \cdot a + 2 \cdot 0 & 1 \cdot 0 + 2 \cdot b \ 3 \cdot a + 4 \cdot 0 & 3 \cdot 0 + 4 \cdot b\end{array}\right) = \left(\begin{array}{cc}a & 2b \ 3a & 4b\end{array}\right)\]

Step 2: Calculate BA

Next, compute \(BA\): \[BA = \left(\begin{array}{cc}a & 0 \ 0 & b\end{array}\right)\left(\begin{array}{cc}1 & 2 \ 3 & 4\end{array}\right) = \left(\begin{array}{cc}a \cdot 1 + 0 \cdot 3 & a \cdot 2 + 0 \cdot 4 \ 0 \cdot 1 + b \cdot 3 & 0 \cdot 2 + b \cdot 4\end{array}\right) = \left(\begin{array}{cc}a & 2a \ 3b & 4b\end{array}\right)\]

Step 3: Set AB equal to BA

Now set the matrices \(AB\) and \(BA\) equal and equate their corresponding elements: \[\left(\begin{array}{cc}a & 2b \ 3a & 4b\end{array}\right) = \left(\begin{array}{cc}a & 2a \ 3b & 4b\end{array}\right)\] From this, we get two equations: 1. \(2b = 2a\) 2. \(3a = 3b\)

Step 4: Solve equations

Simplify the equations obtained: 1. \(2b = 2a\) simplifies to \(b = a\) 2. \(3a = 3b\) simplifies to \(a = b\)Both equations confirm \(b = a\). Therefore, \(a\) and \(b\) must be equal, and they can be any natural number \(n\).

Step 5: Determine number of solutions

Since \(a = b\) can be any natural number, there are infinitely many matrices \(B\) satisfying \(AB = BA\). For each natural number \(n\), there exists a matrix \(B = \left(\begin{array}{cc}n & 0 \ 0 & n\end{array}\right)\).

Key Concepts

Matrix Multiplication

Matrix multiplication is a core operation in linear algebra where two matrices are combined to produce a new matrix. It's essential to understand the detailed steps involved to solve matrix equations efficiently. When multiplying two matrices, say matrix \( A \) of dimensions \( m \times n \) and matrix \( B \) of dimensions \( n \times p \), the result is a new matrix \( C \) of dimensions \( m \times p \).

• To compute the elements in matrix \( C \), take the dot product of each row of matrix \( A \) with each column of matrix \( B \).
• Each element in matrix \( C \) is the sum of the products of corresponding elements from a row in \( A \) and a column in \( B \).

In our exercise, we dealt with matrices \( A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix} \) and \( B = \begin{pmatrix} a & 0 \ 0 & b \end{pmatrix} \). By applying the rules of matrix multiplication, we computed \( AB \) and \( BA \) and established conditions under which they are equal. This process is crucial for understanding matrix commutation scenarios.

Natural Numbers

Natural numbers are the fundamental building blocks of arithmetic and mathematics. They are a set of positive integers that include numbers starting from 1, 2, 3, and so on. These numbers are crucial because they define a system that is used to count, calculate, and order quantities.

• In mathematics, natural numbers are often denoted by \( \mathbb{N} \).
• They are used to specify measurements, analyze sums, and compare sizes.
• The concept of natural numbers was applied in our exercise to determine the values for \( a \) and \( b \) such that \( a = b \).

Natural numbers enabled us to conclude that when \( a = b \), \( a \) and \( b \) can be any natural number. Therefore, there are infinitely many solutions where the matrices \( AB = BA \). This connection to natural numbers is vital as it bridges basic arithmetic with more complex algebraic concepts.

Equations of Equality

Equations of equality are mathematical statements indicating that two expressions have the same value. They are foundational in solving problems involving algebraic manipulation and comparison of quantities. In the context of matrices, equating two matrices means that every element in corresponding positions of the matrices must be equal. This principle is crucial for establishing whether two matrices like \( AB \) and \( BA \) in our exercise are commutative, i.e., when their product remains the same irrespective of their order of multiplication.

• In our step-by-step solution, we compared matrices \( AB = \begin{pmatrix} a & 2b \ 3a & 4b \end{pmatrix} \) and \( BA = \begin{pmatrix} a & 2a \ 3b & 4b \end{pmatrix} \).
• By setting each corresponding element equal, we derived equations \( 2b = 2a \) and \( 3a = 3b \).
• Both simplify to the equation \( b = a \), showing a condition for equality. This highlights how solving equations of equality helps in validating conditions where expressions or structures are equivalent.