Problem 76 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If square matrices \(A\) and \(B\) are such that \(A A^{\theta}=A^{\theta} A\), \(B B^{\theta}=B^{\theta} B\) and \(A B^{\theta}=B^{\theta} A\), then \((A B)(A B)^{\theta}\) is equal to (A) \(B^{\theta} A^{\theta} A B\) (B) \(B A^{\theta} A B\) (C) \(B A^{\theta} A B^{\theta}\) (D) None of these

Step-by-Step Solution

Answer

The expression is equal to option (A): \(B^{\theta} A^{\theta} A B\).

Step 1: Understanding the problem

We are given square matrices \(A\) and \(B\) with specific commutation properties involving their transposes. We need to determine which of the given expressions is equal to \((AB)(AB)^\theta\).

Step 2: Expressing the problem

We need to simplify \((AB)(AB)^\theta\). First, recognize that \((AB)^\theta = B^\theta A^\theta\). The expression \((AB)(AB)^\theta\) becomes \(AB(B^\theta A^\theta)\).

Step 3: Multiplying the matrices

Calculate \(AB(B^\theta A^\theta)\). Using associativity, this becomes \(A(BB^\theta)A^\theta\). We need to find how \(A\), \(B\), and their transposes commute to simplify further.

Step 4: Simplifying further using given conditions

Given \(BB^\theta = B^\theta B\), substitute to get \(A(B^\theta B)A^\theta\). Also, given \(AB^\theta = B^\theta A\), we can further simplify.

Step 5: Final simplification

Now, since \(A(B^\theta B)A^\theta = AB^\theta BA^\theta\) and using the commutation \(AB^\theta = B^\theta A\), this simplifies to \(B^\theta A^\theta AB\).

Step 6: Finding the correct option

Compare our result \(B^\theta A^\theta AB\) with the given options. It matches option (A): \(B^\theta A^\theta AB\).

Key Concepts

Square matrices

Square matrices are special types of matrices where the number of rows and columns are equal. This format can often reveal interesting properties, especially in advanced mathematics.
Square matrices are denoted as such because they form a perfect square grid when drawn out. For example, a 3x3 matrix has 3 rows and 3 columns, and it looks like a complete square when visualized.
In terms of operations, square matrices are quite versatile. They can be added and multiplied by one another, and special calculations like finding determinants and inverses are possible when these matrices are square. This is not always possible with non-square matrices.
In the original exercise, matrices \(A\) and \(B\) are mentioned as square matrices. This means that they can easily be multiplied by themselves or by each other under the conditions stated in the problem, like commutation of certain properties.

Matrix multiplication

Matrix multiplication is an essential concept in linear algebra, involving the combination of two matrices to form a new one. It's a bit different from simple multiplication of numbers, because order matters—a lot.
When multiplying two matrices \(A\) and \(B\), the product is defined only if the number of columns in matrix \(A\) matches the number of rows in matrix \(B\). For square matrices, this condition is naturally satisfied, since the number of rows and columns are equal.
One thing to remember is that matrix multiplication is not commutative. This means \(A \times B\) does not generally equal \(B \times A\). This property is crucial in operations involving multiple matrices.

• Matrix multiplication is associative: \((A \times B) \times C = A \times (B \times C)\).
• It is also distributive: \(A \times (B + C) = A \times B + A \times C\).

In the exercise, matrix multiplication was used to simplify the expression \((AB)(AB)^{\theta}\), carefully taking into account the order and properties of commutative matrices.

Transpose properties

Transposing a matrix involves swapping its rows and columns. This is done by flipping the matrix over its diagonal. The notation \(A^{\theta}\) represents the transpose of matrix \(A\).
Transpose properties often simplify calculations. For example,

• The transpose of a product of matrices is the reverse product of their transposes: \((AB)^{\theta} = B^{\theta} A^{\theta}\).
• The transpose of a transpose brings you back to the original matrix: \((A^{\theta})^{\theta} = A\).

As highlighted in the exercise, understanding transpose properties is crucial for simplifying and solving problems involving multiple matrices and their products. The given conditions that \(A A^{\theta} = A^{\theta} A\) and \(B B^{\theta} = B^{\theta} B\) use these properties to manipulate and simplify the expressions involved.
By applying the concept of the transpose, we can further explore the symmetry and commutation relationships between matrices which aid in deriving the final solution of the exercise.