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

Question

If \(\mathrm{A}\) is a \(3 \times 3\) non-singular matrix such that \(A A^{\prime}=\) \(A^{\prime} A\) and \(B=A^{-1} A^{\prime}\), then \(B B^{\prime}\) equals \(\quad\) [2014] (A) \(I+B\) (B) I (C) \(B^{-1}\) (D) \(\left(B^{-1}\right)^{\prime}\)

Step-by-Step Solution

Answer

BB' equals the identity matrix I.

Step 1: Understand the Problem

We are given that matrix \(A\) is a \(3 \times 3\) non-singular matrix such that \(AA' = A'A\), where \(A'\) denotes the transpose of matrix \(A\). We need to find \(B B'\) where \(B = A^{-1}A'\). We need to determine which of the given options it equals.

Step 2: Simplify the Expression

We are given \(B = A^{-1}A'\). First, find \(B' = (A^{-1}A')'\). Using the property \((XY)' = Y'X'\), we get \(B' = (A')'(A^{-1})'\). Since the transpose of a transpose \((A')'\) is \(A\) and \((A^{-1})' = (A')^{-1}\), this simplifies to \(B' = A(A')^{-1}\).

Step 3: Calculate BB'

Substitute \(B = A^{-1}A'\) and \(B' = A(A')^{-1}\) to find \(BB'\):\[ B B' = (A^{-1}A')(A(A')^{-1}) = A^{-1} (A'A) (A')^{-1} \]Given that \(A'A = AA'\), replace \(A'A\) with \(AA'\) to get:\[ B B' = A^{-1} (AA') (A')^{-1} \]Since \(AA'\) and \((A')^{-1}\) are inverses, this simplifies to:\[ B B' = A^{-1}A = I \] where \(I\) is the identity matrix.

Key Concepts

Matrix Transpose

The transpose of a matrix is an operation that flips the matrix over its diagonal. This means that the rows of the original matrix become columns in the transposed matrix and vice versa.
For a matrix denoted as \( A \), its transpose is represented as \( A' \) or sometimes \( A^T \).
The operation has some important properties:

• The transpose of the transpose of a matrix returns the original matrix: \((A')' = A\).
• When dealing with the transpose of a product of matrices, the order of multiplication reverses: \((XY)' = Y'X'\).
• A square matrix becomes symmetric if \( A = A' \), meaning its transpose is equal to itself.

Understanding these properties is crucial when working with transpose in higher-level linear algebra problems. They help simplify expressions and provide shortcuts in calculations.

Matrix Inverse

A matrix inverse is a concept in linear algebra where a matrix, when multiplied by its inverse, results in the identity matrix. For a given matrix \( A \), its inverse is represented as \( A^{-1} \).
This is similar to how the reciprocal of a number works in arithmetic, where multiplying a number by its reciprocal gives 1.
The product of a matrix and its inverse can be expressed as:

• \( AA^{-1} = I \) and \( A^{-1}A = I \), where \( I \) is the identity matrix.

Not all matrices have an inverse. For a matrix to have an inverse, it must be non-singular, meaning it must have a non-zero determinant. In problems involving matrix inverses, it's also important to understand:

• The inverse of a product is the product of the inverses in reverse order: \((AB)^{-1} = B^{-1}A^{-1}\).
• The inverse of the transpose is the transpose of the inverse: \((A^{-1})' = (A')^{-1}\).

These properties simplify the process of finding and using inverses in various computational problems.

Identity Matrix

An identity matrix is a special type of square matrix where all diagonal elements are 1, and all non-diagonal elements are 0.
It plays a crucial role in matrix operations as it acts like the number 1 in arithmetic operations.
When a matrix is multiplied by the identity matrix, it remains unchanged:

• \( AI = A \) and \( IA = A \) for any matrix \( A \) of compatible dimensions.

The identity matrix is pivotal when discussing matrix inverses. If \( A \) is non-singular and \( A^{-1} \) is its inverse, their product yields the identity matrix:

• \( AA^{-1} = I \) and \( A^{-1}A = I \).

Understanding the identity matrix is fundamental in solving systems of equations and in computer calculations where matrices are used for transformations and other operations.