Problem 4 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(A\) is an involutory matrix and \(I\) is unit matrix of the same order then, \((I-A)(I+A)=\) (A) 0 (B) \(A\) (C) \(I\) (D) \(2 A\)

Step-by-Step Solution

Answer

(I - A)(I + A) = 0. So, the answer is (A) 0.

Step 1: Understanding Involutory Matrices

An involutory matrix is defined as a matrix that is its own inverse. This means for an involutory matrix \( A \), \( A^2 = I \), where \( I \) is the identity matrix of the same order as \( A \).

Step 2: Expression Analysis

We need to simplify the expression \((I - A)(I + A)\). Recognize this as a difference of squares: \((x-y)(x+y) = x^2 - y^2\). Here, let \( x = I \) and \( y = A \).

Step 3: Applying Difference of Squares

Applying the difference of squares formula to our expression, we have: \((I - A)(I + A) = I^2 - A^2\).

Step 4: Simplifying \(I^2\)

The identity matrix \( I \) always satisfies \( I^2 = I \). This follows because multiplying the identity matrix by itself still leaves the identity matrix.

Step 5: Simplifying \(A^2\)

Given that \( A \) is involutory, \( A^2 = I \).

Step 6: Final Calculation

Substituting the simplified expressions back, we have: \((I - A)(I + A) = I - I = 0\). Therefore, the result is 0.

Key Concepts

Identity Matrix

The identity matrix, often denoted by the symbol \( I \), plays a pivotal role in linear algebra due to its unique properties. An identity matrix is a special type of square matrix where all the elements of the principal diagonal are ones, and all other elements are zeros. For example, a 2x2 identity matrix looks like this:

• \( I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \)

This matrix acts as the multiplicative identity in matrix multiplication, which means that for any matrix \( A \) of corresponding dimensions:

• \( AI = A \) and \( IA = A \)

The property of the identity matrix helps simplify many matrix expressions. Furthermore, like the number 1 in regular arithmetic, multiplying any matrix by an identity matrix leaves the original matrix unchanged. This is crucial when dealing with matrix equations and finding inverses.

Matrix Inverses

Matrix inverses are important in solving linear equations and understanding matrix transformations. If a matrix \( A \) has an inverse, then there exists another matrix, denoted as \( A^{-1} \), such that:

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

Where \( I \) is the identity matrix. Not all matrices have inverses; a matrix must be square and have a non-zero determinant to be invertible. One interesting case is involutory matrices. An involutory matrix is its own inverse, which means for such a matrix \( A \):

• \( A^2 = I \)

This characteristic makes involutory matrices particularly fascinating, as they demonstrate the concept of matrices being able to return to the identity matrix with simple operations. Understanding this concept is key when analyzing expressions like the exercise's \((I - A)(I + A)\).

Difference of Squares

The difference of squares is a simple yet powerful algebraic identity that is given by:

• \((x - y)(x + y) = x^2 - y^2 \)

This formula allows us to simplify and manipulate expressions that look like products of binomials, where one is a subtraction and the other an addition of the same terms. In matrices, this is particularly useful for expressions involving the identity and involutory matrices, as shown in the original exercise.
To apply this principle to matrices, we let \( x = I \) (the identity matrix) and \( y = A \) (the involutory matrix), resulting in:

• \((I - A)(I + A) = I^2 - A^2 \)

With the identity \( I^2 = I \) and the involutory property \( A^2 = I \), this simplifies to \( I - I = 0 \). This demonstrates how algebraic identities can be extended and applied within the realm of matrices to simplify and solve problems.