Problem 13 - Algebra for JEE ADVANCED

Question

Matrix \(A\) such that \(A^{2}=2 A-I\), where \(I\) is the identity matrix, then for \(n \geq 2, A^{n}\) is equal to a. \(2^{n-1} A-(n-1) I\) b. \(2^{n-1} A-I\) c. \(n A-(n-1) I\) d. \(n A-I\)

Step-by-Step Solution

Answer

Choice (a), \(2^{n-1} A - (n-1) I\), is correct.

Step 1: Understanding the Given Equation

Start with the equation given in the problem: \( A^2 = 2A - I \). This equation can be rewritten as \( A^2 - 2A + I = 0 \), meaning that the operator polynomial \( A \) satisfies this equation.

Step 2: Verify the Base Case for Induction

We hypothesize that \( A^n = 2^{n-1}A - (n-1)I \) is correct for \( n = 2 \). Testing the base case: \( A^2 = 2A - I \), which is true according to the given problem.

Step 3: Inductive Step

Assume for some \( k \geq 2 \), the equality \( A^k = 2^{k-1}A - (k-1)I \) holds. Then, we seek to prove it for \( n = k + 1 \), which is \( A^{k+1} \). Multiply both sides of \( A^k = 2^{k-1}A - (k-1)I \) by \( A \): \( A^{k+1} = A(A^k) = A(2^{k-1}A - (k-1)I) \).

Step 4: Simplifying the Expression

Simplify \( A^{k+1} = (2^{k-1}A^2 - (k-1)A) \). Substitute \( A^2 = 2A - I \) into the expression: \( A^{k+1} = 2^{k-1}(2A - I) - (k-1)A \). This simplifies to \( 2^k A - 2^{k-1}I - (k-1)A \).

Step 5: Collect Like Terms

Combine like terms to simplify: \( 2^k A - (k-1)A \) gives \( (2^k - (k-1))A \) and the \( I \) terms yield \( -2^{k-1}I \). Thus, \( A^{k+1} = 2^k A - 2^{k-1} I - (k-1)A = (2^k - (k-1))A - 2^{k-1}I \).

Step 6: Observation and Conclusions

Upon checking carefully, the result \( (2^k A - (k-1)A - 2^{k-1}I) \) matches the form \( 2^{k}A - kI + I = 2^{k}A - (k-1)I \). Recognizing that this is consistent across substitutions, verify that \( A^n = 2^{n-1}A - (n-1)I \) holds true for all \( n \). Thus, (a) \( 2^{n-1} A - (n-1) I \) is correct.

Key Concepts

Identity Matrix

One fundamental concept in matrix algebra is the identity matrix, often denoted as "I". The identity matrix is a square matrix where all the diagonal elements are 1s, and all other elements are 0s. Its primary property is that when any matrix is multiplied by the identity matrix, the original matrix is unchanged. This property is similar to how multiplying a number by 1 yields the number itself.

• For a 2x2 identity matrix: \[ I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \]

In the given problem, the identity matrix "I" plays a crucial role in the equation \( A^2 = 2A - I \). It helps transform matrix equations and ensures properties similar to those in traditional algebra where 0 is the additive identity.

Induction Proof

Mathematical induction is a method used to prove statements that are asserted to hold for all natural numbers. The method involves two main steps: proving the base case and the inductive step. The base case verifies if the assertion holds for the initial number, often \( n = 1 \) or \( n = 2 \).

• Base Case: Here, we check if the hypothesis is true for \( n=2 \), which turns out to be correct, matching the equation \( A^2 = 2A - I \).
• Inductive Step: Assume the hypothesis is true for some \( n=k \), and then prove it for \( n=k+1 \).

In this problem, the induction shows how a matrix power formula can be generalized from one case to every higher case, proving \( A^n = 2^{n-1}A - (n-1)I \) holds for all \( n \) greater than or equal to 2. Inductive reasoning is a powerful tool in mathematics that establishes a pattern's truth through logical steps.

Polynomial Operators

In mathematics, polynomial operators are expressions that involve matrices and constants combined using addition, subtraction, and multiplication. They mirror the concept of polynomials with real numbers. Each term in a polynomial operator can involve different powers of a matrix.
Consider the exercise where \( A^2 = 2A - I \). This equation demonstrates a polynomial operator with the matrix \( A \) and the identity matrix \( I \). The equation can be thought of as a quadratic involving the operator \( A \), similar to solving \( x^2 - 2x + 1 = 0 \) for a real number \( x \).
Understanding polynomial operators allows one to manipulate these matrix forms much like algebraic expressions, facilitating tasks like matrix exponentiation and proofs.

Matrix Exponentiation

Matrix exponentiation involves raising a matrix to a power, meaning multiplying the matrix by itself a certain number of times. This is foundational in exploring dynamic systems, transformations, and patterns in matrices. For example, \( A^n \) signifies the matrix \( A \) multiplied by itself \( n \) times.
During the exercise, matrix exponentiation is explored in terms of polynomial operators to derive formulas for \( A^n \). Having \( A^2 = 2A - I \) as a starting point, the task is to generalize \( A^n \) using patterns derived from multiplying \( A \'s\) to itself iteratively, guided by induction.
The derived formula \( A^n = 2^{n-1}A - (n-1)I \) simplifies the process of matrix exponentiation by providing a direct way to compute powers without recursive multiplication. This approach is particularly useful in computing powers of matrices quickly and efficiently.