Chapter 5: Problem 28
Matrix \(A\) is such that \(A^{2}=2 A-I\), where \(I\) is unit matrix then for \(n \geq 2, A^{n}=\) (A) \(n A-(n-1) I\) (B) \(n A-I\) (C) \(2^{n-1} A-(n-1) I\) (D) \(2^{n-1} A-I\)
Short Answer
Expert verified
Option (C): \( 2^{n-1}A - (n-1)I \).
Step by step solution
01
Identify the given relationship
We are given that matrix \( A \) satisfies the equation \( A^2 = 2A - I \), where \( I \) is the identity matrix. This equation is crucial as it provides a recursive relation between matric powers.
02
Check basis for induction
We check for \( n=2 \), \( A^2 = 2A - I \). This matches the given equation. This is our base case for further induction steps.
03
Assume for k-th term
Assume that for some integer \( k \geq 2 \), the pattern \( A^k = 2^{k-1}A - (k-1)I \) holds true. This is the induction hypothesis.
04
Inductive Step
We need to show that \( A^{k+1} = 2^kA - kI \). Using the induction hypothesis: \[ A^{k+1} = A^k \cdot A \]\[ = (2^{k-1}A -(k-1)I)A \] \[ = 2^{k-1}A^2 - (k-1)A \] Substitute the given \( A^2 \) relation: \[ = 2^{k-1}(2A - I) - (k-1)A \]\[ = 2^kA - 2^{k-1}I - (k-1)A \] \[ = (2^k -k+1)A - 2^{k-1}I \] Since \( 2^{k-1}(A^2) = 2^k A - 2^{k-1} I \), conclude that it holds \( A^{k+1} = 2^k A - kI \).
05
Conclusion of Induction
By completing the inductive step, we have shown that the expression \( A^n = 2^{n-1}A - (n-1)I \) holds for all \( n \geq 2 \).
06
Match with Options
The pattern \( A^n = 2^{n-1}A - (n-1)I \) matches with option (C), thus the answer is (C).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Induction
Induction is a powerful mathematical technique used to prove statements for all natural numbers. It works in two main steps: the base case and the inductive step. - **Base Case**: This is where we verify that the statement holds true for the initial value, commonly when the variable is the smallest integer under consideration.- **Inductive Step**: In this step, we assume the statement holds for some specific integer, often denoted as \( k \). Based on this assumption, we prove that the statement also holds for \( k+1 \).In our exercise, induction helps us verify the pattern of matrix powers. By checking that the base case \( A^2 = 2A - I \) is true, we can assume some pattern holds for matrix power \( A^k \). Then, we use this assumption to show that if it is true for \( A^k \), it is also true for \( A^{k+1} \). This completes the induction process, confirming our hypothesis for all \( n \geq 2 \).
Identity Matrix
The identity matrix, denoted as \( I \), is a special type of matrix with 1s on the diagonal and 0s elsewhere. It acts as the multiplicative identity in the realm of matrices, much like the number 1 in arithmetic. - When any matrix \( A \) is multiplied by an identity matrix \( I \) of the compatible size, the result is the matrix \( A \) itself. This property is crucial in equations like our given \( A^2 = 2A - I \).- Additionally, since \( I \) maintains the structure of the matrix multiplication, it simplifies many processes in proofs and calculations, such as establishing patterns that involve matrices, like recursive relations.The identity matrix serves as a reference point in our solution, helping define when and how matrices transform. This utility is reflected in expressing matrix powers that reuse patterns involving \( I \).
Recursive Relation
Recursive relations describe sequences where each term is defined as a function of preceding terms. Such relations are invaluable in mathematics for defining sequences based on simpler calculations. - A recursive relation breaks down a complex problem by reducing it to simpler subproblems.- In our context, the recursive relation originates from the given equation \( A^2 = 2A - I \). This equation sets up how we can compute subsequent powers of matrix \( A \).For the third step in our exercise, knowing that \( A^k = 2^{k-1}A - (k-1)I \) sets up a base to effectively anticipate the next power \( A^{k+1} \). By substituting this relation into subsequent calculations, we efficiently build up the chain of matrix powers, utilizing the recursive logic to handle transitions from one power of \( A \) to the next. This logic helps in identifying patterns in matrix exponentiation and validates our final solution, matching option (C).