Problem 28
Question
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) \(2^{n-1} A-I\)
Step-by-Step Solution
Verified Answer
The correct expression is (A) \(n A - (n-1) I\).
1Step 1: Understanding the given equation
The equation given is \(A^2 = 2A - I\), where \(A\) is a matrix and \(I\) is the identity matrix. We need to use this equation to express \(A^n\), for \(n \geq 2\).
2Step 2: Rewriting the equation
From the equation \(A^2 = 2A - I\), we can rearrange it into a recurrence form: \(A^2 = 2A - I\). We aim to derive higher powers of \(A\) based on this relation.
3Step 3: Calculating further powers of A using induction
Assume that \(A^k = kA - (k-1)I\) holds true for some \(k\), and we will prove it for \(k + 1\). Then \(A^{k+1} = A \cdot A^k = A(kA - (k-1)I) = kA^2 - (k-1)A.\)
4Step 4: Substituting the given equation into the induction step
Using \(A^2 = 2A - I\) in \(A^{k+1} = kA^2 - (k-1)A\), we have \(A^{k+1} = k(2A - I) - (k-1)A = 2kA - kI - (k-1)A.\)
5Step 5: Simplifying the expression
Simplify \(A^{k+1} = (2k - (k-1))A - kI\), which becomes \((k + 1)A - kI\). This confirms \(A^{k+1} = (k+1)A - kI\) holds true.
6Step 6: Concluding with the derived formula
Since the induction holds, the formula for \(A^n\) is \(A^n = nA - (n-1)I\) for \(n \geq 2\).
Key Concepts
Recurrence RelationInductive ProofIdentity MatrixMatrix Algebra
Recurrence Relation
A recurrence relation expresses each term of a sequence as a function of its preceding terms. This is a crucial concept in sequences involving matrix exponentiation, where the power of a matrix can be expressed in relation to its previous powers.
In our problem, we have the equation:
To derive the higher powers of \( A \), we continue to apply this pattern, ensuring every new expression includes the previous powers, precisely as in a sequence governed by recurrence relations.
In our problem, we have the equation:
- Given: \( A^2 = 2A - I \).
To derive the higher powers of \( A \), we continue to apply this pattern, ensuring every new expression includes the previous powers, precisely as in a sequence governed by recurrence relations.
Inductive Proof
Inductive proof is a method used to prove statements about all natural numbers. To prove a statement by induction, you typically follow these steps:
This logic of induction is integral in proving patterns like the one derived in the exercise.
- **Base Case:** Verify the statement holds for the initial value, often \( n=2 \) for problems involving sequences starting from this point.
- **Inductive Step:** Assume it holds for some natural number \( k \), and then demonstrate it holds for \( k+1 \).
- We know \( A^2 = 2A - I \), aligning with our proposed form, confirming the base case.
This logic of induction is integral in proving patterns like the one derived in the exercise.
Identity Matrix
The identity matrix, denoted by \( I \), is like the number 1 in matrix algebra. It possesses unique properties that simplify matrix equations. It is a square matrix with ones on the diagonal and zeros elsewhere.
- **Property:** \( A \cdot I = A \).
- **Relevance:** The equation \( A^2 = 2A - I \) entails subtracting this identity matrix, which impacts the expressions for subsequent powers of \( A \).
Matrix Algebra
Matrix algebra is the foundation of numerous operations and transformations in mathematics, providing a framework for manipulating matrices through addition, multiplication, scalar products, and more.
- **Matrix Multiplication:** Matrices can be multiplied, following specific rules, and this operation is non-commutative.
- **Application:** The equation \( A^2 = 2A - I \) and calculating \( A^{k+1} \) are done using these principles.
Other exercises in this chapter
Problem 26
If \(a b c=p\) and \(A=\left[\begin{array}{lll}a & b & c \\ c & a & b \\ b & c & a\end{array}\right]\) such that \(A A^{\prime}=I\), then \(a, b, c\) are the ro
View solution Problem 27
If \(A\) is a singular matrix, then \(\operatorname{adj} A\) is (A) non-singular (B) singular (C) symmetric (D) not defined
View solution Problem 29
For each real number \(x\) such that \(-1
View solution Problem 30
The inverse of a skew-symmetric matrix of odd order is (A) a symmetric matrix (B) a skew-symmetric matrix (C) diagonal matrix (D) does not exist
View solution