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Chapter 5: Problem 55

If \(A\) and \(B\) are two matrices such that \(A B=B A\), then \(\forall n \in N\) (A) \(A^{n} B=B A^{n}\) (B) \((A B)^{n}=A^{n} B^{n}\) (C) \((A+B)^{n}={ }^{n} C_{0} A^{n}+{ }^{n} C_{1} A^{n-1} B+{ }^{n} C_{2} A^{n-2} B^{2}+\ldots+\) \({ }^{n} C_{n} B^{n}\) (D) \(A^{2 n}-B^{2 n}=\left(A^{n}-B^{n}\right)\left(A^{n}+B^{n}\right)\)

Short Answer

Expert verified
True options: (A), (C), (D); False option: (B).

Step by step solution

01

Understand Commutativity

Since \(AB = BA\), matrices \(A\) and \(B\) commute with each other. This means any multiplication with these matrices in succession maintains the order of multiplication.
02

Evaluate Option (A)

We need to show \(A^n B = B A^n\) for all natural numbers \(n\).\ We use induction. For \(n = 1\), \(A B = B A\) is given. Assume true for \(n = k\): \(A^k B = BA^k\). For \(n = k + 1\), \(A^{k+1} B = A^k (A B) = A^k (B A) = (A^k B)A = (B A^k)A = B A^{k+1}\). Thus, \(A^n B = B A^n\) is proven by induction.
03

Evaluate Option (B)

We need to check if \((AB)^n = A^n B^n\). In general, \((AB)^n = ABAB...AB\), an \(n\)-term product, is typically not equal to \(A^n B^n\). While \(AB = BA\), rearranging \(A^n B^n = A\cdot A^{n-1} B \cdot B^{n-1} = A A^{n-1} B B^{n-1}\), needs multiplication rearrangement; hence option (B) is false.
04

Evaluate Option (C) Using Binomial Expansion

The expansion \((A + B)^n = { }^{n}C_{0}A^n + { }^{n}C_{1}A^{n-1}B + { }^{n}C_{2}A^{n-2}B^2 + \ldots + { }^{n}C_{n}B^n\) is the binomial expansion for matrices, valid when \(A\) and \(B\) commute, because the order of multiplication is preserved.
05

Evaluate Option (D) Using Difference of Squares

The expression \(A^{2n} - B^{2n} = (A^n - B^n)(A^n + B^n)\) follows from the identity \(x^2 - y^2 = (x-y)(x+y)\). Since \(A^n\) and \(B^n\) commute, \(A^{2n} - B^{2n} = (A^n)^2 - (B^n)^2 = (A^n - B^n)(A^n + B^n)\). Thus, (D) is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Multiplication
To begin understanding the problem, we need to grasp the concept of matrix multiplication. Unlike simple arithmetic multiplication, the order of matrix multiplication matters. Given two matrices, \(A\) and \(B\), their product \(AB\) might not be equal to \(BA\). However, if \(AB = BA\), these matrices are said to commute.
  • Matrix multiplication involves computing the dot product of rows from the first matrix with columns of the second matrix.
  • The result will be a new matrix where the number of rows is from the first matrix and the number of columns from the second matrix.
Matrices need to be of compatible sizes: the number of columns in the first matrix must equal the number of rows in the second.When two matrices commute, they exhibit special properties that can simplify many proofs, such as those required in this exercise.
Induction Proof
Mathematical induction is a valuable method to prove statements about natural numbers. It involves two main steps: the base case and the induction step. For our exercise, we're using induction to prove properties about matrix commutativity and powers.
  • Base Case: Begin by proving that the property holds for an initial value, usually \(n=1\).
  • Inductive Step: Assume the property holds for an arbitrary natural number \(k\), and then show it holds for \(k+1\).
For instance, given that \(AB = BA\) (commutativity), we showed through induction that \(A^nB = BA^n\) by assuming the case for \(n=k\) and proving it for \(n=k+1\).Using induction reinforces the truth about patterns or sequences that span infinitely many cases.
Binomial Expansion
In our exercise, we explored the binomial expansion in the context of matrix operations. Binomial expansion relates to expanding expressions raised to a power, represented by the formula:\[(A+B)^n = ^nC_0 A^n + ^nC_1 A^{n-1}B + ^nC_2 A^{n-2}B^2 + \ldots + ^nC_n B^n\]
  • This is very similar to the binomial theorem in algebra and works when matrices \(A\) and \(B\) commute, asserting that the order of terms doesn't affect the result.
  • The coefficients \(^nC_k\) are the same as those in simple algebraic expansion, representing the number of combinations.
Binomial expansion allows us to tackle complex powers of sums of matrices elegantly and effectively.
Difference of Squares
The difference of squares is a classic polynomial identity, expressed as:\[x^2 - y^2 = (x-y)(x+y)\]In our matrix setting, this identity is adopted to show the expression:\[A^{2n} - B^{2n} = (A^n - B^n)(A^n + B^n)\]
  • It rests upon the assumption that \(A^n\) and \(B^n\) commute, ensuring the same order applies when these matrices are multiplied or expanded.
  • The identity simplifies expressions involving the powers of matrices, especially when proving equivalences or solving equations related to matrix commutativity.
Applying such a foundational identity in a matrix context exemplifies how mathematical principles can transcend different algebraic structures.
Matrix Powers
Matrix powers involve multiplying a matrix by itself multiple times. Understanding matrix powers is vital when working with sequences or series involving matrices.
  • A matrix \(A\) raised to the power \(n\), denoted \(A^n\), is the matrix product of \(A\), repeated \(n\) times.
  • For commuting matrices \(AB=BA\), properties regarding powers and multiplication become particularly useful in simplifying expressions.
Matrix powers often appear in solutions related to geometric transformations, Markov chains, and dynamic systems. Recognizing how commute and power relations affect matrix computations is key for advancing in linear algebra.

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Most popular questions from this chapter

Which of the following is correct? (A) If \(A\) is a symmetric matrix, then \(A^{n}\) is symmetric, \(n \in N\) (B) If \(A\) is a skew-symmetric matrix then \(A^{n}\) is symmetric if \(n\) is even, \(\overline{n \in N}\) (C) If \(A\) is a skew-symmetric matrix then \(A^{n}\) is skew-symmetric if \(n\) is odd, \(n \in N\) (D) All of these

Let \(A\) and \(B\) be two symmetric matrices of order \(3 .\) Statement 1: \(A(B A)\) and \((A B) A\) are symmetric matrices. Statement \(2: A B\) is symmetric matrix if matrix multiplication of \(A\) with \(B\) is commutative. (A) Statement 1 is false, Statement 2 is true (B) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1 (C) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 (D) Statement 1 is true, Statement 2 if false

If \(A^{-1}=\left[\begin{array}{rrr}1 & 0 & -2 \\ -2 & 1 & 0 \\ -1 & 1 & 0\end{array}\right]\), then (A) \(|A|=2\) (B) \(a d j \cdot A=\left[\begin{array}{ccc}\frac{1}{2} & 0 & -1 \\ -1 & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{1}{2} & 0\end{array}\right]\) (C) \(|a d j, A|=4\) (D) \(\left|A^{\prime}\right|=\frac{1}{2}\)

Let \(A=\left[\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\\ 0 & 0 & 5\end{array}\right]\), If \(\left|A^{2}\right|=25\) then \(|\alpha|\) equals (A) 5 (B) 1 (C) \(1 / 5\) (D) 5

For each real number \(x\) such that \(-1
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