Math Problem Statement

In each case either prove the assertion or give an example showing that it is false. a. If A ≠ 0 is a square matrix, then A is invertible. b. If A and B are both invertible, then A + B is invertible. c. If A and B are both invertible, then (A⁻¹B)ᵀ is invertible. d. If A⁴ = 3I, then A is invertible. e. If A² = A and A ≠ 0, then A is invertible.

Solution

Let's work through the problems one by one:

Exercise 2.4.9: In each case, either prove the assertion or give an example showing that it is false.

a. If A0A \neq 0 is a square matrix, then AA is invertible.

False. A square matrix being non-zero does not guarantee invertibility. For example, consider the matrix: A=(1000)A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} This is a square matrix, and it is not the zero matrix, but it is not invertible because its determinant is 0 (a requirement for invertibility is that the determinant must be non-zero).

b. If AA and BB are both invertible, then A+BA + B is invertible.

False. The sum of two invertible matrices is not always invertible. Consider the following counterexample: A=(1001),B=(1001)A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} Both AA and BB are invertible, but: A+B=(1001)+(1001)=(0000)A + B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} The resulting matrix is the zero matrix, which is not invertible.

c. If AA and BB are both invertible, then (A1B)T(A^{-1}B)^T is invertible.

True. If both AA and BB are invertible, then their product is also invertible. We know that: (A1B)T=BT(A1)T(A^{-1}B)^T = B^T (A^{-1})^T Both BTB^T and (A1)T(A^{-1})^T are invertible since the transpose of an invertible matrix is also invertible. Therefore, the transpose (A1B)T(A^{-1}B)^T is invertible.

d. If A4=3IA^4 = 3I, then AA is invertible.

True. If A4=3IA^4 = 3I, then: A4=3I    13A4=IA^4 = 3I \implies \frac{1}{3} A^4 = I So, A4A^4 is a scalar multiple of the identity matrix, meaning that AA must be invertible (since we can multiply by 134\frac{1}{\sqrt[4]{3}} and still have a valid inverse matrix).

e. If A2=AA^2 = A and A0A \neq 0, then AA is invertible.

False. This statement is generally false. Consider the matrix: A=(1000)A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} Clearly: A2=(1000)=AA^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = A and A0A \neq 0. However, AA is not invertible because its determinant is 0.


Do you want more details or have any further questions?

Here are five related questions to expand your understanding:

  1. What are the conditions for a matrix to be invertible?
  2. Can the sum of more than two invertible matrices be invertible?
  3. How does the property of the transpose affect invertibility?
  4. Can a matrix with non-zero determinant always be invertible?
  5. How does the rank of a matrix relate to its invertibility?

Tip: Always check the determinant of a matrix when determining if it is invertible!

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Inverses
Determinants
Matrix Transpose

Formulas

A matrix A is invertible if det(A) ≠ 0.
For matrices A and B, (A⁻¹B)ᵀ = Bᵀ(A⁻¹)ᵀ.
If A⁴ = 3I, then A is invertible because A^4 can be reduced to a scalar multiple of the identity matrix.

Theorems

Invertibility criterion: A matrix is invertible if its determinant is non-zero.
The transpose of an invertible matrix is also invertible.

Suitable Grade Level

University Level (Linear Algebra)