Math Problem Statement

Prove or disprove the following statements: (a) If B = A^-1, then A = B^-1. (b) The sum of two invertible matrices need not be an invertible matrix. (c) The sum of two singular matrices of the same order cannot be invertible.

Solution

Let's go through the problem step by step:

(a) Prove or disprove: If B=A1B = A^{-1}, then A=B1A = B^{-1}.

Proof:

If B=A1B = A^{-1}, then we know that: AB=AA1=IA B = A A^{-1} = I where II is the identity matrix. This is the definition of the inverse matrix, meaning that AA and BB are inverses of each other. Therefore, it also follows that: BA=BA1=IB A = B A^{-1} = I This implies that A=B1A = B^{-1}, since BB is the inverse of AA, meaning the statement is true.

(b) Prove or disprove: The sum of two invertible matrices need not be an invertible matrix.

Disproof by example:

Let’s take two invertible matrices: 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, since their determinants are non-zero (det(A)=1\det(A) = 1 and det(B)=1\det(B) = -1). However, their sum is: 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 (since its determinant is zero). Therefore, this statement is true—the sum of two invertible matrices does not have to be invertible.

(c) Prove or disprove: The sum of two singular matrices of the same order cannot be invertible.

Proof:

A singular matrix is one that is not invertible, meaning its determinant is zero. However, it’s possible for the sum of two singular matrices to be invertible. Consider the following example: A=(1111),B=(1000)A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix} Both matrices AA and BB are singular since det(A)=0\det(A) = 0 and det(B)=0\det(B) = 0. Now, consider their sum: A+B=(1111)+(1000)=(0111)A + B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} + \begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} The determinant of A+BA + B is: det(A+B)=(0)(1)(1)(1)=1\det(A + B) = (0)(1) - (1)(1) = -1 Since det(A+B)0\det(A + B) \neq 0, A+BA + B is invertible. Therefore, the statement is false—the sum of two singular matrices can be invertible.


Let me know if you would like more details on any of the proofs or concepts. Here are 5 related questions for further exploration:

  1. How do you compute the inverse of a 2×22 \times 2 matrix?
  2. Can the product of two singular matrices be invertible?
  3. Under what conditions is the sum of two invertible matrices also invertible?
  4. How does matrix addition affect the rank of the resulting matrix?
  5. What properties of matrices ensure that their sum is singular?

Tip: When working with matrix inverses, always check if the matrix is invertible by calculating the determinant. If it's zero, the matrix is singular and non-invertible.

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Math Problem Analysis

Mathematical Concepts

Matrix Inversion
Singular and Invertible Matrices
Matrix Addition
Matrix Determinants

Formulas

A * A^-1 = I (identity matrix)
Det(A + B)

Theorems

Inverse Matrix Theorem
Properties of Matrix Addition
Determinant of Matrix

Suitable Grade Level

Undergraduate Linear Algebra