Math Problem Statement

Suppose A, B, C are 3x3 matrices such that A^-1, B^-1, and C^-1 are given. Then solve ((1/2 A)^2 C(2B)^T)^-1.

Solution

We are tasked with solving the expression:

((12A)2C(2B)T)1\left( \left( \frac{1}{2} A \right)^2 C (2B)^T \right)^{-1}

We are provided with the inverses of the matrices A1A^{-1}, B1B^{-1}, and C1C^{-1}, but we do not need to find AA, BB, or CC explicitly. Instead, we'll use matrix properties and inverses to simplify the expression.

Step-by-Step Approach:

  1. Rewrite the Expression:

    ((12A)2C(2B)T)1\left( \left( \frac{1}{2} A \right)^2 C (2B)^T \right)^{-1}

    Using matrix properties:

    • (12A)2=14A2\left( \frac{1}{2} A \right)^2 = \frac{1}{4} A^2
    • (2B)T=2BT(2B)^T = 2B^T

    This simplifies the expression to:

    = \left( \frac{1}{2} A^2 C B^T \right)^{-1}$$
  2. Apply the Inverse:

    The inverse of a product of matrices (XYZ)1=Z1Y1X1(XYZ)^{-1} = Z^{-1} Y^{-1} X^{-1}, so applying the inverse:

    (12A2CBT)1=2(BT)1C1(A2)1\left( \frac{1}{2} A^2 C B^T \right)^{-1} = 2 (B^T)^{-1} C^{-1} (A^2)^{-1}

  3. Simplify Further:

    • (BT)1=(B1)T(B^T)^{-1} = (B^{-1})^T (since the inverse of a transpose is the transpose of the inverse),
    • (A2)1=(A1)2(A^2)^{-1} = (A^{-1})^2 (since the inverse of A2A^2 is the square of A1A^{-1}).

    Thus, the expression becomes:

    2(B1)TC1(A1)22 (B^{-1})^T C^{-1} (A^{-1})^2

Conclusion:

The final simplified expression is:

2(B1)TC1(A1)22 (B^{-1})^T C^{-1} (A^{-1})^2

You can now substitute the given values of A1A^{-1}, B1B^{-1}, and C1C^{-1} to compute the final result.

Would you like the explicit calculation with the matrix values? Let me know if you'd like more details on any step!

Related Questions:

  1. What is the inverse of a product of two matrices?
  2. How do you compute the transpose of a matrix product?
  3. Why does (A2)1=(A1)2(A^2)^{-1} = (A^{-1})^2 hold?
  4. How can the inverse of a scalar multiple of a matrix be computed?
  5. How do matrix inverses interact with transposes?

Tip:

Always remember that the inverse of a product of matrices is the product of the inverses in reverse order!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Matrix Operations
Matrix Inverses
Matrix Transpose
Scalar Multiplication in Matrices

Formulas

(A^2)^{-1} = (A^{-1})^2
(B^T)^{-1} = (B^{-1})^T
Inverse of a product: (XYZ)^{-1} = Z^{-1} Y^{-1} X^{-1}
Scalar multiplication: (kA)^2 = k^2 A^2

Theorems

Inverse of a Matrix Product
Transpose of Matrix Products
Matrix Inverse and Transpose Properties

Suitable Grade Level

Advanced Undergraduate Level (Grades 12-College)