Math Problem Statement

Given that matrices A and B are nonsingular, prove that \( \begin{pmatrix} A & 0 \\ C & B \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ -B^{-1}CA^{-1} & B^{-1} \end{pmatrix} \).

Solution

This problem involves proving the inverse of a block matrix, given that AA and BB are nonsingular (invertible) matrices. The problem states:

Given matrices AA and BB as nonsingular (non-zero determinants), we need to show that: (A0CB)1=(A10B1CA1B1).\begin{pmatrix} A & 0 \\ C & B \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ -B^{-1}CA^{-1} & B^{-1} \end{pmatrix}.

Solution Outline

For a block matrix of the form M=(A0CB)M = \begin{pmatrix} A & 0 \\ C & B \end{pmatrix}, where AA and BB are invertible, the inverse can be derived using the formula: M1=(A0CB)1=(A10B1CA1B1).M^{-1} = \begin{pmatrix} A & 0 \\ C & B \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ -B^{-1}CA^{-1} & B^{-1} \end{pmatrix}.

To verify this, we can multiply MM and M1M^{-1} to check if we get the identity matrix:

  1. Calculate MM1M \cdot M^{-1}: (A0CB)(A10B1CA1B1).\begin{pmatrix} A & 0 \\ C & B \end{pmatrix} \begin{pmatrix} A^{-1} & 0 \\ -B^{-1}CA^{-1} & B^{-1} \end{pmatrix}.

  2. Multiply the matrices component-wise:

    • The top-left block: AA1+0B1CA1=I.A \cdot A^{-1} + 0 \cdot -B^{-1}CA^{-1} = I.

    • The top-right block: A0+0B1=0.A \cdot 0 + 0 \cdot B^{-1} = 0.

    • The bottom-left block: CA1+BB1CA1=0.C \cdot A^{-1} + B \cdot -B^{-1}CA^{-1} = 0.

    • The bottom-right block: C0+BB1=I.C \cdot 0 + B \cdot B^{-1} = I.

  3. Result: After performing these calculations, we find that the product MM1M \cdot M^{-1} is indeed the identity matrix II, thus confirming that (A0CB)1=(A10B1CA1B1).\begin{pmatrix} A & 0 \\ C & B \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ -B^{-1}CA^{-1} & B^{-1} \end{pmatrix}.

This completes the proof.

Would you like additional details on any part of the calculation?

Related Questions

  1. What conditions are necessary for a block matrix to have an inverse?
  2. How does the presence of zero entries in the matrix simplify the inversion process?
  3. What changes in the formula if C=0C = 0?
  4. Can we generalize this inversion technique for other block matrix forms?
  5. What is the practical significance of inverting block matrices in applications?

Tip

For large matrices, breaking them into blocks can simplify computations, especially in cases where some blocks are zeros.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Inversion
Block Matrix

Formulas

Block matrix inverse formula: \( \begin{pmatrix} A & 0 \\ C & B \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ -B^{-1}CA^{-1} & B^{-1} \end{pmatrix} \)

Theorems

Inverse of block matrices

Suitable Grade Level

Undergraduate Level