1148: For Commutative Ring, Inverse of Transpose of Matrix Is Transpose of Inverse of Matrix

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description/proof of that for commutative ring, inverse of transpose of matrix is transpose of inverse of matrix

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Topics

About: matrices space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of transpose of matrix.
• The reader admits the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order.

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Target Context

• The reader will have a description and a proof of the proposition that for any commutative ring, the inverse of the transpose of any invertible matrix is the transpose of the inverse of the matrix.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the commutative rings }\}$
$M$: $\in \{\text{ the }m\times m\text{ invertible }R\text{ matrices }\}$
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Statements:
${M^t}^{-1}={M^{-1}}^t$
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2: Proof

Whole Strategy: Step 1: see that ${M^{-1}}^t{M}^t=I$ and see that ${M^t{M}^{-1}}^t=I$.

Step 1:

${M^{-1}}^t{M}^t=(M{M}^{-1}{)}^t$, by the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order, $=I^t=I$.

${M^t{M}^{-1}}^t=(M^{-1}M{)}^t$, by the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order, $=I^t=I$.

That means that ${M^t}^{-1}={M^{-1}}^t$.

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References

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