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Procedure for computing an inverse matrix逆行列ぎゃくぎょうれつ

date2026-07-14document_iddoc_e94951a4026c5a7c233a09b03d3d38f9description逆行列を掃き出し法で計算する手順を、拡大行列 (A|I) から (I|A^{-1}) へ変形する操作として説明する講義である。prerequisites行基本変形 / 単位行列 / 行列の積type講義content_typelecturestatusactiverelateddata/lecture/math/linear-algebra/inverse-matrix-basics.lecture.n.md / data/lecture/math/linear-algebra/elementary-row-operations.lecture.n.md / data/lecture/math/linear-algebra/row-echelon-and-reduced-row-echelon-forms.lecture.n.md / data/lecture/math/linear-algebra/determinant-computation-rules.lecture.n.md / data/exercise/math/linear-algebra/elementary-operations-and-linear-systems.exercise.n.md
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1Introduction

The standard computation of A-1 uses row reduction on the augmented matrix拡大行列かくだいぎょうれつ [AI]. Row operations record reversible changes to the equations, and when the left side becomes I, the right side has become A-1.

2Terms and definitions

Let A be a square matrix正方行列せいほうぎょうれつ. A matrix A-1 satisfying

AA-1=A-1A=I

is called the inverse matrix逆行列ぎゃくぎょうれつ of A.

For computation, form the augmented matrix拡大行列かくだいぎょうれつ

(AI)

and aim, by elementary row operations行基本変形ぎょうきほんへんけい, for

(AI)(IA-1).

3Strategy

Check whether A can be transformed into I. If it can, applying the same row operations to the right block I gives A-1. If it cannot, then A is not invertible可逆かぎゃく.

data/lecture/math/linear-algebra/inverse-matrix-basics.lecture.n.md

4Intuitive explanation

An elementary row operation行基本変形ぎょうきほんへんけい corresponds to multiplying by an invertible matrix可逆行列かぎゃくぎょうれつ on the left. If the combined row operation is E and it changes A into I, then

EA=I.

Therefore E=A-1. Applying the same operations to the right-side I reveals E, so the right block becomes A-1.

5Precise explanation

Applying row operations to

\left(\begin{array}{c|c} A&I \end{array}\right)

is the same as multiplying on the left by an invertible matrix. If the product of the elementary matrices is E, then

\left(\begin{array}{c|c} A&I \end{array}\right) \longmapsto \left(\begin{array}{c|c} EA&E \end{array}\right).

If the left block becomes I, then EA=I. For square matrices, the existence of a left inverse implies invertibility, and E=A-1. Hence

\left(\begin{array}{c|c} A&I \end{array}\right) \longmapsto \left(\begin{array}{c|c} I&A^{-1} \end{array}\right).

The key facts are that elementary row operations are invertible left multiplications and that the identity matrix is the neutral element for composition.

data/lecture/math/linear-algebra/elementary-row-operations.lecture.n.md data/lecture/math/linear-algebra/identity-zero-and-transpose-matrices.lecture.n.md

6Why the procedure works

Each row operation is left multiplication by an invertible可逆かぎゃく elementary matrix. If the combined product of all row-operation matrices is E, then

E[AI]=[EAE].

When EA=I, we have E=A-1. Thus the right block is exactly the inverse.

7Worked example

Let

A=(1237).

Start with the augmented matrix拡大行列かくだいぎょうれつ

\left(\begin{array}{cc|cc} 1&2&1&0\\ 3&7&0&1 \end{array}\right).

Subtract three times the first row from the second row:

\left(\begin{array}{cc|cc} 1&2&1&0\\ 0&1&-3&1 \end{array}\right).

Then subtract two times the second row from the first row:

\left(\begin{array}{cc|cc} 1&0&7&-2\\ 0&1&-3&1 \end{array}\right).

Therefore

A-1=(7-2-31).

8Failure check

If a row of zeros appears on the left before the identity can be formed, the columns of A are dependent and A is not invertible可逆かぎゃく. In that case the augmented computation must stop with the conclusion that no inverse exists.

9Common misunderstandings

  • Row operations must be applied to both blocks of [AI].
  • The right block is not the inverse until the left block is exactly I.
  • Column operations are not the standard inverse algorithm unless variable changes are also tracked.

10Scope

This procedure computes the inverse matrix of a square matrix正方行列せいほうぎょうれつ. For rank-deficient matrices or rectangular matrices, one uses the pseudoinverse matrix擬似逆行列ぎじぎゃくぎょうれつ or the least squares method最小二乗法さいしょうにじょうほう instead.

11Final forms

[PARSE ERROR: Undefined("Command(\"boxed\")")][AI][IA-1]
[PARSE ERROR: Undefined("Command(\"boxed\")")]TheleftblockmustreducetoI.
[PARSE ERROR: Undefined("Command(\"boxed\")")]IftheleftblockfailstobecomeI,noinverseexists[PARSE ERROR: Undefined("RBrace")]
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