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Echelon forms and elimination - Basic Exercisesmd 4e31541
exercise/math/linear-algebra/echelon-forms-and-gaussian-elimination.exercise.n.md

Echelon forms階段形かいだんけい and elimination - Basic Exercises

date2026-06-06document_iddoc_ca4c6b1a246b65429f5f693c4483a658description階段形、簡約階段形、掃き出し法、逆行列計算を、ピボットと場合分けに注目して確認する演習である。prerequisites階段形と簡約階段形 / 連立一次方程式と掃き出し法 / 逆行列の計算手順type問題演習content_typeexercisestatusactiverelateddata/lecture/math/linear-algebra/row-echelon-and-reduced-row-echelon-forms.lecture.n.md / data/lecture/math/linear-algebra/linear-systems-and-gaussian-elimination.lecture.n.md / data/lecture/math/linear-algebra/computing-inverse-matrices.lecture.n.md / data/exercise/math/linear-algebra/elementary-operations-and-linear-systems.exercise.n.md
mathlinear-algebraexerciserrefeliminationinverse-matrix
data/lecture/math/linear-algebra/row-echelon-and-reduced-row-echelon-forms.lecture.n.md data/lecture/math/linear-algebra/linear-systems-and-gaussian-elimination.lecture.n.md data/lecture/math/linear-algebra/computing-inverse-matrices.lecture.n.md

1Exercise plan

An echelon form階段形かいだんけい is a form for reading leading variables主変数しゅへんすう, free variables自由変数じゆうへんすう, and contradictory rows矛盾行むじゅんぎょう from pivotピボット positions. In elimination, confirm that a pivot is nonzero before dividing by it.


2Problem 1

Transform the following augmented matrix拡大係数行列かくだいけいすうぎょうれつ to row echelon form行階段形ぎょうかいだんけい and reduced row echelon form簡約行階段形かんやくぎょうかいだんけい, and read off the solution set解集合かいしゅうごう.

\left(\begin{array}{ccc|c} 1&2&-1&0\\ 2&4&-2&0\\ 0&1&3&4 \end{array}\right)

2.1Answer example

\left(\begin{array}{ccc|c}1&2&-1&0\\2&4&-2&0\\0&1&3&4\end{array}\right) \to \left(\begin{array}{ccc|c}1&2&-1&0\\0&0&0&0\\0&1&3&4\end{array}\right) \to \left(\begin{array}{ccc|c}1&2&-1&0\\0&1&3&4\\0&0&0&0\end{array}\right).

This is row echelon form. Applying R1R1-2R2 gives

\left(\begin{array}{ccc|c}1&0&-7&-8\\0&1&3&4\\0&0&0&0\end{array}\right).

Let x3=t. Then

(x1x2x3)=(-840)+t(7-31)(tR).

2.2Explanation

Row echelon form is made by forward elimination. Reduced row echelon form also zeros out entries above pivot columns. The number of solutions is determined by pivots and contradictory rows.


3Problem 2

For

\left(\begin{array}{ccc|c} 1&0&2&3\\ 0&1&-1&1\\ 0&0&a-1&b \end{array}\right),

classify values of a,b giving a unique solution, infinitely many solutions, or no solution.

3.1Answer example

If a1, the third row is (a-1)z=b, and z=b/(a-1) is determined. Hence there is a unique solution一意解いちいかい.

If a=1, the third row is 0z=b. When b=0, z is a free variable自由変数じゆうへんすう and there are infinitely many solutions. When b0, the row is contradictory and there is no solution.

3.2Explanation

Before dividing by a symbolic expression, split cases according to whether the denominator can be zero. Here the relevant expression is a-1.


4Problem 3

For

A(a)=(a110),

eliminate (A(a)I) and find A(a)-1.

4.1Answer example

Using a as a pivot would be problematic when a=0. Instead, swap rows first:

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

Thus

A(a)-1=(011-a).

4.2Explanation

This method never divides by a, so it works even for a=0. Only quantities actually placed in denominators require case splits.


5Problem 4

For

A=(1224),

eliminate (AI) and explain why A-1 does not exist.

5.1Answer example

\left(\begin{array}{cc|cc}1&2&1&0\\2&4&0&1\end{array}\right) \to \left(\begin{array}{cc|cc}1&2&1&0\\0&0&-2&1\end{array}\right).

A zero row appears on the left, so the left side cannot become I. Therefore A is not invertible可逆かぎゃく.

5.2Explanation

For inverse computation by elimination, the left side must become the identity matrix単位行列たんいぎょうれつ. A zero row on the left means an input direction has collapsed.


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