Elementary operations基本変形きほんへんけい and systems of linear equations連立一次方程式れんりついちじほうていしき - Basic Exercises

date2026-07-14document_iddoc_d65904d689ae98215ae62570ca47b693description行基本変形・列基本変形・拡大係数行列を、何を保存し何を変えるかに注目して確認する入口演習である。prerequisites連立一次方程式と拡大係数行列 / 行基本変形の基本 / 列基本変形の基本 / 連立一次方程式と掃き出し法type問題演習content_typeexercisestatusactive

mathlinear-algebraexerciserow-operationcolumn-operation

data/lecture/math/linear-algebra/linear-systems-and-augmented-matrices.lecture.n.md data/lecture/math/linear-algebra/elementary-row-operations.lecture.n.md data/lecture/math/linear-algebra/elementary-column-operations.lecture.n.md data/lecture/math/linear-algebra/linear-systems-and-gaussian-elimination.lecture.n.md

1Exercise plan

Elementary row operations行基本変形ぎょうきほんへんけい transform equations into equivalent同値どうち equations, so they preserve the solution set解集合かいしゅうごう. Elementary column operations列基本変形れつきほんへんけい preserve column space列空間れつくうかん and rank階数かいすう, but change the meaning of the unknowns未知数みちすう.

This file checks what each operation preserves and what it changes.

2Problem 1

Write the system

\begin{cases} x+2y=5,\\ 3x+4y=11 \end{cases}\begin{cases} x+2y=5,\\ 3x+4y=11 \end{cases}

as an augmented matrix拡大係数行列かくだいけいすうぎょうれつ and solve it by elementary row operations行基本変形ぎょうきほんへんけい.

2.1Answer example

\left(\begin{array}{cc|c}1&2&5\\3&4&11\end{array}\right) \to \left(\begin{array}{cc|c}1&2&5\\0&-2&-4\end{array}\right) \to \left(\begin{array}{cc|c}1&2&5\\0&1&2\end{array}\right) \to \left(\begin{array}{cc|c}1&0&1\\0&1&2\end{array}\right).\left(\begin{array}{cc|c}1&2&5\\3&4&11\end{array}\right) \to \left(\begin{array}{cc|c}1&2&5\\0&-2&-4\end{array}\right) \to \left(\begin{array}{cc|c}1&2&5\\0&1&2\end{array}\right) \to \left(\begin{array}{cc|c}1&0&1\\0&1&2\end{array}\right).

Therefore $x = 1$ and $y = 2$ .

2.2Explanation

Row operations swap equations, multiply an equation by a nonzero constant, or add a multiple of one equation to another. Each has an inverse, so the solution set is unchanged.

3Problem 2

Decide whether the system represented by

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

has a solution解かい.

3.1Answer example

The second row represents $0 x + 0 y = 3$ , which is impossible. Therefore no solution exists.

3.2Explanation

Distinguish a zero row零行れいぎょう from a contradictory row矛盾行むじゅんぎょう. A zero right-hand side adds no condition; a nonzero right-hand side creates contradiction.

4Problem 3

List the three types of elementary row operations行基本変形ぎょうきほんへんけい and state what they preserve.

4.1Answer example

The three types are row exchange, multiplication of a row by a nonzero constant, and adding a constant multiple of one row to another. All preserve the solution set解集合かいしゅうごう.

4.2Explanation

Multiplication by 0 is not an elementary row operation because it erases information and cannot be undone.

5Problem 4

For

A = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}),

perform $C_{2} \leftarrow C_{2} - 2 C_{1}$ . Find the new matrix and explain what changes for the column space and for $A x = b$ .

5.1Answer example

The new matrix is

(\begin{matrix} 1 & 0 \\ 3 & - 2 \end{matrix}) .

A column operation preserves the column space列空間れつくうかん and rank階数かいすう, but it does not preserve the meaning of the unknown variables in $A x = b$ .

5.2Explanation

Column operations replace the generating system of column vectors. The reachable outputs do not change, but the direction assigned to each unknown changes.

6Problem 5

Read the solution set解集合かいしゅうごう from

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

6.1Answer example

Pivots are in columns 1 and 2, so $x_{1}, x_{2}$ are leading variables主変数しゅへんすう and $x_{3}$ is a free variable自由変数じゆうへんすう. Let $x_{3} = t$ . Then

x_{2} = 4 - 3 t, x_{1} = - 8 + 7 t .

Thus

(\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} - 8 \\ 4 \\ 0 \end{matrix}) + t (\begin{matrix} 7 \\ - 3 \\ 1 \end{matrix}) (t \in R) .

6.2Explanation

An echelon form階段形かいだんけい reveals the structure of solutions. Columns without pivots give free variables, and zero rows add no conditions.

7Problem 6

Point out the error:

"Column operations preserve rank階数かいすう, so $A x = b$ and $A F x = b$ have the same solution set解集合かいしゅうごう."

7.1Answer example

Preserving rank is different from preserving the solution set. $A F x = b$ means $A (F x) = b$ . If the original unknown is $z = F x$ , then the equation becomes $A z = b$ . Comparing solutions requires tracking this change of variables未知数変換みちすうへんかん.

7.2Explanation

Row operations change equations equivalently. Column operations change coordinates on the input side, so the preserved object is different.

8Related exercises

data/exercise/math/linear-algebra/echelon-forms-and-gaussian-elimination.exercise.n.md data/exercise/math/linear-algebra/elementary-column-operations-and-variable-changes.exercise.n.md

9Related lectures

data/lecture/math/linear-algebra/linear-systems-and-augmented-matrices.lecture.n.md data/lecture/math/linear-algebra/elementary-row-operations.lecture.n.md data/lecture/math/linear-algebra/elementary-column-operations.lecture.n.md 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

10Proof exercise: row operations preserve solution sets

10.1Problem

Explain why applying row operations to an augmented matrix does not change the solution set of the system.

10.2Answer

Swapping rows only changes equation order. Multiplying a row by a nonzero constant multiplies both sides by a nonzero number and can be undone. Adding a multiple of one row to another adds a linear combination of equations and can also be undone.

10.3Explanation

Row operations are invertible equivalent transformations, so they can make a system easier to solve without changing its solutions.