Determinants行列式ぎょうれつしき and invertibility - Basic Exercises

date2026-06-30document_iddoc_7b1f80887c5c853fb0a761c560be6430description行列式の計算規則・余因子展開・可逆性判定を、体積倍率と操作ごとの変化に注目して確認する基本演習である。prerequisites行列式 / 置換による行列式の定義 / 行列式の計算規則 / 余因子展開と可逆性の判定 / 逆行列の基本type問題演習content_typeexercisestatusactive

mathlinear-algebraexercisedeterminantinvertible

data/lecture/math/linear-algebra/determinant-computation-rules.lecture.n.md data/lecture/math/linear-algebra/determinants-by-permutations.lecture.n.md data/lecture/math/linear-algebra/cofactor-expansion-and-invertibility.lecture.n.md data/lecture/math/linear-algebra/determinants.lecture.n.md

1Exercise plan

The determinant行列式ぎょうれつしき tells how much a square matrix正方行列せいほうぎょうれつ scales volume and whether orientation is preserved or reversed. In computations, track which row or column operations change the determinant and which leave it unchanged.

Use determinant computations to answer structural questions, not only to get a number. A zero determinant means the columns are dependent, the matrix is not invertible, and the associated square linear map collapses volume to zero. A nonzero determinant means invertibility and gives permission to use inverse-matrix arguments.

Track how each operation affects the determinant. Swapping two rows or columns changes the sign, scaling one row or column scales the determinant, and adding a multiple of one row or column to another leaves it unchanged. These rules are often safer than expanding directly, but the operation and its effect must be recorded.

2Problem 1

Find the determinant of

A = (\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix})

and state its geometric meaning.

2.1Answer example

\det A = 2 \cdot 3 - 1 \cdot 5 = 1.

Thus the linear transformation線型変換せんけいへんかん preserves area by factor 1 and does not reverse orientation.

2.2Explanation

For a $2 \times 2$ matrix, the determinant is the signed area of the parallelogram made by the two column vectors列れつベクトル.

3Problem 2

Suppose $\det A = 6$ . How does the determinant change after each operation?

$R_{1} \leftrightarrow R_{2}$
$R_{3} \leftarrow 4 R_{3}$
$R_{2} \leftarrow R_{2} - 5 R_{1}$

3.1Answer example

$- 6$
$24$
$6$

3.2Explanation

Swapping rows reverses sign. Multiplying a row by 4 multiplies the determinant by 4. Adding a multiple of one row to another does not change the determinant.

4Problem 3

Compute

B = (\begin{matrix} 1 & 0 & 2 \\ 3 & 0 & 4 \\ - 1 & 5 & 6 \end{matrix})

by cofactor expansion余因子展開よいんしてんかい along the second column.

4.1Answer example

The second column is $0, 0, 5$ , so

\det B = 5 (- 1)^{3 + 2} \det (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) = - 5 (4 - 6) = 10.

4.2Explanation

Choose a row or column with many zeros to reduce computation, and do not omit the cofactor sign $(- 1)^{i + j}$ .

5Problem 4

Is

C = (\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{matrix})

invertible可逆かぎゃく? Give a reason.

5.1Answer example

No. Since it is triangular,

\det C = 1 \cdot 1 \cdot 0 = 0.

Therefore its inverse matrix逆行列ぎゃくぎょうれつ does not exist.

5.2Explanation

For square matrices, nonzero determinant is equivalent to invertibility. A zero determinant means volume collapses to 0 and input information cannot be recovered uniquely.

6Problem 5

Decide whether this statement is true:

"If a square matrix正方行列せいほうぎょうれつ has two identical rows, then $\det A = 0$ ."

6.1Answer example

True.

6.2Explanation

Two identical rows are linearly dependent一次従属いちじじゅうぞく. Also, swapping them leaves the matrix unchanged but changes the determinant sign, so $\det A = - \det A$ and hence $\det A = 0$ .

7Supplementary problems: determinant computation and invertibility

7.1Problem 6

Compute $\det (\begin{matrix} 4 & 1 \\ 2 & 3 \end{matrix})$ .

7.2Answer example

4 \cdot 3 - 1 \cdot 2 = 10.

7.3Problem 7

What happens to the determinant when two rows are swapped?

7.4Answer example

The sign is reversed.

7.5Problem 8

What happens to the determinant when two columns are swapped?

7.6Answer example

The sign is reversed.

7.7Problem 9

Construct a $2 \times 2$ matrix with determinant 0.

7.8Answer example

$(\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix})$ has determinant 0 because the second row is twice the first.

7.9Problem 10

Compute a $3 \times 3$ determinant with many zeros in the second column by cofactor expansion.

7.10Answer example

For

A = (\begin{matrix} 1 & 0 & 2 \\ 3 & 5 & 1 \\ 0 & 0 & 4 \end{matrix}),

expanding along the second column gives

\det A = 5 (- 1)^{2 + 2} \det (\begin{matrix} 1 & 2 \\ 0 & 4 \end{matrix}) = 20.

7.11Problem 11

Write the cofactor sign pattern for $3 \times 3$ .

7.12Answer example

(\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}) .

7.13Problem 12

Explain why $A^{- 1}$ does not exist if $\det A = 0$ .

7.14Answer example

$\det A = 0$ means the transformation collapses volume to 0. Information is lost, so inputs cannot be recovered uniquely.

7.15Problem 13

Let column vectors $a, b, c$ satisfy $\det (a, b, c) = 5$ . Using multilinearity and alternation, find:

$\det (a + 2 b, b, c)$
$\det (2 a, c, b)$
$\det (a, b, a + c)$

7.16Answer example

\det (a + 2 b, b, c) = \det (a, b, c) + 2 \det (b, b, c) = 5.

\det (2 a, c, b) = 2 \det (a, c, b) = - 2 \det (a, b, c) = - 10.

\det (a, b, a + c) = \det (a, b, a) + \det (a, b, c) = 5.

7.17Explanation

The terms with two equal columns are 0, and swapping columns reverses sign.

7.18Problem 14

For

A = (\begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix}), B = (\begin{matrix} 7 & - 2 \\ - 3 & 1 \end{matrix}),

confirm that $B = A^{- 1}$ by computing $A B$ and $B A$ .

7.19Answer example

A B = I_{2}, B A = I_{2.}

Therefore $B = A^{- 1}$ .

7.20Problem 15

For the same $A$ and $b = (5, 17)^{T}$ , solve $A x = b$ by $x = A^{- 1} b$ and explain uniqueness.

7.21Answer example

x = A^{- 1} b = (\begin{matrix} 7 & - 2 \\ - 3 & 1 \end{matrix}) (\begin{matrix} 5 \\ 17 \end{matrix}) = (\begin{matrix} 1 \\ 2 \end{matrix}) .

If two solutions existed, their difference would satisfy $A (x_{1} - x_{2}) = 0$ . Multiplying by $A^{- 1}$ gives $x_{1} - x_{2} = 0$ , so the solution is unique.

7.22Problem 16

For

A = (\begin{matrix} 1 & 0 \end{matrix}), B = (\begin{matrix} 1 \\ 0 \end{matrix}),

compute $A B$ and $B A$ and explain why a one-sided identity is not enough for an inverse matrix.

7.23Answer example

A B = I_{1},

but

B A = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) \neq I_{2.}

Thus $B$ is not a two-sided inverse of $A$ .

7.24Problem 17

Suppose $\det A = 6$ . How does the determinant change after:

$C_{1} \leftrightarrow C_{3}$
$C_{2} \leftarrow 3 C_{2}$
$C_{3} \leftarrow C_{3} - 5 C_{1}$

7.25Answer example

$- 6$
$18$
$6$

7.26Problem 18

Find the determinant of

A = (\begin{matrix} 0 & 2 & 1 \\ 1 & 3 & 0 \\ 2 & 4 & 1 \end{matrix})

while tracking row-operation changes.

7.27Answer example

Swap $R_{1} \leftrightarrow R_{2}$ , reversing sign:

(\begin{matrix} 0 & 2 & 1 \\ 1 & 3 & 0 \\ 2 & 4 & 1 \end{matrix}) \to (\begin{matrix} 1 & 3 & 0 \\ 0 & 2 & 1 \\ 2 & 4 & 1 \end{matrix}) .

The additions $R_{3} \leftarrow R_{3} - 2 R_{1}$ and $R_{3} \leftarrow R_{3} + R_{2}$ do not change determinant, giving

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

Then $R_{2} \leftarrow \frac{1}{2} R_{2}$ halves the determinant and gives triangular determinant $2$ , so before that scaling it was $4$ . Undoing the initial swap gives

\det A = - 4.

8Related lectures

data/lecture/math/linear-algebra/determinant-computation-rules.lecture.n.md data/lecture/math/linear-algebra/determinants-by-permutations.lecture.n.md data/lecture/math/linear-algebra/cofactor-expansion-and-invertibility.lecture.n.md data/lecture/math/linear-algebra/determinants.lecture.n.md data/lecture/math/linear-algebra/inverse-matrix-basics.lecture.n.md

9Proof exercise: column addition does not change determinant

9.1Problem

Using column-linearity of the determinant and the fact that the determinant is 0 when two columns are equal, prove that $C_{i} \leftarrow C_{i} + λ C_{j}$ does not change the determinant.

9.2Answer

By linearity in the $i$ -th column,

D (\dots, C_{i} + λ C_{j}, \dots, C_{j}, \dots) = D (\dots, C_{i}, \dots, C_{j}, \dots) + λ D (\dots, C_{j}, \dots, C_{j}, \dots) .

The second term is 0 because two columns are equal. Hence the determinant is unchanged.

9.3Explanation

This proof shows how determinant computation rules follow from multilinearity多重線型性たじゅうせんけいせい.