2次正方行列 $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ に対して、$tr(A) = a_{11} + a_{22}$, $det(A) = a_{11}a_{22} - a_{12}a_{21}$ と定義する。2次正方行列A, Bとスカラーcについて、以下の等式の真偽を判定し、根拠を述べる。 1. $tr(cA) = c \cdot tr(A)$

はい、承知いたしました。以下に問題の解答を示します。

1. 問題の内容

2次正方行列 $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ に対して、$tr(A) = a_{11} + a_{22}$, $det(A) = a_{11}a_{22} - a_{12}a_{21}$ と定義する。2次正方行列A, Bとスカラーcについて、以下の等式の真偽を判定し、根拠を述べる。

1. $tr(cA) = c \cdot tr(A)$

2. $tr(A+B) = tr(A) + tr(B)$

3. $tr(AB) = tr(A) \cdot tr(B)$

4. $tr(AB) = tr(BA)$

5. $det(cA) = c \cdot det(A)$

6. $det(A+B) = det(A) + det(B)$

7. $det(AB) = det(A) \cdot det(B)$

8. $det(AB) = det(BA)$

2. 解き方の手順

以下、各等式の真偽を判定する。

1. $tr(cA) = c \cdot tr(A)$

$cA = \begin{pmatrix} ca_{11} & ca_{12} \\ ca_{21} & ca_{22} \end{pmatrix}$

$tr(cA) = ca_{11} + ca_{22} = c(a_{11} + a_{22}) = c \cdot tr(A)$

よって真。

2. $tr(A+B) = tr(A) + tr(B)$

$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$, $B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}$

$A+B = \begin{pmatrix} a_{11}+b_{11} & a_{12}+b_{12} \\ a_{21}+b_{21} & a_{22}+b_{22} \end{pmatrix}$

$tr(A+B) = a_{11}+b_{11} + a_{22}+b_{22} = (a_{11}+a_{22}) + (b_{11}+b_{22}) = tr(A) + tr(B)$

よって真。

3. $tr(AB) = tr(A) \cdot tr(B)$

$AB = \begin{pmatrix} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} \end{pmatrix}$

$tr(AB) = a_{11}b_{11} + a_{12}b_{21} + a_{21}b_{12} + a_{22}b_{22}$

$tr(A) \cdot tr(B) = (a_{11}+a_{22})(b_{11}+b_{22}) = a_{11}b_{11}+a_{11}b_{22}+a_{22}b_{11}+a_{22}b_{22}$

一般には $tr(AB) \ne tr(A) \cdot tr(B)$. 例えば $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$とすると、$tr(AB) = 0$, $tr(A)tr(B) = 1 \cdot 1 = 1$.

よって偽。

4. $tr(AB) = tr(BA)$

$BA = \begin{pmatrix} b_{11}a_{11}+b_{12}a_{21} & b_{11}a_{12}+b_{12}a_{22} \\ b_{21}a_{11}+b_{22}a_{21} & b_{21}a_{12}+b_{22}a_{22} \end{pmatrix}$

$tr(BA) = b_{11}a_{11} + b_{12}a_{21} + b_{21}a_{12} + b_{22}a_{22} = a_{11}b_{11} + a_{12}b_{21} + a_{21}b_{12} + a_{22}b_{22} = tr(AB)$

よって真。

5. $det(cA) = c \cdot det(A)$

$cA = \begin{pmatrix} ca_{11} & ca_{12} \\ ca_{21} & ca_{22} \end{pmatrix}$

$det(cA) = (ca_{11})(ca_{22}) - (ca_{12})(ca_{21}) = c^2(a_{11}a_{22} - a_{12}a_{21}) = c^2 det(A)$

よって$det(cA) = c^2det(A)$なので、一般には偽。

6. $det(A+B) = det(A) + det(B)$

例えば$A= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, $B=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$のとき$det(A)=1, det(B)=1, A+B=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}, det(A+B)=4$なので、$det(A+B) \ne det(A)+det(B)$.

よって偽。

7. $det(AB) = det(A) \cdot det(B)$

$det(AB) = (a_{11}b_{11}+a_{12}b_{21})(a_{21}b_{12}+a_{22}b_{22}) - (a_{11}b_{12}+a_{12}b_{22})(a_{21}b_{11}+a_{22}b_{21})$

$= a_{11}b_{11}a_{21}b_{12} + a_{11}b_{11}a_{22}b_{22} + a_{12}b_{21}a_{21}b_{12} + a_{12}b_{21}a_{22}b_{22} - a_{11}b_{12}a_{21}b_{11} - a_{11}b_{12}a_{22}b_{21} - a_{12}b_{22}a_{21}b_{11} - a_{12}b_{22}a_{22}b_{21}$

$det(A)det(B) = (a_{11}a_{22} - a_{12}a_{21})(b_{11}b_{22} - b_{12}b_{21}) = a_{11}a_{22}b_{11}b_{22} - a_{11}a_{22}b_{12}b_{21} - a_{12}a_{21}b_{11}b_{22} + a_{12}a_{21}b_{12}b_{21}$

上式と下式を比較すると一致する。

よって真。

8. $det(AB) = det(BA)$

$det(BA) = det(B)det(A) = det(A)det(B)$ なので、7より $det(AB) = det(BA)$

よって真。

3. 最終的な答え

1. 真

2. 真

3. 偽

4. 真

5. 偽

6. 偽

7. 真

8. 真