3次正方行列 $A$ に対して、$\det(A - xI_3) = -x^3 + \operatorname{tr}(A)x^2 - \operatorname{tr}(\tilde{A})x + \det(A)$ が成り立つことを証明せよ。ここで、$\tilde{A}$ は $A$ の余因子行列を表し、$I_3$ は3次の単位行列、$tr$はトレースを表す。

1. 問題の内容

3次正方行列 $A$ に対して、$\det(A - xI_3) = -x^3 + \operatorname{tr}(A)x^2 - \operatorname{tr}(\tilde{A})x + \det(A)$ が成り立つことを証明せよ。ここで、$\tilde{A}$ は $A$ の余因子行列を表し、$I_3$ は3次の単位行列、$tr$はトレースを表す。

2. 解き方の手順

$A = (a_{ij})$ を3次正方行列とする。

$A - xI_3 = \begin{pmatrix} a_{11}-x & a_{12} & a_{13} \\ a_{21} & a_{22}-x & a_{23} \\ a_{31} & a_{32} & a_{33}-x \end{pmatrix}$

とおくと、

$\det(A - xI_3)$ は $x$ の3次多項式として表される。

$\det(A - xI_3) = (a_{11}-x)((a_{22}-x)(a_{33}-x) - a_{23}a_{32}) - a_{12}(a_{21}(a_{33}-x) - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - (a_{22}-x)a_{31})$

$= (a_{11}-x)(a_{22}a_{33} - a_{22}x - a_{33}x + x^2 - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{21}x - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31} + a_{31}x)$

$= a_{11}a_{22}a_{33} - a_{11}a_{22}x - a_{11}a_{33}x + a_{11}x^2 - a_{11}a_{23}a_{32} - a_{22}a_{33}x + a_{22}x^2 + a_{33}x^2 - x^3 + a_{23}a_{32}x - a_{12}a_{21}a_{33} + a_{12}a_{21}x + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} + a_{13}a_{31}x$

$= -x^3 + (a_{11} + a_{22} + a_{33})x^2 + (-a_{11}a_{22} - a_{11}a_{33} - a_{22}a_{33} + a_{23}a_{32} + a_{12}a_{21} + a_{13}a_{31})x + (a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31})$

ここで、$\operatorname{tr}(A) = a_{11} + a_{22} + a_{33}$.

$\det(A) = a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} + a_{12}a_{23}a_{31} - a_{12}a_{21}a_{33} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31}$.

余因子行列 $\tilde{A}$ の対角成分は、

$\tilde{a}_{11} = a_{22}a_{33} - a_{23}a_{32}$

$\tilde{a}_{22} = a_{11}a_{33} - a_{13}a_{31}$

$\tilde{a}_{33} = a_{11}a_{22} - a_{12}a_{21}$

したがって、

$\operatorname{tr}(\tilde{A}) = \tilde{a}_{11} + \tilde{a}_{22} + \tilde{a}_{33} = a_{22}a_{33} - a_{23}a_{32} + a_{11}a_{33} - a_{13}a_{31} + a_{11}a_{22} - a_{12}a_{21}$

よって、

$\det(A - xI_3) = -x^3 + \operatorname{tr}(A)x^2 - \operatorname{tr}(\tilde{A})x + \det(A)$.

3. 最終的な答え

$\det(A - xI_3) = -x^3 + \operatorname{tr}(A)x^2 - \operatorname{tr}(\tilde{A})x + \det(A)$