## 每週問題 May 19, 2014

Let $A$ be a $3\times 3$ nonsingular matrix. Show that

$\displaystyle A^{-1}=\frac{1}{\det A}\left(A^2-(\hbox{trace}A)A+\frac{(\hbox{trace}A)^2-\hbox{trace}(A^2)}{2}I\right)$.

\displaystyle \begin{aligned} p(t)&=\begin{vmatrix} a_{11}-t&a_{12}&a_{13}\\ a_{21}&a_{22}-t&a_{23}\\ a_{31}&a_{32}&a_{33}-t \end{vmatrix}\\ &=-t^3+(a_{11}+a_{22}+a_{33})t^2-\left(\begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix}+\begin{vmatrix} a_{11}&a_{13}\\ a_{31}&a_{33} \end{vmatrix}+\begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}\right)t+\begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix}\\ &=-t^3+(\hbox{trace}A)t^2-\left(\begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix}+\begin{vmatrix} a_{11}&a_{13}\\ a_{31}&a_{33} \end{vmatrix}+\begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}\right)t+\det A.\end{aligned}

$\lambda_1,\lambda_2,\lambda_3$$A$ 的特徵值。因為特徵值是特徵多項式 $p(t)$ 的根，

\displaystyle \begin{aligned} p(t)&=-(t-\lambda_1)(t-\lambda_2)(t-\lambda_3)\\ &=-t^3+(\lambda_1+\lambda_2+\lambda_3)t^2-(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)t+\lambda_1\lambda_2\lambda_3.\end{aligned}

$\displaystyle \lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3=\frac{(\lambda_1+\lambda_2+\lambda_3)^2-(\lambda_1^2+\lambda_2^2+\lambda_3^2)}{2}=\frac{(\hbox{trace}A)^2-\hbox{trace}(A^2)}{2}$

$\displaystyle p(A)=-A^3+(\hbox{trace}A)A^2-\frac{(\hbox{trace}A)^2-\hbox{trace}(A^2)}{2}A+(\det A)I=0$

$\displaystyle A^{-1}=\frac{1}{\det A}\left(A^2-(\hbox{trace}A)A+\frac{(\hbox{trace}A)^2-\hbox{trace}(A^2)}{2}I\right)$

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