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二階矩陣det(A+B)有公式

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tommywong posted 2022-10-23 22:58 |Read mode
Matrix and Determinant

Let $a,b,c,d,t$ is real number.
if $ A = \begin{bmatrix}
a & b \\
c&d
\end{bmatrix}$ and $det(A) = t \neq0$
and $det(A+t^{2}A^{-1})=0$
Find $det(A-t^{2}A^{-1})?$

Sum identity for 2×2 matrices

$\det(A+B)=\det(A)+\det(B)+\operatorname{tr}(A)\operatorname{tr}(B)-\operatorname{tr}(AB)$
$A=B,~\operatorname{tr}(A^2)=\operatorname{tr}(A)^2-2\det(A)$

$\det(A+t^2 A^{-1})=0$
$\det(A^2+t^2 I)=t^2+t^4+2t^2\operatorname{tr}(A^2)-t^2\operatorname{tr}(A^2)=0$
$(t-1)^2+\operatorname{tr}(A)^2=0$
As $\det(A),\operatorname{tr}(A)\in \mathbb{R}$
$t=1,~\operatorname{tr}(A)=0,~\operatorname{tr}(A^2)=-2$

$\det(A-t^2 A^{-1})=\det(A^2-t^2 I)$
$=t^2+t^4-2t^2\operatorname{tr}(A^2)+t^2\operatorname{tr}(A^2)=4$
现充已死,エロ当立。
维基用户页:https://zh.wikipedia.org/wiki/User:Tttfffkkk
Notable algebra methods:https://artofproblemsolving.com/community/c728438
《方幂和及其推广和式》 数学学习与研究2016.

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