每週問題 May 12, 2014

Let $A$ be a $2\times 2$ matrix. It is known that $\text{trace}A=2$ and $\text{trace}(A^2)=10$. Determine the determinant of $A$.

$A$ 有特徵值 $\lambda_1$$\lambda_2$，則 $\text{trace}A=\lambda_1+\lambda_2$$\det A=\lambda_1\lambda_2$。使用 $\text{trace}(A^2)=\lambda_1^2+\lambda_2^2$，可得 $(\text{trace}A)^2=\lambda_1^2+\lambda_2^2+2\lambda_1\lambda_2=\text{trace}(A^2)+2\det A$。由此解出 $\det A=\frac{1}{2}((\text{trace}A)^2-\text{trace}(A^2))$，故 $\det A=\frac{1}{2}(2^2-10)=-3$

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