下面二次型中正定的是（ ）。
A: $f(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{2}$;
B: $f(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{2}+2x_{1}x_{2}+6x_{3}^{2}$;
C: $f(x_{1},x_{2},x_{3})=4x_{1}^{2}+3x_{2}^{2}+6x_{3}^{2}-x_{1}x_{2}-x_{1}x_{3}$;
D: $f(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+2x_{2}x_{3}$.
A: $f(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{2}$;
B: $f(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{2}+2x_{1}x_{2}+6x_{3}^{2}$;
C: $f(x_{1},x_{2},x_{3})=4x_{1}^{2}+3x_{2}^{2}+6x_{3}^{2}-x_{1}x_{2}-x_{1}x_{3}$;
D: $f(x_{1},x_{2},x_{3})=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+2x_{2}x_{3}$.
举一反三
- 以下变换$\cal{A}$是线性变换的有（ ）。 A: $R^{3}$上变换：$\cal{A}(x_{1},x_{2},x_{3})=(x_{1},x_{3},x_{2}+1)$ B: $R^{3}$上变换：$\cal{A}(x_{1},x_{2},x_{3})=(\mid x_{1}\mid ,x_{3},x_{2})$ C: $R[x]$上变换：$\cal{A}(f(x))=f(x+3)$ D: $R[x]$上变换：$\cal{A}(f(x))=f(x+1)-f(x)$
- 对线性空间$R^{2}$中以下函数$f$，不是线性函数的有 ( ). A: $f(x_{1},x_{2})=4x_{1}+x_{2}log_{3}8$ B: $f(x_{1},x_{2})=x_{1}+4x_{2}+4$ C: $f(x_{1},x_{2})=x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}$ D: $f(x_{1},x_{2})=sin (x_{1})+cos( x_{2})$
- 以下集合对于所指的线性运算构成实数域上线性空间的有 ( ）。 A: $R^{2}$上定义加法，数乘如下：$$(x_{1},x_{2})+(y_{1},y_{2})=(x_{1}+y_{1},0），k(x,y)=(kx,0)$$ B: $R^{2}$上定义加法，数乘如下：$$(x_{1},x_{2})+(y_{1},y_{2})=(x_{1}+y_{1},x_{2}），k(x,y)=(kx,y)$$ C: 平面上不平行于$X$ 轴的向量全体，关于向量的加法与数量乘法 D: $R^{2}$上定义加法，数乘如下：$$(x_{1},x_{2})+(y_{1},y_{2})=(x_{1}+y_{1},x_{2}+y_{2}+x_{1}y_{1})），$$$$k(x,y)=(kx,ky+\frac{k(k-1)}{2}x^{2})$$
- 为求方程${x^3} - {x^2} - 1 = 0$在${x_0} = 1.5$附近的一个根，以下迭代格式收敛的是： A: ${x_{k + 1}} = 1 + {1 \over {x_k^2}}$ B: ${x_{k + 1}} = 1 - {1 \over {x_k^2}}$ C: ${x_{k + 1}} = \root 3 \of {x_k^2 - 1} $ D: ${x_{k + 1}} = {1 \over {\sqrt {{x_k} - 1} }}$
- 设总体X~N($\mu,{\sigma}^2$),$\mu,{\sigma}^2$未知，$x_{1},x_{2},...,x_{n} $ 是来自该总体的样本，记$\overline x=\frac{1}{n}\sum\limits_{i = 1}^{n}{x_{i}}$,则对假设检验$ H_{0}:u=u_{0},H_{1}:u!=u_{0}$的拒绝域为()