题目
单选题(共6题,30.0分)在R³中取两组基:(Ⅰ)alpha_(1)=(1,0,1)^T,alpha_(2)=(1,1,0)^T,alpha_(3)=(0,1,1)^T;(Ⅱ)beta_(1)=(1,0,3)^T,beta_(2)=(2,2,2)^T,beta_(3)=(-1,1,4)^T,28.(5.0分)则基(Ⅰ)到基(Ⅱ)的过渡矩阵是()A }2&1&11&1&21&1&3B }1&2&11&1&0
单选题(共6题,30.0分)
在R³中取两组基:
(Ⅰ)$\alpha_{1}=(1,0,1)^{T},\alpha_{2}=(1,1,0)^{T},\alpha_{3}=(0,1,1)^{T}$;
(Ⅱ)$\beta_{1}=(1,0,3)^{T},\beta_{2}=(2,2,2)^{T},\beta_{3}=(-1,1,4)^{T}$,
28.(5.0分)则基(Ⅰ)到基(Ⅱ)的过渡矩阵是()
A $\begin{bmatrix}2&1&1\\1&1&2\\1&1&3\end{bmatrix}$
B $\begin{bmatrix}1&2&1\\1&1&0\end{bmatrix}$
题目解答
答案
设基(Ⅰ)的矩阵为 $A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$,基(Ⅱ)的矩阵为 $B = \begin{bmatrix} 1 & 2 & -1 \\ 0 & 2 & 1 \\ 3 & 2 & 4 \end{bmatrix}$。求解 $B = A P$,即 $P = A^{-1} B$。
计算 $A^{-1}$:
$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix}$
进行矩阵乘法:
$P = A^{-1} B = \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \\ 0 & 2 & 1 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 1 & -2 \\ 1 & 1 & 3 \end{bmatrix}$
答案: $\boxed{A}$