Tính định thức: $D= \begin{vmatrix} a^{2} &\left ( a+1 \right )^{2} &\left ( a+2 \right )^{2} &\left ( a+3 \right )^{2} \\ b^{2} & \left ( b+1 \right )^{2} &\left ( b+2 \right )^{2} &\left ( b+3 \right )^{2} \\ c^{2}& \left ( c+1 \right )^{2} & \left ( c+2 \right )^{2} & \left ( c+3 \right )^{2}\\ d^{2} &\left ( d+1 \right )^{2} &\left ( d+2 \right )^{2} & \left ( d+3 \right )^{2} \end{vmatrix}$
Định thức ma trận vuông cấp 4
#1
Posted 30-10-2018 - 04:34
#2
Posted 30-10-2018 - 21:30
Tính định thức: $D= \begin{vmatrix} a^{2} &\left ( a+1 \right )^{2} &\left ( a+2 \right )^{2} &\left ( a+3 \right )^{2} \\ b^{2} & \left ( b+1 \right )^{2} &\left ( b+2 \right )^{2} &\left ( b+3 \right )^{2} \\ c^{2}& \left ( c+1 \right )^{2} & \left ( c+2 \right )^{2} & \left ( c+3 \right )^{2}\\ d^{2} &\left ( d+1 \right )^{2} &\left ( d+2 \right )^{2} & \left ( d+3 \right )^{2} \end{vmatrix}$
$D= \begin{vmatrix} a^{2} &\left ( a+1 \right )^{2} &\left ( a+2 \right )^{2} &\left ( a+3 \right )^{2} \\ b^{2} & \left ( b+1 \right )^{2} &\left ( b+2 \right )^{2} &\left ( b+3 \right )^{2} \\ c^{2}& \left ( c+1 \right )^{2} & \left ( c+2 \right )^{2} & \left ( c+3 \right )^{2}\\ d^{2} &\left ( d+1 \right )^{2} &\left ( d+2 \right )^{2} & \left ( d+3 \right )^{2} \end{vmatrix}$
$=(a-b)(b-c)(c-d)(d-a).\begin{vmatrix} a+b & a+b+2 & a+b+4 & a+b+6 \\ b+c & b+c+2 & b+c+4 & b+c+6 \\ c+d & c+d+2 & c+d+4 & c+d+6 \\ a+d & a+d+2 & a+d+4 & a+d+6 \end{vmatrix}=(a-b)(b-c)(c-d)(d-a) \begin{vmatrix} -2 & -2 & -2 & 6 \\ -2 & -2 & -2 & 6 \\ -2 & -2 & -2 & 6 \\ -2 & -2 & -2 & 6 \end{vmatrix}=0$
Edited by anhquannbk, 30-10-2018 - 21:33.
- lenamhvtc likes this
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users