Tính định thức sau: detA=$\left[\begin{array}{ccc}1&1/2&1/3\\1/2&1/3&1/4\\1/3&1/4&1/5\end{array}\right]$ Hãy tính detB= detA.det(2A).det(4A).det(16A)? Tính det(10B) và so sánh detA

Đáp án:

•$\left[\begin{array}{ccc}1&1/2&1/3\\1/2&1/3&1/4\\1/3&1/4&1/5\end{array}\right]$ 

=$\frac{1}{2160}$

•det(10B)=$10^{3}$ .detB$^{T}$=1000.detB=0,00009634

•detB<detA
Giải thích các bước giải:

Ta có:

detA

=$\left[\begin{array}{ccc}1&1/2&1/3\\1/2&1/3&1/4\\1/3&1/4&1/5\end{array}\right]$ 

=$\frac{1}{6}$.$\frac{1}{12}$.$\frac{1}{60}$.$\left[\begin{array}{ccc}6&3&2\\6&4&3\\20&15&12\end{array}\right]$ 

=$\frac{1}{2160}$.$\left[\begin{array}{ccc}3&3&2\\3&4&3\\10&15&12\end{array}\right]$ 

=$\frac{1}{2160}$.$\left[\begin{array}{ccc}3&3&2\\0&1&1\\10&15&12\end{array}\right]$ 

=$\frac{1}{2160}$.$\left[\begin{array}{ccc}3&3&-1\\0&1&0\\10&15&-3\end{array}\right]$ 

=$\frac{1}{2160}$.$\left[\begin{array}{ccc}3&-1\\10&-3\end{array}\right]$ 

=$\frac{1}{2160}$ 

=>detA=$\frac{1}{2160}$ 

  det2A=$2^{3}$.detA$^{T}$=8.detA=$\frac{1}{270}$ 

  det4A=$4^{3}$.detA$^{T}$=64.detA=$\frac{4}{135}$ 

  det16A=$16^{3}$.detA$^{T}$=4096.detA=$\frac{256}{135}$ 

=>detB=0,00000009634

=>det(10B)=$10^{3}$ .detB$^{T}$=1000.detB=0,00009634

Vậy detB<detA