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
