For the matrices A and B, verify that (AB)′=B'A' where
I. A=\(\begin{bmatrix}1\\-4\\3\end{bmatrix}\),B=\(\begin{bmatrix}-1&2&1\end{bmatrix}\)
II. A= \(\begin{bmatrix}0\\1\\2\end{bmatrix}\),B=\(\begin{bmatrix}1&5&7\end{bmatrix}\)
I. A=\(\begin{bmatrix}1\\-4\\3\end{bmatrix}\),B=\(\begin{bmatrix}-1&2&1\end{bmatrix}\)
AB =\(\begin{bmatrix}1\\-4\\3\end{bmatrix}\)\(\begin{bmatrix}-1&2&1\end{bmatrix}\)
so (AB)'=\(\begin{bmatrix}-1&4&-3\\4&-8&6\\1&-4&3\end{bmatrix}\)
Now A'=\(\begin{bmatrix}-1&4&3\end{bmatrix}\),B'=\(\begin{bmatrix}-1\\2\\1\end{bmatrix}\)
so B'A'=\(\begin{bmatrix}-1&4&3\end{bmatrix}\)\(\begin{bmatrix}-1\\2\\1\end{bmatrix}\)
=\(\begin{bmatrix}-1&4&-3\\4&-8&6\\1&-4&3\end{bmatrix}\)
Hence we have verified that (AB)′=B'A'
II. A=\(\begin{bmatrix}0\\1\\2\end{bmatrix}\),B=\(\begin{bmatrix}1&5&7\end{bmatrix}\)
AB=\(\begin{bmatrix}0\\1\\2\end{bmatrix}\)\(\begin{bmatrix}1&5&7\end{bmatrix}\)=\(\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}\)
so (AB)'=\(\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\)
Now A'=\(\begin{bmatrix}0&1&2\end{bmatrix}\),B'=\(\begin{bmatrix}1\\5\\7\end{bmatrix}\)
B'A'=\(\begin{bmatrix}1\\5\\7\end{bmatrix}\)\(\begin{bmatrix}0&1&2\end{bmatrix}\)=\(\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\)
Hence we have verified that (AB)′=B'A'
A settling chamber is used for the removal of discrete particulate matter from air with the following conditions. Horizontal velocity of air = 0.2 m/s; Temperature of air stream = 77°C; Specific gravity of particle to be removed = 2.65; Chamber length = 12 m; Chamber height = 2 m; Viscosity of air at 77°C = 2.1 × 10\(^{-5}\) kg/m·s; Acceleration due to gravity (g) = 9.81 m/s²; Density of air at 77°C = 1.0 kg/m³; Assume the density of water as 1000 kg/m³ and Laminar condition exists in the chamber.
The minimum size of particle that will be removed with 100% efficiency in the settling chamber (in $\mu$m is .......... (round off to one decimal place).