If A=\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\),show that A2-5A+7I=0.Hence find A-1.
Given A=\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)
A2=A.A =\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\) =\(\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}\)=\(\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)
therefore LHS=A2-5A+7I
\(\Rightarrow\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)-5\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)+7\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
=\(\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)-\(\begin{bmatrix}15&5\\-5&10\end{bmatrix}\)+\(\begin{bmatrix}7&0\\0&7\end{bmatrix}\)
=\(\begin{bmatrix}-7&0\\0&-7\end{bmatrix}\)+\(\begin{bmatrix}7&0\\0&7\end{bmatrix}\) =0
=RHS
So A2-5A+7I=0
therefore A. A-5A=-7I
\(\Rightarrow\) A. A(A-1)-5A(A-1)=-7I(A-1) [post multiplying by A-1 as IAI≠0]
\(\Rightarrow\) A(AA-1)-5I=-7I(A-1)
\(\Rightarrow\) AI-5I=-7A-1
\(\Rightarrow\) A-1=-\(\frac{1}{7}\)(A-5I)
\(\Rightarrow\) A-1=\(\frac{1}{7}\)(5I-A)
=\(\frac{41}{7}\) \(\bigg(\)\(\begin{bmatrix}5&0\\0&5\end{bmatrix}\)-\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)\(\bigg)\)=\(\frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}\)
\(\therefore\) A-1=\(\frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}\)
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).