Using Cofactors of elements of second row, evaluate △=\(\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}\)
The given determinant is \(\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}\)
We have:
M21=\(\begin{vmatrix}3&8\\2&3\end{vmatrix}\)=9-16=-7
∴A21=cofactor of a21=(−1)2+1 M21=7
M22 =\(\begin{vmatrix}5&8\\1&3\end{vmatrix}\)=15-8=7
∴A22=cofactor of a22=(−1)2+2 M22=7
M23=\(\begin{vmatrix}5&3\\1&2\end{vmatrix}\)=10-3=7
∴A23 = cofactor of a23 = (−1)2+3 M23 = −7
We know that ∆ is equal to the sum of the product of the elements of the second row
with their corresponding cofactors.
∴ ∆ = a21A21 + a22A22 + a23A23 = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7
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).