GATE CE 2005

With the given order we can say that order of matrices are as follows:
\begin{aligned} X^{T} &\rightarrow 3\times 4 \\ Y & \rightarrow 4\times 3 \\ X^{T}Y & \rightarrow 3\times 3 \\ \left ( X^{T}Y \right )^{-1} & \rightarrow 3\times 3 \\ P & \rightarrow 2\times 3 \\ P^{T} & \rightarrow 3\times 2 \\ P\left ( X^{T}Y \right )^{-1}P^{T} & \rightarrow \left ( 2\times 3 \right )\left ( 3\times 3 \right )\left ( 3\times 2 \right ) \\ & \rightarrow 2\times 2 \\ \therefore\;\; \left ( P\left ( X^{T}Y \right )^{-1}P^{T} \right )^{T} & \rightarrow 2\times 2 \end{aligned}

In an over determined system having more equations than variables, it is necessary to have consistent unique solution, by definition.

(A) is true since
\frac{Z_{1}}{Z_{2}}=\frac{Z_{1}\bar{Z_{2}}}{Z_{2}\bar{Z_{2}}}=\frac{Z_{1}\bar{Z_{2}}}{\left | Z_{2} \right |^{2}}
(B) is true by triangle inequality of complex number.
(C) is not true since \left | Z_{1}-Z_{2} \right |\geq \left | Z_{1} \right |-\left | Z_{2} \right |
(D) is true since
\begin{aligned}\left | Z_{1}+Z_{2} \right |^{2}&=\left ( Z_{1}+Z_{2} \right )\left ( \overline{Z_{1}+Z_{2}} \right ) \\ &=\left ( Z_{1}+Z_{2} \right )\left ( \bar{Z_{1}}+\bar{Z_{2}} \right ) \\ &=Z_{1}\bar{Z_{1}}+Z_{2}\bar{Z_{2}}+Z_{2}\bar{Z_{1}}+Z_{1}\bar{Z_{2}} \;\;...(i)\\ \left | Z_{1}-Z_{2} \right |^{2}&=\left ( Z_{1}+Z_{2} \right )\left ( \overline{Z_{1}-Z_{2}} \right ) \\ &=\left ( Z_{1}-Z_{2} \right )\left ( \bar{Z_{1}}-\bar{Z_{2}} \right ) \\ &=Z_{1}\bar{Z_{1}}+Z_{2}\bar{Z_{2}}-Z_{2}\bar{Z_{1}}-Z_{1}\bar{Z_{2}}\;\;...(ii)\end{aligned}
Adding (i) and (ii) we get,
\left | Z_{1}+Z_{2} \right |^{2}+\left | Z_{1}-Z_{2} \right |^{2}=2Z_{1}\bar{Z_{2}}+2Z_{2}\bar{Z_{2}} =2\left | Z_{1} \right |^{2}+2\left | Z_{2} \right |^{2}

A, B, C are true
(D) is not true since in a negatively skewed distribution, mode \gt median \gt mean

Taking moment equilibrium about the centre, we get
\begin{aligned} \tau_{y x} \times \frac{d}{2}+\tau_{y x} \times \frac{d}{2}&=\tau_{x y} \times \frac{d}{2}+\tau_{x y} \times \frac{d}{2} \\ \therefore \quad \tau_{x y}&=\tau_{y x} \end{aligned}

In case of plane strain condition, the components
of strain tensor at a point are \epsilon_{x}, \in_{y} \text{ and }\phi_{x y} .
Here we have three unknowns so we require 3 equations to find them and these unknowns can be find by the equations of longitudinal strain,
\epsilon_{1}=\epsilon_{x} \cos ^{2} \theta_{1}+\epsilon_{y} \sin ^{2} \theta_{1}+\frac{\phi_{x y}}{2} \sin 2 \theta_{1}
\epsilon_{2}=\epsilon_{x} \cos ^{2} \theta_{2}+\epsilon_{y} \sin ^{2} \theta_{2}+\frac{\phi_{x y}}{2} \sin 2 \theta_{2}
\epsilon_{3}=\epsilon_{x} \cos ^{2} \theta_{3}+\epsilon_{y} \sin ^{2} \theta_{3}+\frac{\phi_{x y}}{2} \sin 2 \theta_{3}
Therefore to find out component of strain tensor in plain strain condition, we measure longitudinal strains \left(\epsilon_{1}, \epsilon_{2}\text{ and }\epsilon_{3}\right) along any three arbitrary direction.

Degree of kinematic indeterminacy or degree of freedom,
\begin{aligned} D_{k} &=3 j-r_{e}-m \\ &=3 \times 4-3-1 \\ &=8 \end{aligned}