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Consider a 3 × 3 real symmetric matrix S such that two of its eigen values are a ≠ 0, b≠ 0 with respective eigen vectors \[\left[ {\begin{array}{*{20}{c}} {{{\text{x}}_1}} \\ {{{\text{x}}_2}} \\ {{{\text{x}}_3}} \end{array}} \right

Correct Answer: ,\left.\,   If a ≠ b then x<sub>1</sub>y<sub>1</sub> + x<sub>2</sub>y<sub>2</sub> + x<sub>3</sub>y<sub>3</sub> equals, <p><span>A.</span> a
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] Option D ]

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

Let the eigen values of a 2 × 2 matrix A be 1, -2 with eigen vectors x1 and x2 respectively. Then the eigen values and eigen vectors of the matrix A2 - 3A + 4$$I$$ would, respectively, be

The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?

The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with

$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$     $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$     $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$     is simplified to :

Consider the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}\left( {\text{t}} \right)}}{{{\text{dt}}}} + {\text{y}}\left( {\text{t}} \right) = \delta \left( {\text{t}} \right)$$      with $${\left. {{\text{y}}\left( {\text{t}} \right)} \right|_{{\text{t}} = 0}} = - 2$$   and $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 0.$$
The numerical value of $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}}$$   is

Consider a system governed by the following equations:
$$\frac{{{\text{d}}{{\text{x}}_1}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_2}\left( {\text{t}} \right) - {{\text{x}}_1}\left( {\text{t}} \right);\,\frac{{{\text{d}}{{\text{x}}_2}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_1}\left( {\text{t}} \right) - {{\text{x}}_2}\left( {\text{t}} \right)$$
The initial conditions are such that $${{\text{x}}_1}\left( 0 \right)

The quark content of $$\sum {^ + } ,\,{K^ - },\,{\pi ^ - }$$   and p is indicated: $$\left| {\sum {^ + } } \right\rangle = \left| {uus} \right\rangle ;\,\left| {{K^ + }} \right\rangle = \left| {s\overline u } \right\rangle ;\,\left| \pi \right\rangle = \left| d \right\rangle ;\,\left| p \right\rangle = \left| {uud} \right\rangle $$
In the process, $${\pi ^ - } + p \to {K^ - } + \sum {^ + } ,$$     considering strong interactions only, which of the following statements is true?

Let the function
\[{\text{f}}\left( \theta \right) = \left| {\begin{array}{*{20}{c}} {\sin \theta }&{\cos \theta }&{\tan \theta } \\ {\sin \left( {\frac{\pi }{6}} \right)}&{\cos \left( {\frac{\pi }{6}} \right)}&{\tan \left( {\frac{\pi }{6}} \right)} \\ {\sin \left( {\frac{\pi }{3}} \right)}&{\cos \left( {\frac{\pi }{3}} \right)}&{\tan \left( {\frac{\pi }{3}} \right)} \end{array}} \right|\