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This is a multiple choice type question as classified in NCERT Exemplar

Correct option (ii)

The solubility of salt depends on the lattice and hydration energy both. By increasing lattice energy and decreasing hydration energy solubility decreases. In case of biE, both the ions are small, thus the compound has high lattice energy and is insoluble in water. If both the cations and anions are comparable in size, then stability is maximum and solubility is minimum

Let A = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill -3\hfill & \hfill 0\hfill & \hfill -5\hfill \\ \hfill 2\hfill & \hfill 5\hfill & \hfill 0\hfill \end{array}\right]$

We write A = IA

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill -3\hfill & \hfill 0\hfill & \hfill -5\hfill \\ \hfill 2\hfill & \hfill 5\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{lll}1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill -3+3(1)\hfill & \hfill 0+3(3)\hfill & \hfill -5+3(-2)\hfill \\ \hfill 2-2(1)\hfill & \hfill 5-2(3)\hfill & \hfill 0-2(-2)\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0+3(1)\hfill & \hfill 1+0\hfill & \hfill 0+0\hfill \\ \hfill 0-2(1)\hfill & \hfill 0-0\hfill & \hfill 1-0\hfill \end{array}\right]$ A. $\left(\begin{array}{ll}{R}_2\to R_2+3R_1\hfill & R_3\to R_3-2R_1\hfill \end{array}\right)$

$=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 9\hfill & \hfill -11\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill .4\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\to \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 4p\hfill \\ \hfill 0\hfill & \hfill 9\hfill & \hfill -11\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A. (R2        R3)

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 9\hfill & \hfill -11\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A. (R2   (-1) R2)

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0-0\hfill & \hfill 9-9(1)\hfill & \hfill -11-9(-4)\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 3-9(2)\hfill & \hfill 1-0\hfill & \hfill 0-9(-1)\hfill \end{array}\right]$ A. (R3     R3--> 4R2)

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 25\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill -15\hfill & \hfill 1\hfill & \hfill 9\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$ A. $\left(R_3\to \frac{1}{25}{R}_3\right)$

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 0+0\hfill & \hfill 1+0\hfill & \hfill -4+4(1)\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=$ $=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill 2+4\left(\frac{3}{5}\right)\hfill & \hfill 0+4\left(\frac{1}{25}\right)\hfill & \hfill -1+4\times \left(\frac{9}{25}\right)\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \end{array}\right]$ A. (R2     R2 + 4R3)

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 1+0\hfill & \hfill 3+0\hfill & \hfill -2+2(1)\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1+2\left(-\frac{3}{5}\right)\hfill & \hfill 0+2\left(\frac{1}{25}\right)\hfill & \hfill 0+2\left(\frac{9}{23}\right)\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \end{array}\right]$ A. (R1  R1 + 2R3)

$\Rightarrow \left[\begin{array}{lll}1\hfill & 3\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 1+0\hfill & \hfill 3-3(1)\hfill & \hfill 0+0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=$ $\left[\begin{array}{ccc}\hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.-3\left(\frac{-2}{5}\right)\hfill & \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$25$}\right.-3\left(\frac{4}{25}\right)\hfill & \hfill \frac{18}{25}-3\left(\frac{11}{25}\right)\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 25\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \end{array}\right]$ A. (R1   R1 3R2)

$\Rightarrow \left[\begin{array}{lll}1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill \raisebox{1ex}{$-10$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$-15$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \end{array}\right]$ A.

∴A^-1 = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$25$}\right.\hfill \end{array}\right]$

MBA is a course which teaches students the art of Leadership. Better or not, every college/university offers different advantages and disadvantages to students. Which is better or not depends on the students and what they are searching for in a college/ university ersity. 

The MBA specialisations offered by Adikavi Nannaya University are as follows: 

• MBA in Tourism and Hospitality
• MBA

Candidates can get admission to various campuses for the MBA programme that are affiliated by the university. The admission process remains entrance based for all the campuses. 

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (iii)On moving down the group as the size increases, the nuclear charge attraction decreases. The effect of increasing size outweighs the increasing nuclear charge. Thus, ionization enthalpy decreases. The order of decreasing ionization enthalpy is,

Li>Na>K>Rb

Let A = $\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -3\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill 2\hfill \end{array}\right]$

We write, A = IA

$\Rightarrow \left[\begin{array}{ccc}\hfill 2\hfill & \hfill -3\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill -3\hfill & \hfill 3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ A. $\left[R_3\leftrightarrow R_1\right)$

$\Rightarrow \left[\begin{array}{ccc}\hfill 3-2\hfill & \hfill -2-2\hfill & \hfill 2-3\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill -3\hfill & \hfill 3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 0-0\hfill & \hfill 0-1\hfill & \hfill 1-0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ A. (R1   R1 -->R2)

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill -3\hfill & \hfill 3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 10\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ A.

$\Rightarrow \left|\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill -1\hfill \\ \hfill 2-2(1)\hfill & \hfill 2+2(-4)\hfill & \hfill 3-2(-1)\hfill \\ \hfill 2-2(1)\hfill & \hfill -3+2(-4)\hfill & \hfill 3-2(-1)\hfill \end{array}\right|=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0-2(0)\hfill & \hfill 1-2(-1)\hfill & \hfill 0-2(1)\hfill \\ \hfill 1-2(0)\hfill & \hfill 0-2(-1)\hfill & \hfill 0-2(1)\hfill \end{array}\right]$ A.

$(\begin{array}{l}{R}_2\to R_2-2R_1\hfill \end{array}$

$\begin{array}{l}{R}_3\to R_3-2R_1\hfill \end{array})$

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 10\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 5\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -2\hfill \end{array}\right]$ A.

$\bcancel{1:\frac{x^2}{36}}+\bcancel{\frac{y^2}{16}=1}$

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill -1\hfill \\ \hfill 0-0\hfill & \hfill 10-5\hfill & \hfill 5-5\hfill \\ \hfill 0\hfill & \hfill 5\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0-1\hfill & \hfill 3-2\hfill & \hfill -2-(-2)\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -2\hfill \end{array}\right]$ A. (R2   R2 --->R3)

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 5\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 5\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -2\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$ A. $\left(\begin{array}{cc}\hfill R_2\to \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.R_2\hfill & \hfill R_3\to \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.R_3\hfill \end{array}\right)$

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0-0\hfill & \hfill 1-1\hfill & \hfill 1-0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill \\ \hfill 1/5-\left(\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)\hfill & \hfill \frac{2}{5}-\frac{1}{5}\hfill & \hfill 0\left(\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)-0\hfill \end{array}\right]$ A. (R3       R3 -->R2)

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 1+0\hfill & \hfill -4+0\hfill & \hfill -1+1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill -1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 1+\left(\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)\hfill \\ \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$ A. (R1      R1 + R3)

$\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{ccc}\hfill 1+0\hfill & \hfill -4+4(1)\hfill & \hfill 0+0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=$ $\left[\begin{array}{ccc}\hfill \frac{2}{5}+4\left(-\frac{1}{5}\right)\hfill & \hfill -\frac{4}{5}+4\left(\frac{1}{5}\right)\hfill & \hfill \frac{3}{5}+4(0)\hfill \\ \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$ A. (R1      R1 + 4R2)

$\Rightarrow \left[\begin{array}{lll}1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$ A.

∴A^-1 = $\left[\begin{array}{ccc}\hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

BeCl₂ is covalent in nature due to the high polarising power of beryllium (+2) ion. On moving down the group, as the size increases, covalent nature decreases. Therefore, beryllium chloride being most covalent is soluble in ethanol.

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$

We write, A = IA.

$\Rightarrow \left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A. $\left(R_1\to \frac{1}{2}{R}_1\right)$

$\to \left[\begin{array}{cc}\hfill 1\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 4-4(1)\hfill & \hfill 2-4\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \\ \hfill 0-4\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\hfill & \hfill 1-4\left(\raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\hfill \end{array}\right]$ A. (R2       R2 --->4R1)

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]$ A.

∴A^-1 does not exit

Students who wish to pursue MMS at Mumbai colleges do not necessarily have to possess work experience. Most colleges admit freshers. However, some colleges may give priority or preference to students with at least 2 years of experience. 

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

As the size of the metal increases, the basic character of the hydroxide increases down the group. The solubility of hydroxide increases, and hydroxide of Be and Mg is almost insoluble.

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -3\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$

We write A = IA.

$\Rightarrow \left[\begin{array}{cc}\hfill 2\hfill & \hfill -3\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A .

$\Rightarrow \left[\bcancel{\begin{array}{cc}\hfill 2+2(-1)\hfill & \hfill -3+2\bcancel{(2)}\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}}\right]=\left[\bcancel{\begin{array}{cc}\hfill 1+2(0)\hfill & \hfill 0+2(1)\hfill \\ \hfill \bcancel{0}\hfill & \hfill 1\hfill \end{array}}\right]$ A $\left(\bcancel{R_1\to R_1}+\bcancel{2R_2}\right)$