Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.1 Question 4 Subquestion (iv) Maths Textbook Solution.

Answer: $A= \begin{bmatrix} 2 &4.5 &8 \\ 4.5& 8& 12.5 \end{bmatrix}$
Given: Here given that matrix of order $2\times 3$
$A= \left [ a_{ij} \right ]_{2\times 3}$
Here we have to construct the matrix according to $a_{ij}= \frac{\left ( i+j \right )^{2}}{2}$
Hint: Adding row and column element and squaring then divide by 2
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 3}$ and given that $a_{ij}= \frac{\left ( i+j \right )^{2}}{2}$
So, the elements in a $2\times 3$ matrix are $a_{11},a_{12},a_{13},a_{21},a_{22},a_{23}$
$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{\left ( 1+1 \right )^{2}}{2}=\frac{4}{2} = 2\\a_{12}=\frac{\left ( 1+2 \right )^{2}}{2}=\frac{\left ( 3 \right )^{2}}{2} = \frac{9}{2}= 4.5\\a_{13}=\frac{\left ( 1+3 \right )^{2}}{2}= \frac{\left ( 4 \right )^{2}}{2}= \frac{16}{2}= 8$                 $\! \! \! \! \! \! \! \! \! a_{21}=\frac{\left ( 2+ 1 \right )^{2}}{2}= \frac{\left ( 3 \right )^{2}}{2}= \frac{9}{2}= 4.5\\a_{22}=\frac{\left ( 2+ 2 \right )^{2}}{2} = \frac{\left ( 4 \right )^{2}}{2}= \frac{16}{2}= 8\\a_{23}=\frac{\left ( 2+ 3 \right )^{2}}{2} = \frac{\left ( 5 \right )^{2}}{2}= \frac{25}{2}= 12.5$
Substituting these values in Matrix $A$, we get
$A= \begin{bmatrix} 2 &4.5 &8 \\ 4.5& 8& 12.5 \end{bmatrix}$
Hence this is the required answer.