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

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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.

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