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Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.1 Question 6 Subquestion (ii) Maths Textbook Solution.

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Answer:   A= \begin{bmatrix} 0 &-1 &-2 &-3\\ 1 &0 &-1 &-2 \\ 2 &1 &0 &-1 \end{bmatrix}

Given: a_{ij}= \left ( i-j \right )
             Here we have to construct 3\times 4  matrix according to \left ( i-j \right )

Hint: Find the sum of i and j  for each element.

Solution: Here a_{ij}= \left ( i-j \right )

                  Let A= \left [ a_{ij} \right ]_{3\times 4}

So, A= \begin{bmatrix} a_{11} &a_{12} &a_{13} & a_{14}\\ a_{21} &a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} &a_{34} \end{bmatrix}_{3\times 4}         

\! \! \! \! \! \! \! \! \! a_{11}= 1-1= 0\\a_{12}= 1-2= -1\\a_{13}= 1-3=-2\\a_{14}= 1-4= -3             \! \! \! \! \! \! \! \! \! a_{21}= 2-1= 1\\a_{22}= 2-2= 0\\a_{23}= 2-3=-1\\a_{24}= 2-4= -2           \! \! \! \! \! \! \! \! \! a_{31}= 3-1= 2\\a_{32}= 3-2= 1\\a_{33}= 3-3=0\\a_{34}= 3-4= -1

Substituting these values in Matrix A , we get

A= \begin{bmatrix} 0 &-1 &-2 &-3\\ 1 &0 &-1 &-2 \\ 2 &1 &0 &-1 \end{bmatrix}

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