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

Answer: $A= \begin{bmatrix} 1 &\frac{1}{2} &0 &\frac{1}{2}\\ \\\frac{5}{2} &2 &\frac{3}{2} &1 \\\\ 4 &\frac{7}{2} &3 &\frac{5}{2} \end{bmatrix}$

Given: $a_{ij}= \frac{1}{2}\left | -3i+j \right |$
Here we have to construct $3\times 4$  matrix according to  $a_{ij}= \frac{1}{2}\left | -3i+j \right |$

Hint: We have to find all the elements of matrix according to  $\frac{1}{2}\left | -3i+j \right |$

Solution: Here  $a_{ij}= \frac{1}{2}\left | -3i+j \right |$

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

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

$\! \! \! \! \! \! \! \! \! a_{11}= \frac{1}{2} \left | -3\times 1 +1\right |= \frac{1}{2}\times 2= 1\\\\a_{12}=\frac{1}{2}\left | -3\times 1+2 \right |= \frac{1}{2}\times 1= \frac{1}{2}\\\\a_{13}=\frac{1}{2}\left | -3\times 1+3 \right |= \frac{1}{2} \times 0= 0\\\\a_{14}=\frac{1}{2}\left | -3\times 1+4 \right | = \frac{1}{2}\times 1= \frac{1}{2}$         $\! \! \! \! \! \! \! \! \! a_{21}\frac{1}{2}\left | -3\times 2+1 \right |= \frac{1}{2}\times 5= \frac{5}{2}\\\\a_{22}\frac{1}{2}\left | -3\times 2+2 \right |= \frac{1}{2}\times 4= 2\\\\a_{23}\frac{1}{2}\left | -3\times 2+3 \right |= \frac{1}{2}\times 3= \frac{3}{2}\\\\a_{24}\frac{1}{2}\left | -3\times 2+4 \right |= \frac{1}{2}\times 2= 1$          $\! \! \! \! \! \! \! \! \! a_{31}= \frac{1}{2}\left | -3\times 3+1 \right |= \frac{1}{2}\times 8= 4\\\\a_{32}= \frac{1}{2}\left | -3\times 3+2 \right |= \frac{1}{2}\times7 = \frac{7}{2}\\\\a_{33}= \frac{1}{2}\left | -3\times 3+3 \right |= \frac{1}{2}\times 6= 3\\\\a_{34}= \frac{1}{2}\left | -3\times 3+4 \right |= \frac{1}{2}\times 5= \frac{5}{2}$

Substituting these values in Matrix $A$ , we get

$A= \begin{bmatrix} 1 &\frac{1}{2} &0 &\frac{1}{2}\\ \\\frac{5}{2} &2 &\frac{3}{2} &1 \\\\ 4 &\frac{7}{2} &3 &\frac{5}{2} \end{bmatrix}$