MCQ
Construct a $3 \times 2$ matrix whose elements are given by   $a_{i j}=\frac{1}{2}|i-3 j|$.
  • A
    $A\, = \,\left[ {\begin{array}{*{20}{c}}
      1&{\frac{3}{2}} \\ 
      {\frac{1}{2}}&2 \\ 
      0&{\frac{3}{2}} 
    \end{array}} \right]$
  • B
    $A\, = \,\left[ {\begin{array}{*{20}{c}}
      1&{\frac{5}{2}} \\ 
      {\frac{1}{2}}&2 \\ 
      0&{\frac{1}{2}} 
    \end{array}} \right]$
  • $A\, = \,\left[ {\begin{array}{*{20}{c}}
      1&{\frac{5}{2}} \\ 
      {\frac{1}{2}}&2 \\ 
      0&{\frac{3}{2}} 
    \end{array}} \right]$
  • D
    $A\, = \,\left[ {\begin{array}{*{20}{c}}
      1&{\frac{5}{2}} \\ 
      {\frac{1}{2}}&2 \\ 
      0&{\frac{5}{2}} 
    \end{array}} \right]$

Answer

Correct option: C.
$A\, = \,\left[ {\begin{array}{*{20}{c}}
  1&{\frac{5}{2}} \\ 
  {\frac{1}{2}}&2 \\ 
  0&{\frac{3}{2}} 
\end{array}} \right]$
c
In general a $3 \times 2$ matrix is given by $A\, = \,\left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\ 
  {{a_{21}}}&{{a_{22}}} \\ 
  {{a_{31}}}&{{a_{32}}} 
\end{array}} \right]$

Now                 ${a_{ij}} = \frac{1}{2}|i - 3j|,$  $i=1,\,2,\,3$ and $j=1,\,2$

therefore          ${a_{11}} = \frac{1}{2}|1 - 3 \times 1|\, = \,1$

                         ${a_{12}} = \frac{1}{2}|1 - 3 \times 2|\, = \,\frac{5}{2}$

                         ${a_{21}} = \frac{1}{2}|2 - 3 \times 1|\, =\,\frac{1}{2}$

                         ${a_{22}} = \frac{1}{2}|2 - 3 \times 2|\, = \,2$

                         ${a_{31}} = \frac{1}{2}|3 - 3 \times 1|\, = \,0$

                         ${a_{32}} = \frac{1}{2}|3 - 3 \times 2|\, = \,\frac{3}{2}$

Hence the required matrix is given by   $A\, = \,\left[ {\begin{array}{*{20}{c}}
  1&{\frac{5}{2}} \\ 
  {\frac{1}{2}}&2 \\ 
  0&{\frac{3}{2}} 
\end{array}} \right]$

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