Correct option: A.$A=\left[\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{array}\right]$
a
In general, a $3 \times 4$ matrix is given by $A=\left[\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}\end{array}\right]$
Given : $a_{ij}=\frac{1}{2}|-3 i+j|$, $i=1,\,2,\,3$ and $j=1,\,2,\,3,\,4$
Thus, we have
${a_{11}} = \frac{1}{2}| - 3 \times 1 + 1| = $ $\frac{1}{2}| - 3 + 1| = \frac{1}{2}| - 2| = \frac{2}{2} = 1$
${a_{21}} = \frac{1}{2}| - 3 \times 2 + 1| = $ $\frac{1}{2}| - 6 + 1| = \frac{1}{2}| - 5| = \frac{5}{2}$
${a_{31}} = \frac{1}{2}| - 3 \times 3 + 1| = $ $\frac{1}{2}| - 9 + 1| = \frac{1}{2}| - 8| = 4$
${a_{12}} = \frac{1}{2}| - 3 \times 1 + 2| = $ $\frac{1}{2}| - 3 + 2| = \frac{1}{2}| - 1| = \frac{1}{2}$
${a_{22}} = \frac{1}{2}| - 3 \times 2 + 2| = $ $\frac{1}{2}| - 6 + 2| = \frac{1}{2}| - 4| = \frac{4}{2} = 2$
${a_{32}} = \frac{1}{2}| - 3 \times 3 + 2| = $ $\frac{1}{2}| - 9 + 2| = \frac{1}{2}| - 7| = \frac{7}{2}$
$a_{13}=\frac{1}{2}|-3 \times 1+3|=\frac{1}{2}|-3+3|=0$
${a_{23}} = \frac{1}{2}| - 3 \times 2 + 3| = $ $\frac{1}{2}| - 6 + 3| = \frac{1}{2}| - 3| = \frac{3}{2}$
${a_{33}} = \frac{1}{2}| - 3 \times 3 + 3| = $ $\frac{1}{2}| - 9 + 3| = \frac{1}{2}| - 6| = \frac{6}{2} = 3$
${a_{14}} = \frac{1}{2}| - 3 \times 1 + 4| = $ $\frac{1}{2}| - 3 + 4| = \frac{1}{2}|1| = \frac{1}{2}$
${a_{24}} = \frac{1}{2}| - 3 \times 2 + 4| = $ $\frac{1}{2}| - 6 + 4| = \frac{1}{2}| - 2| = \frac{2}{2} = 1$
${a_{34}} = \frac{1}{2}| - 3 \times 3 + 4| = $ $\frac{1}{2}| - 9 + 4| = \frac{1}{2}| - 5| = \frac{5}{2}$
Therefore, the required matrix is $A=\left[\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{array}\right]$