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    $A=\left[\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{array}\right]$