Question
Construct a 3 × 4 matrix, whose elements are given by:$\text{a}_{\text{ij}}=\frac{1}{2} \left|-3{\text{i}+\text{j}}\right| $

Answer

$\text{Let A}=\left[\text {a}_{\text {ij}}\ \text {be required}\ 3\times4\ {\text {matrix where}}\ {\text a_{\text {ij} }}={\frac{1}{2}}\left|-3{\text{i+j}}\right|\right] $ $\therefore\ \text a_{11}=\frac{1}{2}\left|{-3+1}\right|=\frac{1}{2}(2)=1, $ $\text a_{12}=\frac{1}{2}\left|{-3+2}\right|=\frac{1}{2}(1)=\frac{1}{2} $$\text a_{13}=\frac{1}{2}\left|{-3+3}\right|=\frac{1}{2}(0)=0, $
$\text a_{14}=\frac{1}{2}\left|{-3+4}\right|=\frac{1}{2}(1)=\frac {1}{2} $ $\text a_{21}=\frac{1}{2}\left|{-6+1}\right|=\frac{1}{2}(5)=\frac {5}{2}, $$\text a_{22}=\frac{1}{2}\left|-6+2 \right|=\frac {1}{2}(4)=2 $
$\text a_{23}=\frac{1}{2}\left|-6+3\right|=\frac {1}{2}(3)=\frac {3}{2}, $ $\text a_{24}=\frac{1}{2}\left|-6+4\right|=\frac{1}{2}|-2|=\frac{1}{2}\times2=1 $$\text a_{31}=\frac{1}{2}\left|-9+1\right|=\frac{1}{2}(8)=4, $
$\text a_{32}=\frac{1}{2}\left|-9+2\right|=\frac{1}{2}(7)=\frac{7}{2} $
$\text a_{33}=\frac{1}{2}\left|-9+3\right|=\frac{1}{2}(6)=3,$ $\text a_{34}=\frac{1}{2}\left|-9+4\right|=\frac {1}{2}(5)=\frac{5}{2} $ $\therefore\ \text{A}=\begin{bmatrix}1 & \frac{1}{2}&0&\frac{1}{2}\\ \frac{5}{2}&2 &\frac{3}{2} & 1\\ 4&\frac{7}{2}&3& \frac{5}{2}\end{bmatrix}$

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