Construct a 3 4 matrix A = [a_ ij ] whose element a_ ij are given by: a_ij=1 2 |-3i+j|

Here, A=(a_ij)_ 3 4 = a_ 11 &a_ 12 &a_ 13 &a_ 14 _ 21 &a_ 22 &a_ 23 &a_ 24 _ 31 &a_ 32 &a_ 33 &a_ 34 (1) a_ 11 |-3(1)+1|=1 2 |-2|=1, a_ 12 =1 2 |-3(1)+2|=1 2 |-1|=1 2 , a_ 13 =1 2 |-3(1)+3|=1 2 |0|=0 2 =0, a_ 14 =1 2 |-3(1)+4|=1 4 |1|=1 2 a_ 21 =1 2 |-3(2)+1|=1 2 |-5|=5 2 , a_ 22 =1 2 |-3(2)+2|=1 2 |-4|=2, a_ 23 =1 2 |-3(2)+3|=1 2 |-3|=3 2 , a_ 24 =1 2 |-3(2)+4|=1 2 |-2|=1 a_ 31 =1 2 |-3(3)+1|=1 2 |-8|=4, a_ 32 =1 2 |-3(3)+2|=1 2 |-7|=7 2 , a_ 33 =1 2 |-3(3)+3|=1 2 |-6|=3 and a_ 34 =1 2 |-3(3)+4|=1 2 |-5|=5 2 So, the required matrix is 1&1 2 &0&1 2 5 2 &2&3 2 &1 4&7 2 &3&5 2 .

Construct a 3 4 matrix A = [a_ ij ] whose element a_ ij are given…

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Question

Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by$: \text{a}_\text{ij}=\frac{1}{2}|-3\text{i}+\text{j}|$

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Answer

Here, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$\text{a}_{11}|-3(1)+1|=\frac{1}{2}|-2|=1,$ $\text{a}_{12}=\frac{1}{2}|-3(1)+2|=\frac{1}{2}|-1|=\frac{1}{2},$
$\text{a}_{13}=\frac{1}{2}|-3(1)+3|=\frac{1}{2}|0|=\frac{0}{2}=0,$ $\text{a}_{14}=\frac{1}{2}|-3(1)+4|=\frac{1}{4}|1|=\frac{1}{2}$
$\text{a}_{21}=\frac{1}{2}|-3(2)+1|=\frac{1}{2}|-5|=\frac{5}{2},$ $\text{a}_{22}=\frac{1}{2}|-3(2)+2|=\frac{1}{2}|-4|=2,$
$\text{a}_{23}=\frac{1}{2}|-3(2)+3|=\frac{1}{2}|-3|=\frac{3}{2},$ $\text{a}_{24}=\frac{1}{2}|-3(2)+4|=\frac{1}{2}|-2|=1$
$\text{a}_{31}=\frac{1}{2}|-3(3)+1|=\frac{1}{2}|-8|=4,$ $\text{a}_{32}=\frac{1}{2}|-3(3)+2|=\frac{1}{2}|-7|=\frac{7}{2},$
$\text{a}_{33}=\frac{1}{2}|-3(3)+3|=\frac{1}{2}|-6|=3$ and $\text{a}_{34}=\frac{1}{2}|-3(3)+4|=\frac{1}{2}|-5|=\frac{5}{2}$
So, the required matrix is $\begin{bmatrix}1&\frac{1}{2}&0&\frac{1}{2}\\\frac{5}{2}&2&\frac{3}{2}&1\\4&\frac{7}{2}&3&\frac{5}{2}\end{bmatrix}.$

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