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Algebra of Matrices — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices2 Marks
Question
Construct a $3 × 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by:
$\text{a}_\text{ij}=\frac{1}{2}|-3\text{i}+\text{j}|$
✓

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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