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Matrices — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMatrices2 Marks
Question
Construct a 3$\times$4 matrix, whose elements are given by aij $ = \frac{1}{2}\left| { - 3i + j} \right|$
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Answer

3$\times$4 matrix is given by $A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}} \\ {{a_{21}}} \\ {{a_{31}}} \end{array}\begin{array}{*{20}{c}} {{a_{12}}} \\ {{a_{22}}} \\ {{a_{32}}} \end{array}\begin{array}{*{20}{c}} {{a_{13}}} \\ {{a_{23}}} \\ {{a_{33}}} \end{array}\begin{array}{*{20}{c}} {{a_{14}}} \\ {{a_{24}}} \\ {{a_{34}}} \end{array}} \right]$
Here, ${a_{ij}} = \frac{1}{2}\left| { - 3i + j} \right|$
$\therefore \;{a_{11}} = \frac{1}{2}| - 3 \times 1 + 1| = \frac{1}{2}| - 3 + 1|$ $= \frac{1}{2}| - 2| = \frac{2}{2} = 1$
${a_{21}} = \frac{1}{2}| - 3 \times 2 + 1| = \frac{1}{2}| - 6 + 1| = \frac{1}{2}| - 5| = \frac{5}{2}$
${a_{31}} = \frac{1}{2}| - 3 \times 3 + 1| = \frac{1}{2}| - 9 + 1|$ $ = \frac{1}{2}| - 8| = \frac{8}{2} = 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 $A = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{5}{2}} \\ 4 \end{array}\;\;\begin{array}{*{20}{c}} {\frac{1}{2}} \\ 2 \\ {\frac{7}{2}} \end{array}\;\;\begin{array}{*{20}{c}} 0 \\ {\frac{3}{2}} \\ 3 \end{array}\;\;\begin{array}{*{20}{c}} {\frac{1}{2}} \\ 1 \\ {\frac{5}{2}} \end{array}} \right]$

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