Question

Construct a $3 \times 4$ matrix, whose elements are given by
$
  \left( i \right){a_{ij}} = \dfrac{1}{2}\left| { - 3i + j} \right| \\
  \left( {ii} \right){a_{ij}} = 2i - j \\
$

Answer 1

Hint- Assume a $3 \times 4$ matrix
$ = \left[ {\begin{array}{*{20}{c}}
  {{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]$
and find the value of every element by substituting the suitable values of i & j.

$\left( i \right){a_{ij}} = \dfrac{1}{2}\left| { - 3i + j} \right|$, in $3 \times 4$ matrix, number of rows and column are 3 and 4 respectively.
So, $i$ is for row and $j$ is for column, therefore
$i = 1,2,3{\text{ ;}} j = 1,2,3,4$
As you know modulus of any negative number is positive for examples$\left| { - a} \right| = a$, so, we use this property to evaluate the values of $\left( {{a_{11}},{a_{12}},....................,{a_{34}}} \right)$
$
  {a_{11}} = \dfrac{1}{2}\left| { - 3 + 1} \right| = \dfrac{{\left| { - 2} \right|}}{2} = \dfrac{2}{2} = 1 \\
  {a_{12}} = \dfrac{1}{2}\left| { - 3 + 2} \right| = \dfrac{{\left| { - 1} \right|}}{2} = \dfrac{1}{2} \\
  {a_{13}} = \dfrac{1}{2}\left| { - 3 + 3} \right| = \dfrac{{\left| 0 \right|}}{2} = \dfrac{0}{2} = 0 \\
  {a_{14}} = \dfrac{1}{2}\left| { - 3 + 4} \right| = \dfrac{{\left| 1 \right|}}{2} = \dfrac{1}{2} \\
  {a_{21}} = \dfrac{1}{2}\left| { - 3 \times 2 + 1} \right| = \dfrac{{\left| { - 5} \right|}}{2} = \dfrac{5}{2} \\
  {a_{22}} = \dfrac{1}{2}\left| { - 3 \times 2 + 2} \right| = \dfrac{{\left| { - 4} \right|}}{2} = \dfrac{4}{2} = 2 \\
  {a_{23}} = \dfrac{1}{2}\left| { - 3 \times 2 + 3} \right| = \dfrac{{\left| { - 3} \right|}}{2} = \dfrac{3}{2} \\
  {a_{24}} = \dfrac{1}{2}\left| { - 3 \times 2 + 4} \right| = \dfrac{{\left| { - 2} \right|}}{2} = \dfrac{2}{2} = 1 \\
  {a_{31}} = \dfrac{1}{2}\left| { - 3 \times 3 + 1} \right| = \dfrac{{\left| { - 8} \right|}}{2} = \dfrac{8}{2} = 4 \\
  {a_{32}} = \dfrac{1}{2}\left| { - 3 \times 3 + 2} \right| = \dfrac{{\left| { - 7} \right|}}{2} = \dfrac{7}{2} \\
  {a_{33}} = \dfrac{1}{2}\left| { - 3 \times 3 + 3} \right| = \dfrac{{\left| { - 6} \right|}}{2} = \dfrac{6}{2} = 3 \\
  {a_{34}} = \dfrac{1}{2}\left| { - 3 \times 3 + 4} \right| = \dfrac{{\left| { - 5} \right|}}{2} = \dfrac{5}{2} \\
$
So, the required $3 \times 4$ matrix is,
$ = \left[ {\begin{array}{*{20}{c}}
  1&{\dfrac{1}{2}}&0&{\dfrac{1}{2}} \\
  {\dfrac{5}{2}}&2&{\dfrac{3}{2}}&1 \\
  4&{\dfrac{7}{2}}&3&{\dfrac{5}{2}}
\end{array}} \right]$


$\left( {ii} \right){a_{ij}} = 2i - j$
${a_{ij}} = 2i - j$, in $3 \times 4$ matrix, number of rows and columns are 3 and 4 respectively.
So, $i$ is for row and $j$ is for column, therefore
$i = 1,2,3{\text{ ;}}j = 1,2,3,4$
Now, evaluate the values of $\left( {{a_{11}},{a_{12}},....................,{a_{34}}} \right)$
$
  {a_{11}} = 2 \times 1 - 1 = 1 \\
  {a_{12}} = 2 \times 1 - 2 = 0 \\
  {a_{13}} = 2 \times 1 - 3 = - 1 \\
  {a_{14}} = 2 \times 1 - 4 = - 2 \\
  {a_{21}} = 2 \times 2 - 1 = 3 \\
  {a_{22}} = 2 \times 2 - 2 = 2 \\
  {a_{23}} = 2 \times 2 - 3 = 1 \\
  {a_{24}} = 2 \times 2 - 4 = 0 \\
  {a_{31}} = 2 \times 3 - 1 = 5 \\
  {a_{32}} = 2 \times 3 - 2 = 4 \\
  {a_{33}} = 2 \times 3 - 3 = 3 \\
  {a_{34}} = 2 \times 3 - 4 = 2 \\
$
So, the required $3 \times 4$ matrix
$ = \left[ {\begin{array}{*{20}{c}}
  1&0&{ - 1}&{ - 2} \\
  3&2&1&0 \\
  5&4&3&2
\end{array}} \right]$
So, these are the required matrices.

Note: In these types of questions always remember that in an $m \times n$ matrix, the number of rows and columns are m and n respectively. Calculate all the elemental values of the matrix according to the given condition.