Question

Construct $(3 \times 2)$ the matrix whose elements are given by ${a_n} = \dfrac{1}{2}\left[ {i - 3j} \right]$

Answer 1

Hint: Matrix is a set of numbers arranged in rows and columns so as to form a rectangular array. The number of $\text{Row}\times \text{Columns}$ are called the element or entries of the matrix. The number of rows and columns of the matrix is called the order or its dimension of the matrix. Only a zero matrix has zero rank.

Complete step-by-step solution:
Given,
${a_{ij}} = \dfrac{1}{2}\left[ {i - 3j} \right]$
The order of the matrix is $[3 \times 2]$
This means number of rows, $i = 3$
Number of columns, $j = 2$
The matrix will be
\[\therefore {a_{3 \times 2}} = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}} \\
  {{a_{31}}}&{{a_{32}}}
\end{array}} \right]\]
Now the to solve for the matrix
As given
${a_{ij}} = \dfrac{1}{2}\left[ {i - 3j} \right]$
Put
$ \Rightarrow {a_{11}} = \dfrac{1}{2}\left[ {1 - 3 \times 1} \right]$
\[ \Rightarrow {a_{11}} = \dfrac{1}{2}\left( { - 2} \right)\]
\[ \Rightarrow {a_{11}} = - 1\]
$ \Rightarrow {a_{12}} = \dfrac{1}{2}\left[ {1 - 3 \times 2} \right]$
\[ \Rightarrow {a_{12}} = \dfrac{1}{2}\left( { - 5} \right)\]
\[ \Rightarrow {a_{12}} = - \dfrac{5}{2}\]
$ \Rightarrow {a_{21}} = \dfrac{1}{2}\left[ {2 - 3 \times 1} \right]$
\[ \Rightarrow {a_{21}} = \dfrac{1}{2}\left( { - 1} \right)\]
\[ \Rightarrow {a_{21}} = - \dfrac{1}{2}\]
$ \Rightarrow {a_{22}} = \dfrac{1}{2}\left[ {2 - 3 \times 2} \right]$
\[ \Rightarrow {a_{22}} = \dfrac{1}{2}\left( { - 4} \right)\]
\[ \Rightarrow {a_{22}} = - 2\]
$ \Rightarrow {a_{31}} = \dfrac{1}{2}\left[ {3 - 3 \times 1} \right]$
\[ \Rightarrow {a_{31}} = \dfrac{1}{2}\left( 0 \right)\]
\[ \Rightarrow {a_{31}} = 0\]
$ \Rightarrow {a_{32}} = \dfrac{1}{2}\left[ {3 - 3 \times 2} \right]$
\[ \Rightarrow {a_{32}} = \dfrac{1}{2}\left( { - 3} \right)\]
\[ \Rightarrow {a_{32}} = - \dfrac{3}{2}\]
As we know
$\therefore {a_{3 \times 2}} = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}} \\
  {{a_{31}}}&{{a_{32}}}
\end{array}} \right]$
Put these values in matrix and we get
\[\therefore {a_{3 \times 2}} = \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - \dfrac{5}{2}} \\
  { - \dfrac{1}{2}}&{ - 2} \\
  0&{ - \dfrac{3}{2}}
\end{array}} \right]\]

Note: When we do multiplication in a matrix we have to keep things in mind: the number of columns of the first matrix is equal to the rows of the second matrix. And the result we get the same number of rows in the first matrix and the column in the second matrix. Determinant is the type of square matrix.