Question

How do you find the determinant of a $2 \times 3$ matrix?

Answer 1

Hint: Row elements are horizontal lines and columns are vertical elements like $(2,1)$ is the row one element.
To know the matrix $2 \times 3$ if a determinant exists or not, we will first need to know about the determinant definition.
Also, if there is an inverse exists matrix is known as the non-singular because of the determinant non-zero.
Square matrix means, same order matrix-like$2 \times 2,3 \times 3$.

Complete step-by-step solution:
The definition of the determinant says;
Let us take n-order determinant of a square matrix \[A = {\text{ }}[{a_{ij}}]\] , with plus and minus signs alternating, where the entries \[{a_{11}},{a_{12}},{\text{ }} \ldots ,{a_{1n}}\] are from the first row of A.
In symbols;
\[det\left( A \right){\text{ }} = {\text{ }}\left| A \right|{\text{ }} = {a_{11}}det{A_{11}} - {a_{12}}det{A_{12}} + \ldots + {\left( { - 1} \right)^{1 + n}}{a_{1n}}de{t_{1n}}\]
$ = \sum a _{(j = 1)}^n{a_1}_jdet{A_1}_j$
Since $2 \times 3$ can be expressed as $\left( {\begin{array}{*{20}{c}}
  a&f&d \\
  b&c&e
\end{array}} \right)$ two rows and three columns and which is not the square matrix.
Hence if we need to find the determinant of the matrix then the matrix must be same order like $1 \times 1,2 \times 2,....n \times n$
But from the given, we clearly see that $2 \times 3$ is not the square matrix and hence we cannot find its determinant.
Also, if a matrix with non-zero determinant then we can able to find its inverse matrix
Additional information:
If for example take a $2 \times 2$ matrix which is $A = \left( {\begin{array}{*{20}{c}}
  2&1 \\
  4&3
\end{array}} \right)$then we can able to find its ${A^{ - 1}}$
If we see $\left| A \right| = \left| {\begin{array}{*{20}{c}}
  2&1 \\
  4&3
\end{array}} \right| = (2 \times 3) - (1 \times 4)$
$ \Rightarrow 6 - 4$
$ \Rightarrow 2$
Thus, the determinant of a matrix $A$ is non zero, that is $\left| A \right| = 2$
Hence $\left| A \right| \ne 0$(determinant will do not equal to zero)
Thus, now we can able to find ${A^{ - 1}}$ the matrix $A = \left( {\begin{array}{*{20}{c}}
  2&1 \\
  4&3
\end{array}} \right)$ by applying ${A^{ - 1}}$= $\dfrac{1}{{\left| A \right|}} \times adjA$

Note: A matrix is nonsingular ($\left| A \right| \ne 0$) then we are able to find its inverse form.
If it singular ($\left| A \right| = 0$) then we cannot find its inverse form.
Also, in a matrix nonsingular matrix = invertible matrix.
If the order of the elements is not equal (not the same size), then it is called a non-square matrix.
$1 \times 2,5 \times 7$ and determinant not possible.