Question

If we have a matrix $\vartriangle =\left| \begin{matrix}
   5 & 3 & 8 \\
   2 & 0 & 1 \\
   1 & 2 & 3 \\
\end{matrix} \right|$, then write the minor of the element ${{a}_{23}}$.
(a) 7
(b) 13
(c) -2
(d) 4

Answer 1

Hint: In order to solve these types of question, we need to understand the element position in a matrix. The element position in a matrix can be understand as $\text{A=}\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right]$. Hence, we can find the minor of any element. The minor of any element ${{a}_{ij}}$ is defined as the sub determinant formed by deleting the $i^{th}$ row and $j^{th}$ column from the given matrix. This is similar in case of finding the minor of any element of a determinant. So, to solve this question, firstly we need to locate the element in the given determinant. After locating the element, we need to form a sub – determinant by deleting the entire row and column where the element is located. Finally, we need to find the value of the sub – determinant which gives us the required value of minor for any element in a determinant.

Complete step-by-step solution:
Here, we are given $\vartriangle =\left| \begin{matrix}
   5 & 3 & 8 \\
   2 & 0 & 1 \\
   1 & 2 & 3 \\
\end{matrix} \right|................(i)$. We need to find the minor of ${{a}_{23}}$ element .
The element position in a matrix can be understand as $\text{A=}\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right]$. Similarly, the element position in a determinant can be understand as $\text{A=}\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|................(ii)$.
In order to find the minor of ${{a}_{23}}$ element, firstly we need to locate the element ${{a}_{23}}$. According to element position in a matrix or determinant, the ${{a}_{23}}$ element from equation (i) and (ii) is ${{a}_{23}}=1$.
So, we have to delete the rows and columns associated with the element ${{a}_{23}}$, i.e., 1. After deleting the rows and column associated with the element ${{a}_{23}}$, i.e., 1, we get,
$\text{Minor of }{{\text{a}}_{23}}=\left| \begin{matrix}
   5 & 3 \\
   1 & 2 \\
\end{matrix} \right|................(iii)$
Now, solving the above obtained sub – determinant in equation (iii), we get,
$\Rightarrow \text{Minor of }{{\text{a}}_{23}}=\left( 5\times 2 \right)-\left( 3\times 1 \right)$
$\Rightarrow \text{Minor of }{{\text{a}}_{23}}=10-3$
$\therefore \text{Minor of }{{\text{a}}_{23}}=7$
Hence, the minor of element ${{a}_{23}}$ is 7.
Thus, option (a) is the correct option.

Note: In these types of the equation, students often make mistakes while deleting the rows and columns associated with the element whose minor is to be found. Students must be careful to identify which of the elements will remain after deleting the rows and columns. Besides, students must understand the element position in a matrix or determinant. This will help to easily identify the element and delete the unrequired rows and columns and the respective elements contained by them.