Question

If we have a matrix as \[\Delta =\left( \begin{matrix}
   1 & 2 & 3 \\
   2 & 0 & 1 \\
   5 & 3 & 8 \\
\end{matrix} \right)\], then write the minor of elements \[{{a}_{22}}\]

Answer 1

Hint: Firstly we will represent our given matrix into a general 3x3 matrix from where we will learn that where is \[{{a}_{22}}\] and other similar elements then we will eliminate that row and that column in which \[{{a}_{22}}\] is present then form a matrix out of remaining elements present in the matrix now after getting a 2x2 new matrix of remaining elements we just have to find the determinant of that 2x2 matrix and that determinant is the answer.

Complete step-by-step solution:
Given a 3x3 matrix \[\Delta =\left( \begin{matrix}
   1 & 2 & 3 \\
   2 & 0 & 1 \\
   5 & 3 & 8 \\
\end{matrix} \right)\] and we are asked to find the minor of elements \[{{a}_{22}}\]
Firstly we can write any 3x3 matrix \[\Delta =\left( \begin{matrix}
   1 & 2 & 3 \\
   2 & 0 & 1 \\
   5 & 3 & 8 \\
\end{matrix} \right)\] as \[\left( \begin{matrix}
   1 & 2 & 3 \\
   2 & 0 & 1 \\
   5 & 3 & 8 \\
\end{matrix} \right)=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)\] now we can clearly see all elements of matrix and there representation now we get that \[{{a}_{22}}\] is the middle element which is 0
Second step now eliminate the row and elements column of \[{{a}_{22}}\]or 0 containing. So, after eliminating row and 0 containing column our matrix will look like
\[\left( \begin{matrix}
   1 & {} & 3 \\
   {} & {} & {} \\
   5 & {} & 8 \\
\end{matrix} \right)\], which is a 2x2 matrix, now we just have to find the determinant of this 2x2 matrix which is given by \[\det \left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)={{a}_{11}}\times {{a}_{22}}-{{a}_{12}}\times {{a}_{21}}\]
So, the determinant of our equation \[\left( \begin{matrix}
   1 & {} & 3 \\
   {} & {} & {} \\
   5 & {} & 8 \\
\end{matrix} \right)\] will be given as \[8\times 1-5\times 3=8-15\]
Which comes out be \[-7\]
Hence minor of elements \[{{a}_{22}}\] is \[-7\]

Note: If in this case we have been given 2x2 matrix instead of 3x3 and now we are asked to find minor of the matrix, we will learn by taking an example
\[\Delta =\left( \begin{matrix}
   1 & 3 \\
   4 & 2 \\
\end{matrix} \right)\] here if we have to find minor of \[{{a}_{22}}\]
\[\Delta =\left( \begin{matrix}
   1 & 3 \\
   4 & 2 \\
\end{matrix} \right)=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)\] now we got that \[{{a}_{22}}=2\] so after eliminating that row and that column in which \[{{a}_{22}}\] is present ,element remaining will be 1
hence 1 will be minor of \[{{a}_{22}}\] in matrix \[\Delta =\left( \begin{matrix}
   1 & 3 \\
   4 & 2 \\
\end{matrix} \right)\]