
Find the co-factors of the elements of $A = \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right]$
Answer
508.2k+ views
Hint:
We are the matrix $A = \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right]$. We will find the co-factors of the elements as ${\left( { - 1} \right)^{i + j}}{M_{ij}}$, where ${M_{ij}}$is the corresponding minor of the element. Substitute the values of $i$, the row of the element and $j$, the column of the element for each element of the given matrix to find the corresponding co-factors.
Complete step by step solution:
We have to determine the co-factors of $A = \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right]$
The cofactors of the matrix $\left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right]$, then the co-factors can be calculated as ${\left( { - 1} \right)^{i + j}}{M_{ij}}$, where ${M_{ij}}$is the corresponding minor of the element.
Minor of an element can be obtained by deleting the row and column of that particular element, the element left after deleting the row and column is minor in case of $2 \times 2$ matrix.
Substitute the values of $i$, the row of the element and $j$, the column of the element of the given matrix.
Then, the co-factor corresponding to ${a_{11}} = 1$ is ${\left( { - 1} \right)^{1 + 1}}{M_{11}} = {\left( { - 1} \right)^2}4 = 4$
Similarly, the co-factor corresponding to ${a_{12}} = 2$ is ${\left( { - 1} \right)^{1 + 2}}{M_{12}} = {\left( { - 1} \right)^3}3 = - 3$
The co-factor corresponding to ${a_{21}} = 3$ is ${\left( { - 1} \right)^{2 + 1}}{M_{21}} = {\left( { - 1} \right)^3}2 = - 2$
And the co-factor of the element ${a_{11}} = 1$ is ${\left( { - 1} \right)^{2 + 2}}{M_{22}} = {\left( { - 1} \right)^4}1 = 1$
Thus the co-factor of the element 1 is 4, 2 is $ - 3$, 3 is $ - 2$ and 4 is 1.
Note:
The number of minors and co-factors of a matrix is the number of the elements of the matrix. We use co-factors to calculate the adjoint of the matrix and hence the inverse of the matrix. Minor of an element can be obtained by deleting the row and column of that particular element, we take the determinant of the elements left after deleting the row and column is the minor of that element.
We are the matrix $A = \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right]$. We will find the co-factors of the elements as ${\left( { - 1} \right)^{i + j}}{M_{ij}}$, where ${M_{ij}}$is the corresponding minor of the element. Substitute the values of $i$, the row of the element and $j$, the column of the element for each element of the given matrix to find the corresponding co-factors.
Complete step by step solution:
We have to determine the co-factors of $A = \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right]$
The cofactors of the matrix $\left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right]$, then the co-factors can be calculated as ${\left( { - 1} \right)^{i + j}}{M_{ij}}$, where ${M_{ij}}$is the corresponding minor of the element.
Minor of an element can be obtained by deleting the row and column of that particular element, the element left after deleting the row and column is minor in case of $2 \times 2$ matrix.
Substitute the values of $i$, the row of the element and $j$, the column of the element of the given matrix.
Then, the co-factor corresponding to ${a_{11}} = 1$ is ${\left( { - 1} \right)^{1 + 1}}{M_{11}} = {\left( { - 1} \right)^2}4 = 4$
Similarly, the co-factor corresponding to ${a_{12}} = 2$ is ${\left( { - 1} \right)^{1 + 2}}{M_{12}} = {\left( { - 1} \right)^3}3 = - 3$
The co-factor corresponding to ${a_{21}} = 3$ is ${\left( { - 1} \right)^{2 + 1}}{M_{21}} = {\left( { - 1} \right)^3}2 = - 2$
And the co-factor of the element ${a_{11}} = 1$ is ${\left( { - 1} \right)^{2 + 2}}{M_{22}} = {\left( { - 1} \right)^4}1 = 1$
Thus the co-factor of the element 1 is 4, 2 is $ - 3$, 3 is $ - 2$ and 4 is 1.
Note:
The number of minors and co-factors of a matrix is the number of the elements of the matrix. We use co-factors to calculate the adjoint of the matrix and hence the inverse of the matrix. Minor of an element can be obtained by deleting the row and column of that particular element, we take the determinant of the elements left after deleting the row and column is the minor of that element.
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