Question

What is the cofactor expansion method to finding the determinant?

Answer 1

Hint: We need to explain the cofactor expansion concept for finding the determinant in the topic of matrices. Let us explain this with a simple example. This is usually a method by splitting the given matrix into smaller components in order to easily calculate the determinant. Cofactor is obtained by considering the minor matrix which is of the order $(n-1)\times (n-1)$ and its formula can be given by $C_{ij}={(-1)}^{i+j}det\left(A_{ij}\right)$ . Here, $A_{ij}$ is the minor matrix.

Complete step by step solution:
In order to answer this question, let us explain the concept of cofactors. Suppose, we are given a matrix as $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and we are required to calculate the determinant using the concept of cofactors, we first consider the minor matrices. Suppose we want the minor matrix $A_{ij},$ we need to remove the elements from the row i and column j and write the remaining matrix.
The above matrix has positions shown by
$\Rightarrow \left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$
Suppose we need the minor matrix $A_{11},$ we need to eliminate the row 1 and column 1 from the given matrix. Eliminating row 1, we remove elements 1 and 2 and eliminating column 1, we remove elements 1 and 3. The remaining elements then form the minor matrix which in this case is
$\Rightarrow A_{11}=\left[4\right]$
Similarly, we calculate the other 3 minor matrices and we obtain them as,
For $A_{12},$ we eliminate row 1 and column 2 and write the remaining elements as,
$\Rightarrow A_{12}=\left[3\right]$
For $A_{21},$ we eliminate row 2 and column 1 and write the remaining elements as,
$\Rightarrow A_{21}=\left[2\right]$
For $A_{22},$ we eliminate row 2 and column 2 and write the remaining elements as,
$\Rightarrow A_{22}=\left[1\right]$
The cofactors can be calculated by using the formula $C_{ij}={(-1)}^{i+j}det\left(A_{ij}\right).$
For the first cofactor i and j are 1 and 1,
$\Rightarrow C_{11}={(-1)}^{1+1}det\left(A_{11}\right)$
We know the determinant of a single element is the element itself. Therefore,
$\Rightarrow C_{11}={(-1)}^{2}det\left(\left[4\right]\right)$
$\Rightarrow C_{11}=4$
Similarly, we calculate the second cofactor by considering i as 1 and j as 2.
$\Rightarrow C_{12}={(-1)}^{1+2}det\left(A_{12}\right)$
We get the power of -1 as and odd number hence, the cofactor will be negative.
$\Rightarrow C_{12}={(-1)}^{3}det\left(\left[3\right]\right)$
$\Rightarrow C_{12}=-3$
Next, we calculate the third cofactor $C_{21}.$
$\Rightarrow C_{21}={(-1)}^{2+1}det\left(A_{21}\right)$
Again, the power of -1 is odd so cofactor is negative.
$\Rightarrow C_{21}={(-1)}^{3}det\left(\left[2\right]\right)$
$\Rightarrow C_{21}=-2$
We calculate the last cofactor by using the same formula for $C_{22}.$
$\Rightarrow C_{22}={(-1)}^{2+2}det\left(A_{22}\right)$
This is positive since the power of -1 is an even number.
$\Rightarrow C_{22}={(-1)}^{4}det\left(\left[1\right]\right)$
$\Rightarrow C_{22}=1$
The determinant can be found from cofactors using the formula $det\left(A\right)=\sum _{i=1}^n{a}_{i1}{C}_{i1}$ . We need to substitute the values here for the terms and cofactors.
$\Rightarrow det\left(A\right)=a_{11}{C}_{11}+a_{21}{C}_{21}$
Substituting the values,
$\Rightarrow det\left(A\right)=1\times 4+3\times -2=4-6$
Subtracting the terms, we get the determinant.
$\Rightarrow det\left(A\right)=-2$
Hence, we have shown how we can calculate the determinant using the cofactor method.

Note: We need to know the basic matrix solving techniques in order to solve such questions. We need to note that the determinant calculation happens only by traversing the matrix along the first row because we have already calculated the cofactors for all the terms below.