# Linear Algebra

## Determinants by Cofactors

We came across the determinant of a $$2\times 2$$ matrix in a previous lesson. That determinant can be useful when to find determinants of larger square matrices. This is known as the cofactor method.

 The determinant of a $$2\times 2$$ matrix $$A=\left[\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right]$$ is defined by $detA=\left|\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right|=det\left[\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right]=a_{11}a_{22}-a_{12}a_{21}$

Example: Find the determinant of $$\left[\begin{array}{cc} 10&-3\\4&-8\end{array}\right]$$

Solution: $$\left|\begin{array}{cc} 10&-3\\4&-8\end{array}\right|=10(-8)-4(-3)=-68$$

 Cofactors of a $$3\times 3$$ matrix Let $$A\in M_{3\times 3}(\mathbb{R})$$ and let $$A(i,j)$$ denote the $$2\times 2$$ submatrix obtained from $$A$$ by deleting the i-th row and the j-th column. Define the $$(i,j)-cofactor of A to be $C_{ij}=(-1)^{i+j}|A(i,j)|$ Example: Let \(A=\left[\begin{array}{ccc} 1&0&3\\ 0&-1&3\\1&-2&3\end{array}\right]$$

Cofactor $$C_{11}$$ is defined by $$C_{11}=(-1)^{1+1}|A(1,1)|$$

This means to remove row 1 and column 1 from the matrix and calculate the determinant of the remaining submatrix $$A(1,1)$$.

$A(1,1)=\left[\begin{array}{cc} -1&3\\-2&3\end{array}\right]$

The determinant $$|A(1,1)|=\left|\begin{array}{cc} -1&3\\-2&3\end{array}\right|=3$$, therefore, the cofactor $$C_{11}=+3$$

Let's calculate $$C_{12}$$, which requires us to remove row 1 and column 2, with submatrix

$A(1,2)=\left[\begin{array}{cc} 0&3\\1&3\end{array}\right]$

The determinant $$|A(1,2)|=\left|\begin{array}{cc} 0&3\\1&3\end{array}\right|=-3$$, therefore, the cofactor $$C_{12}=(-1)^{1+2}(-3)=-(-3)=3$$

Continuing with this process, the determinant $$|A(1,3)|=\left|\begin{array}{cc} 0&-1\\1&-2\end{array}\right|=1$$, therefore, the cofactor $$C_{13}=(-1)^{1+3}(-3)=+(1)=1$$.

There are 9 possible cofactors of matrix $$A$$. Can you find them all?

 Determinant of $$n\times n$$ matrix using Cofactors Let $$A\in M_{n\times n}(\mathbb{R})$$ with $$n>2$$. Let $$A(i,j)$$ denote the $$(n-1)\times (n-1)$$ submatrix obtained from $$A$$ by deleting the i-th row and j-th column.  The determinant of $$A\in M_{n\times n}(\mathbb{R})$$ is defined by  $|A|=a_{k1}C_{k1}+a_{k2}C_{k2}+\ldots+a_{kn}C_{kn}$ where $$k$$ is any row and the $$(i,j)$$-cofactor of $$A$$ is defined to be $C_{ij}=(-1)^{i+j}|A(i,j)|$ The determinant can also be defined along any column $$l$$ $|A|=a_{1l}C_{1l}+a_{2l}C_{2l}+\ldots+a_{nl}C_{nl}$

Example: Find the determinant of $$A=\left[\begin{array}{ccc} 1&0&3\\ 0&-1&3\\1&-2&3\end{array}\right]$$

Solution

We can find the determinant using cofactors of row 1.

\begin{align} |A|&=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}\\ &=1(3)+0(3)+3(1)\\&=6\end{align}

Tips:

• Since any row or column can be chosen. Choosing the row or column with the most 0 elements will decrease steps in calculation.
• You can perform adding or subtracting a row by another first then calculate the determinant.