Chapter 8, Problem 149RE

To determine

Give the explanation for the cofactors of an entry of a matrix and about the cofactors used to find the determinant of the matrix too.

Explanation of Solution

Given information:

Conditions for the inverse matrix.

Calculation:

To define the determinant of a square matrix of order $3\times 3$ or higher, it is convenient to introduce the concepts of minors and cofactors.

If $A$ is square matrix, then the minor $M_{ij}$ of the entry $a_{ij}$ is the determinant of the matrix obtained by deleting the $i\text{th}$ row and $j\text{th}$ column of A . The cofactor $C_{ij}$ of the entry $a_{ij}$ is

  $C_{ij}={\left(-1\right)}^{i+j}{M}_{ij}$

In the sign pattern for cofactors at the left, notice that odd positions ( where $i+j$ is odd) have negative signs and even positions ( where $i+j$ is even) have positive signs.

If $A$ is a square matrix( of order $2\times 2$ or greater)

  $\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& ---\\ a_{21}& a_{22}& a_{23}& ---\\ a_{31}& a_{32}& a_{33}& ---\\ \begin{array}{l}|\\ |\\ |\end{array}& \begin{array}{l}|\\ |\\ |\end{array}& \begin{array}{l}|\\ |\\ |\end{array}& \begin{array}{l}|\\ |\\ |\end{array}\end{array}\right]$ , then the determinant of A is the sum of the entries in any row (or column) of A multiplied by their respective cofactors. For instance, expanding along the first row yields

det

  $\left(A\right)$

  $=\left|A\right|=a_{11}{c}_{11}+a_{12}{c}_{12}+---+a_{1n}{c}_{11n}$

Applying this definition to find a determinant is called expanding by cofactors.

Example$\left|\begin{array}{ccc}1& 1& 4\\ -4& 1& 2\\ 0& 1& -1\end{array}\right|$

Cofactors are (first column and first row).

  $\begin{array}{l}{C}_{11}=1\cdot \left[1\cdot \left(-1\right)-1\cdot 2\right]=-3\\ C_{12}=-1\left[\left(-4\right)\cdot \left(-1\right)-0\cdot 2\right]=-4\\ C_{13}=+4\cdot \left[\left(-4\right)\cdot 1-0\cdot 1\right]=-16\end{array}$

  $\begin{array}{l}\det =1\cdot \left[1\cdot \left(-1\right)-1\cdot 2\right]-1\left[\left(-4\right)\cdot \left(-1\right)-0\cdot 2\right]+4\cdot \left[\left(-4\right)\cdot 1-0\cdot 1\right]\\ =-3-4-16\\ =-23\end{array}$