Answer: The cofactor of an element (a_{ij}) in a matrix is defined as:
[ C_{ij} = (-1)^{i+j} M_{ij} ]
where (M_{ij}) is the minor obtained by deleting the (i^{th}) row and (j^{th}) column.
[
A=
\begin{bmatrix}
2 & 3
5 & 7
\end{bmatrix}
]
Answer:
Element 5 is at position ((2,1)).
Minor:
[ M_{21}=3 ]
Cofactor:
[ C_{21}=(-1)^{2+1}(3) ]
[ C_{21}=(-1)^3(3)=-3 ]
Final Answer: (-3)
[
A=
\begin{bmatrix}
1 & 3
4 & 2
\end{bmatrix}
]
Answer:
Element 3 is at position ((1,2)).
Minor:
[ M_{12}=4 ]
Cofactor:
[ C_{12}=(-1)^{1+2}(4) ]
[ =-4 ]
Final Answer: (-4)
[
A=
\begin{bmatrix}
1 & 2
3 & 4
\end{bmatrix}
]
Answer:
[ C_{11}=(-1)^{1+1}(4)=4 ]
[ C_{12}=(-1)^{1+2}(3)=-3 ]
[ C_{21}=(-1)^{2+1}(2)=-2 ]
[ C_{22}=(-1)^{2+2}(1)=1 ]
Cofactor Matrix:
[
\begin{bmatrix}
4 & -3
-2 & 1
\end{bmatrix}
]
[
A=
\begin{bmatrix}
1 & 2 & 3
4 & 5 & 6
7 & 8 & 9
\end{bmatrix}
]
Answer:
Element 6 is at position ((2,3)).
Delete row 2 and column 3:
[
\begin{bmatrix}
1 & 2
7 & 8
\end{bmatrix}
]
Minor:
[ M_{23}=(1)(8)-(2)(7) ]
[ =8-14=-6 ]
Cofactor:
[ C_{23}=(-1)^{2+3}(-6) ]
[ =(-1)^5(-6)=6 ]
Final Answer: 6
[
A=
\begin{bmatrix}
2 & 1 & 3
4 & 5 & 6
7 & 8 & 9
\end{bmatrix}
]
Answer:
Element 8 is at position ((3,2)).
Delete row 3 and column 2:
[
\begin{bmatrix}
2 & 3
4 & 6
\end{bmatrix}
]
Minor:
[ M_{32}=(2)(6)-(3)(4) ]
[ =12-12=0 ]
Cofactor:
[ C_{32}=(-1)^{3+2}(0)=0 ]
Final Answer: 0
Answer:
The sign pattern is:
[ \begin{bmatrix}
Answer:
[
A=
\begin{bmatrix}
2 & 0
1 & 5
\end{bmatrix}
]
Answer:
[ C_{11}=5 ]
[ C_{12}=-1 ]
[ C_{21}=0 ]
[ C_{22}=2 ]
Cofactor Matrix:
[
\begin{bmatrix}
5 & -1
0 & 2
\end{bmatrix}
]
Answer:
Cofactors are used for: