Cofactor matrix: how is it obtained?

[sigma]

I don't understand how the cofactor of a matrix is obtained. You have a 3x3 matrix with a bunch of 2x2 matrices inside but how the numbers for these 2x2 matrices are obtained is what I don't understand. So I just want to know how to set up the 2x2 matrices inside the 3x3 matrix. If you have a matrix like this, what pattern is being used in the 3x3 matrix with all the smaller 2x2 matrices inside? It doesn't make sense how they got that.
\(\displaystyle \
\L\
\begin{array}{l}
{\rm If }A = \left[ {\begin{array}{*}
1 & 2 & 6 \\
5 & 8 & 7 \\
0 & 3 & 4 \\
\end{array}} \right],{\rm Then cof}(A) = \left[ {\begin{array}{*}
{\left[ {\begin{array}{*}
8 & 7 \\
3 & 4 \\
\end{array}} \right]} & { - \left[ {\begin{array}{*}
5 & 7 \\
0 & 4 \\
\end{array}} \right]} & {\left[ {\begin{array}{*}
5 & 8 \\
0 & 3 \\
\end{array}} \right]} \\
{ - \left[ {\begin{array}{*}
2 & 6 \\
3 & 4 \\
\end{array}} \right]} & {\left[ {\begin{array}{*}
1 & 6 \\
0 & 4 \\
\end{array}} \right]} & { - \left[ {\begin{array}{*}
1 & 2 \\
0 & 3 \\
\end{array}} \right]} \\
{\left[ {\begin{array}{*}
2 & 6 \\
8 & 7 \\
\end{array}} \right]} & { - \left[ {\begin{array}{*}
1 & 6 \\
5 & 7 \\
\end{array}} \right]} & {\left[ {\begin{array}{*}
1 & 2 \\
5 & 8 \\
\end{array}} \right]} \\
\end{array}} \right] \\
= \left[ {\begin{array}{*}
{11} & {- 20} & {15} \\
{10} & 4 & { - 3} \\
{ - 34} & {23} & { - 2} \\
\end{array}} \right] \\
\end{array}
\\)

I know that you alternate the signs +/- between all the 2x2 matrices inside. Also, how do you do this with just a 2x2 matrix? (finding the cofactor of a 2x2 matrix)

 

[galactus]

Determinants, determinants, and more determinants.

To obtain those cofactors from the little submatrices, take the determinants.

\(\displaystyle Det\left(\begin{bmatrix}8&7\\3&4\end{bmatrix}\right)=11\)

And so forth.

Cover the first column and first row and use what matrix stays exposed.

See?. it's \(\displaystyle \begin{bmatrix}8&7\\3&4\end{bmatrix}\)

Keep the first column covered and cover the 2nd row and use what matrix is exposed.

Keep the first column covered and cover the 3rd row and use what matrix is exposed.

See your little matrices now?.

Move to the next column and row unitl you have all 9.

Then take the determinant of each.

Are you trying to find the adjoint?.

 

[sigma]

Yes I was trying to find the adjoint. I understood how to do everything including the determinant it was just that cover the column and row thing that I didn't get. How do you find cofactors for just a 2x2 matrix?

 

[galactus]

For a 2-by-2, do the same thing.

Let's use \(\displaystyle \begin{bmatrix}2&3\\5&1\end{bmatrix}\)

Cover 1st row, 1st column and 1 is left exposed

Cover 1st row, 2nd column and the 5 is left exposed.

Cover 2nd row, 1st column and the 3 is left.

Cover the 2nd row, 2nd column and the 2 is left.

Don't forget to altenate signs and you have:

\(\displaystyle \begin{bmatrix}1&-5\\-3&2\end{bmatrix}\)

Take the transpose to get the adjoint.

\(\displaystyle \text{adjoint}=\begin{bmatrix}1&-3\\-5&2\end{bmatrix}\)

 

[sigma]

How do you find determinants using cofactors? In this example, I now know how to do the cover the row and column for cofactoring, but where do the numbers in front of the det come from?

\(\displaystyle \
\L\
\begin{array}{l}
A = \left[ {\begin{array}{*}
1 & 2 & 6 \\
5 & 8 & 7 \\
0 & 3 & 4 \\
\end{array}} \right] \\
\det (A) = (1)\det \left[ {\begin{array}{*}
8 & 7 \\
3 & 4 \\
\end{array}} \right] - (2)\det \left[ {\begin{array}{*}
5 & 7 \\
0 & 4 \\
\end{array}} \right] + (6)\det \left[ {\begin{array}{*}
5 & 8 \\
0 & 3 \\
\end{array}} \right] \\
= (1)(11) - (2)(20) + (6)(15) = 61 \\
\end{array}
\\)

 

[pka]

How do you find determinants using cofactors? In this example, I now know how to do the cover the row and column for cofactoring, but where do the numbers in front of the det come from?

Expand across the first row! So we use (1), (-2), & (6).
If we expand along the first column we would use: (1), (-5) & (0).
If we expand along the second row we would use: (-5), (8) & (-7).
Note how the signs change: (-1)<SUP>j+k</SUP>a<SUB>j,k</SUB>.