A is a 4x4 matrix. The cofactor = {(-1)^(i+j) }{determinant in a minor}.

Can the determinant in the minor be 2x2 or 3x3?

Thanks(Nod)

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- Dec 18th 2013, 01:33 AMeuleriancofactor
A is a 4x4 matrix. The cofactor = {(-1)^(i+j) }{determinant in a minor}.

Can the determinant in the minor be 2x2 or 3x3?

Thanks(Nod) - Dec 18th 2013, 02:24 AMromsekRe: cofactor
the determinant is just a number. For that matter the (i,j) minor is the determinant of the matrix formed by deleting the ith row and jth column from a square matrix. So the phrase "determinant in a minor" doesn't really make sense. You would just say the (i,j) cofactor is (-1)

^{i+j}(i,j) minor.

The cofactors for a 4x4 matrix will be found by taking the determinants of 3x3 matrices and in general the cofactors of an NxN matrix are found by taking the determinants of (N-1)x(N-1) matrices. - Dec 18th 2013, 02:54 AMeulerianRe: cofactor
I mean, a minor of a matrix A is the determinant of some smaller square matrix, cut down from A by removing one or more of its rows or columns, right?

For example, a 3x3 matrix B,

B=1 2 3

__4 5 6

__7 8 9

("___"due to spacing)

minor at 1,1= 5 6

____________8 9

So, a 4x4 matrix can have minors which cut down 1 row & 1 column(case 1), or 2 vows & 2 columns(case 2), right?

So, minor in case 1 is in a form like a 3x3 matrix in calculating determinant; minor in case 2 like 2x2 matrix, right?

I know minor and cofactor are finally numbers in calculation, I mean the process.

Thanks. - Dec 18th 2013, 07:11 AMromsekRe: cofactor
yes that's correct. the (i,j) minor is just the determinant of the original matrix with the ith row and jth column removed.

oh I see what you're getting at.

Yes. the process of finding the determinant of a matrix is basically recursive. You work further and further down the minors until you end up with a 2x2 matrix who's determinant is easily found.

I should caveat that. The process of finding the determinant of a matrix is recursive if you do it using cofactors. There are other methods where you'd diagonalize, or partially diagonalize the matrix and then take the determinant as the product of the trace elements. There's nothing recursive about that process. - Dec 18th 2013, 07:38 AMDevenoRe: cofactor
If I understand your question correctly, the answer is "sometimes, but usually not". Consider the matrix:

$\displaystyle A = \begin{bmatrix}1&a&b&c\\0&1&d&e\\0&0&f&g\\0&0&h&k \end{bmatrix}$

Expanding by minors along the first column, we get:

$\displaystyle \det(A)$

$\displaystyle = 1\ast\begin{vmatrix}1&d&e\\0&f&g\\0&h&k \end{vmatrix} + (-1)\ast(0)\ast \begin{vmatrix}a&b&c\\0&f&g\\0&h&k \end{vmatrix} + 0\ast \begin{vmatrix}a&b&c\\1&d&e\\0&h&k \end{vmatrix} + (-1)\ast(0)\ast \begin{vmatrix}a&b&c\\1&d&e\\0&f&g \end{vmatrix}$

We don't even have to evaluate the last 3 determinants, since we're just going to multiply whatever we get by 0, so:

$\displaystyle \det(A) = \begin{vmatrix}1&d&e\\0&f&g\\0&h&k \end{vmatrix}$

Expanding THIS matrix by minors along the first column, we get:

$\displaystyle \det(A) = \det\left(\begin{bmatrix}f&g\\h&k \end{bmatrix}\right)$,

so that in this "special case", a 4x4 determinant actually reduces to calculating a 2x2 determinant.

It is worth noting that if a matrix is upper-triangular, the determinant can be calculated by just multiplying the diagonal elements together. This means that row-reduction may be a more efficient way to calculate determinants:

If $\displaystyle A = PB$, where $\displaystyle P$ is a product of elementary row-operation matrices, and $\displaystyle B$ is upper-triangular, then both det(P) and det(B) are fairly easy to calculate:

The determinant of the matrix that swaps two rows is 1.

The determinant of the matrix that multiplies a row by r is r.

The determinant of the matrix that adds r times row j to row i (where i and j are distinct) is 1.

So if you keep track of which row-operations you do when row-reducing, and multiply all of their determinants together, that gives you det(P).

Since det(A) = det(PB) = det(P)det(B), this often saves a LOT of time. - Dec 20th 2013, 05:46 AMeulerianRe: cofactor
Thank you