How to find all the minors and cofactors of the matrix #A=((1, -2, 3), ( 6, 7, -1 ), (-3, 1, 4))#?

1 Answer
Sep 12, 2015

The minors of a matrix are the determinants of the smaller matrices you get when you delete one row and one column of the original matrix. The cofactors of a matrix are the matrices you get when you multiply the minor by the right sign (positive or negative).

Explanation:

There is a minor and a cofactor for every entry in the matrix -- so that's 9 altogether! I'll just go through a few of them.

The minor for the first row and column is called #M_(1,1)#. We calculate it by removing the first row and column of the matrix, so we are left with the matrix:

#((7,-1), (1,4))#

Then we take the determinant of this matrix, which is #(7*4) - (-1*1) = 28 + 1 = 29#. So #M_(1,1) = 29#.

To find the corresponding cofactor, called #A_(1,1)#, we look at the subscripts on the name of the minor -- in our case, #(1,1)#, and we add them together to get 2. Then we raise #-1# to this value, so we get #(-1)^2 = 1#, and we multiply this by the minor. So that gives us #A_(1,1)= 29*1 = 29# as the cofactor.

Let's do one more. To find the minor #M_(2,3)# we delete the second row and third column of the matrix, so we are left with:

#((1, -2),(-3,1))#.

Then we take the determinant of this matrix, which is #(1*1)-(-3*-2) = 1 - 6 = -5#. So #M_(2,3) = -5#.

To find the cofactor #A_(2,3)#, we add the subscripts #(2,3)# to get 5, and raise #-1# to this value, which gets us #(-1)^5 = -1#. So we multiply the minor #-5# by #-1# to get #A_(2,3) = 5#.