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Minors and Cofactors:
  Expanding Down a Column
(page 2 of 3)


    (b) To expand along the third column, I need to find the minors and then the cofactors of the third-column entries: a1,3, a2,3, a3,3, and a4,3.

matrix with 1,3-entry circled, first row and third columns crossed off

matrix with 2,3-entry circled, second row and third columns crossed off

matrix with 3,3-entry circled, third row and third columns crossed off

matrix with 4,3-entry circled, fourth row and third columns crossed off





    Wait a sec... The a2,3-entry of the original matrix is zero. This means that I'll be getting zero for that term when I expand down the column, no matter what the value of the minor M2,3 turns out to be. So I don't really care what the A2,3 cofactor is; I can just put "0" for this entry, because a2,3A2,3 = (0)(A2,3) = 0. In fact, I can ignore each of the last three terms in the expansion down the third column, because the third column's entries (other than the first entry) are all zero.

    So the only computation I care about is one I've already done (for part (a), when expanding across the first row):

      det(A) = a1,3A1,3
                 = (2)(3)

    (c) Comparison: The value of the determinant is the same in each expansion.

In the example above, we expanded by taking the 4-by-4 matrix down to 3-by-3 determinants. But technically, you're "supposed" to go down to 2-by-2 determinants when you "expand" by this method. That is, the above cofactor "should" have been computed using many more steps. Suppose you'd gone across the first row again. The first cofactor "should" have been computed like this:

a1,1-entry selected

minor M1,1

matrix with 1,1-entry selected, and first row and first column crossed out

|| 1 0 -1 || 1 0 -2 || -1 0 3 ||

By crossing out a row and column, a new matrix is formed. Call this new matrix "C". To find the determinant of C, expand along the first row.




1,1-entry circled, with first row and first column crossed off

1,2-entry circled, with first row and second column crossed off

1,3-entry circled, with first row and third column crossed off




(-1)^(1+1) || 0 -2 || 0 3 ||

= (1)2[0 0]
= (1)[0] = 0   

(-1)^(1+2) || 1 -2 || -1 3 ||

= (1)3[3 (2)]
= (1)[1] = 1  

(-1)^(1+3) || 1 0 || -1 0 ||

= (1)4[0 0]
= (1)[0] = 0    

Then the value of det(C) is given by:

    (1)(0) + (0)(1) + (1)(0) = 0 + 0 + 0 = 0

And then:   Copyright Elizabeth Stapel 2006-2011 All Rights Reserved

    A1,1 = (1)2det(C) = (1)(0) = 0

And that's just one cofactor; you still have three to go!

Using this methodology, all determinants can be boiled down to finding 2-by-2 determinants. But since you know how to find 3-by-3 determinants, there is no need to go all the way down to 2-by-2's (unless the instructions specifically require that you do).

As you can see from the previous example, having a "zero-rich" row or column in your determinant can make your life a lot easier. Since you'll get the same value, no matter which row or column you use for your expansion, you can pick a zero-rich target and cut down on the number of computations you need to do. Of course, not all matrices have a zero-rich row or column. But there is a rule that can help:

    If you add a multiple of one row (or column) to another row (or column), the value of the determinant will not change.

In other words, you can do row operations on determinants, creating a row (or column) with lots of zeroes, and you'll still get the right answer. (You can also just multiply rows -- without the adding -- or switch rows, but those operations will change the determinant's value. The changes are annoying to keep track of, so try only to do the row-addition operation.)

Here's an example...

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Cite this article as:

Stapel, Elizabeth. "Minors and Cofactors: Expanding Down a Column." Purplemath. Available from Accessed


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