If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

The determinant when one matrix has a row that is the sum of the rows of other matrices (and every other term is identical in the 3 matrices). Created by Sal Khan.

Want to join the conversation?

• Does the determinant have to be taken along the row being added? In other words if in the example given the determinant was calculated for the first row, although the second rows were the ones added, would you still end up with the sum of the determinants?
I haven't finished checking for a 3 X 3 but I suspect the answer is yes.
• should be the same regardless of row chosen, just find a row that quickest to work with i.e. most zeroes so theres less calculations and less room for error
• Why is this useful? Is it just a lemma for a more substantial proof?
• When you are working with a big matrix and you have already calculated the determinant once, it's much easier to use this lemmas if something modifies your matrix than to have to calculate the determinant all over again.
• (at ) I know that this point came up many times in the previous videos, but I am having hard time understanding how he got the checker board patterns of pluses and minuses. Can you prove it? Please help!
(1 vote)
• You could understand this way:

If A =
|a11 a12 a13|
|a21 a22 a23|
|a31 a32 a33|

then, the determinant is calculated by:-

Expanding along Column 1
(-1)^1+1 * a11(a22 * a33 - a32 * a23) +

(-1) ^1+2 * a12(a21 * a33 - a31 * a23) +

(-1)^1+3 * a13(a21 * a32 - a31 * a22)

Here in (-1)^i+j ,
i + j represents the sum of suffix of the element i.e. a11
So here (-1)^1+1
and similarly for all the columns.
• At when he says you can immediately see that det(Z) = det(X) + det(Y), is there math you can perform on the sigma notation to express that? Can x_j and y_j factor out somehow?
(1 vote)
• Since each x_j and y_j is different, they can't be factored outside the summation symbol. but for each j you can check that:

(x_j+y_j)|A_ij| = x_j|A_ij + y_j|A_ij|, You then have (leaving out some detail) "sum(x_j|A_ij|) + sum(y_j|A_ij|" and "sum(x_j|A_ij + y_j|A_ij|)", which you can verify are equal for each j (for every term in the summation), which makes the summations equal.
(1 vote)
• What is |Aij| in the determinant formula? I can follow the formula, but that term looks strange in general form.
(1 vote)
• The determinant of the submatrix when you leave out the ith row and jth column.
(1 vote)
• At Sal says x_j did he mean x_ij? shouldn't the i still be part of the summation even though he described i as a row separately? Also should the |A_ij| be |(X,Y,Z)_ij| respectively?
(1 vote)
• I agree. "i" can still be a subscript even though it doesn't vary, as he did with "A_ij".

I used "X_ij" for all three, instead of "A_ij". But I think it's good to make them all the same to make it clear that all three are the same.
(1 vote)