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Course: Linear algebra>Unit 2

Lesson 5: Finding inverses and determinants

Formula for 2x2 inverse

Figuring out the formula for a 2x2 matrix. Defining the determinant. Created by Sal Khan.

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• Why do determinants work? Also, where would you use a determinant in real life?
• i'd buy every mathematics textbook you'll ever write
• Sal sure tripped me up with this one! The transformations at the beginning are quite nuanced and actually use elementary matrix multiplication - am I right? .

It took me some time to work through the workings of this for the general case. It was skipped over quickly in the video.

If I am missing the ' elephant in the room ' on this could somebody point it out to me!
• This can be more easily seen by an example: try to zero out:
| 4 8 |
| 2 2 |
To zero out the second row on the first column, you need to do:
4(2) - 2(4) = 0
And then the same thing to the second column, second row:
4(2) - 2(8) = -8
Therefore, the more generalized transformation is:
4(c2) - 2(c1)
That makes sense, right? The c2 and the c1 changes for every column.
However, the 4 and 2, stay the same in each calculation. And what is 4 and 2? It's just 4=a and 2=b, right? That makes sense.
So if you can generalize even more to say that 4=a and 2=b, then the even more generalized equation is:
a(c2) - b(c1)
If that makes sense, then maybe you can think about is as just multiplying the top and bottom and then subtracting it from the product of the bottom and the top. That's the basic idea, if it makes sense to you: (top)*(bottom) - (bottom)*(top) = 0
• Is the determinant useful in any other way besides understanding if we can find the inverse or not?
• Why does Sal cancels out the a's at without admitting that a is different than zero? This would result in division by zero, I guess.
• Sal should have stated explicitly that a is not equal to zero. That said, a is not equal to zero because a is the first vector in the first row, meaning a is a pivot variable in reduced row echelon form, which can't be equal to zero. If a is zero you can easily interchange the first row and the second row. If a is zero, then c certainly is not equal to zero because that would mean the two row vectors (or column vectors if you'd like) would not be linearly independent. If the two rows (or columns) are not linearly independent, the matrix is not invertible. To recap, for matrix A to be invertible, at least one of a and c is non-zero and you can just assume that a is non-zero. Hope that helps.
• I don't know why Sal keeps referring to column vectors when he does row operations...I recognize that a matrix can always be visualized as a bunch of column vectors, but that is not at all helpful when he's about to do a bunch of ROW operations.
• I also find it a huge distraction. He's obviously performing operations on row vectors. It's work enough understanding what he's trying to say without having to decode his language.
• Where does c1 and c2 come from, what must I read to understand the T symbol? Are c1 and c2 scalars?
• I think Sal means row1 and row2 or r1 and r2, not c1 and c2. He leaving row1 unchanged and changing row2, not column c1.
• If the determinant of a square matrix is zero, does that imply that there is NO inverse on that matrix?
• Yes, it does. Let A be any n x n matrix for which det A = 0. Then A is singular (not invertible).

Proof
Suppose A is not singular, and let B denote the inverse of A. That is, if I is the n x n identity matrix, then BA = I. By the product formula for determinants, we have det A = 1 / det B ≠ 0. But this contradicts the fact that det A = 0, and the proof is complete.
• Does this method work to find the inverse of bigger square matrices?
• Yes, it does work. If you augment the matrix with the identity and when you put the new matrix into Reduce Row Echelon from you get the identity on the left side, the right side will be the inverse of the matrix, no matter the dimensions of it.