3,3 of inverse matrix should be 6. Nice job anyway

In general, if a matrix M is larger than 3 by 3 and nonsingular, which means an inverse exists, it is more efficient to factor it into the product of a lower-triangular matrix L with ones on the diagonal, and an upper-triangular matrix U with nonzero "pivotal elements" on the diagonal. That is, M = L*U. Inverses of L and U are generally not computed, but products of their inverses with other matrices are efficiently evaluated recursively using the existing entries in L and U. Check the internet for more detail if interested.

Wonderful thank you. For clarity, could please explain whether augmented matrix approach doesn't work for a 2 by 2 matrix, or whether the 2 by 2 approach is preferred for simplicity?

You are the one all along 🙏💕

Thanks teacher; you are father. ኣመሠግናለሁኝ from Ethiopia 🇪🇹.

At 33:18. Do you multiply or divide the entries as to get the new row for row one?

Mr. Jayson, how am I supposed to know what to multiply and subtract with? I mean, I would simply subtract with a constant, instead of multiplying with a row then add or subtract. Please help me!

Inverse functions 🔥🔥🔥

Nice 👍

"i'm not gonna give you that proof"🤣🤣

Beauty ful new look mr