What you can now do

• §2.5 The inverse. Like changing money and changing it back: A⁻¹ turns every output Ax back into x, so A⁻¹A = I and Ax = b is solved by x = A⁻¹b. A matrix that sends two inputs to the same output has no inverse. That happens exactly when a pivot is zero, or one column is a combination of the others.
• §2.5 The 2 by 2 formula. Swap a and d, flip the signs of b and c, divide by ad − bc. That number is the determinant, the yes-or-no number for a 2 by 2 inverse: ad − bc = 0 means the columns lie on one line and there is no inverse.
• §2.5 Inverse of a product. Socks on, shoes on; shoes off, socks off. AB does B first, so (AB)⁻¹ = B⁻¹A⁻¹. Undoing lesson 2's elimination steps in reverse order gives L.
• §2.5 Gauss-Jordan. Write down every move as you solve the puzzle: do row steps on [A | I] until the left half is I. The right half has recorded the steps, and it reads A⁻¹. A zero row on the left means no inverse. To solve Ax = b, elimination is still faster than computing A⁻¹.
• §2.7 Transposes. Turn the spreadsheet: rows become columns, and an m by n matrix becomes n by m. The dot product is xᵀy, and (AB)ᵀ = BᵀAᵀ. A distance table is symmetric, S = Sᵀ, and AᵀA always is.
• §2.7 Permutations. A new playlist order: P is I with its rows reordered, and PA reorders the rows of A. Sending every row back is the transpose, so P⁻¹ = Pᵀ. Doing the row exchanges first gives PA = LU.

Next lesson: the four subspaces (chapter 3)

What if A has no inverse, or isn't even square? The nullspace, pivot and free columns, the complete solution to Ax = b, and the big picture that ties them together.

Read along

Professor Gilbert Strang, Introduction to Linear Algebra, 5th edition (2016), sections 2.5 and 2.7. Lectures 3 and 5 of MIT 18.06 on MIT OpenCourseWare.