00:01
For this exercise, we're starting off with three n by n matrices.
00:05
We're told that the first matrix a is invertible, and given the equation that a times b equals a times c, our goal is to show that the matrix b must be equal to c.
00:19
Our strategy here is going to be to start with the equation and subtract a .c from both sides.
00:28
So we have first that ab minus ac is equal to zero.
00:33
Then we can left factor a from this left side to get a times b minus c equals 0.
00:42
Next, let's set up some variables here for what b minus c is going to be equal to.
00:46
Let b minus c be equal to this matrix, where the first vector, i'm going to denote it by x1, the second column vector is x2, and so on up until xn, the last column vector.
01:04
Now that we have a times b equals 0 in this equation, we can use the following idea for matrix multiplication.
01:13
We'll say, then, a times the first vector, x1, must be equal to the zero vector.
01:22
That becomes b minus c has the first column x1...