Quote: Now create a +1 in the upper left-hand corner by multiplying the first row by -1: [ 1 -3 3 -1 0 0] [ 0 -6 5 0 1 0]. [-5 -3 1 0 0 1]So would I always use -1 or does that depend on the matrix variables?
Quote: Now create 0's below it by adding a multiple of row 1 to the other rows (in this case 5 times the first row to the third row): [1 -3 3 -1 0 0] [0 -6 5 0 1 0]. [0 -18 16 -5 0 1]where does the 5 come from?
Quote: Now the first column looks fine, so we create a +1 at the (2,2) entry by multiplying the second row by -1/6: [1 -3 3 -1 0 0] [0 1 -5/6 0 -1/6 0]. [0 -18 16 -5 0 1]wheres the -1/6 come from? I haven't gotten to the calculations below yet.
Quote: Now create 0's above and below that 1 by adding multiples of the second row (3 and 18 times it) to the others (first and third): [1 0 1/2 -1 -1/2 0] [0 1 -5/6 0 -1/6 0]. [0 0 1 -5 -3 1] Now the second column is right, and we already have a +1 at the (3,3) entry. Now create 0's above that by adding multiples of row 3 (-1/2 and 5/6 times it) to the rows above it (first and second): [1 0 0 3/2 1 -1/2] [0 1 0 -25/6 -8/3 5/6]. [0 0 1 -5 -3 1] The left half is now a 3x3 identity matrix, so we stop. Then the inverse of the original matrix is the right half, or [ 3/2 1 -1/2] [-25/6 -8/3 5/6] [ -5 -3 1] You can check this by multiplying it times the original matrix, and getting the identity.Thanks
