00:01
In this problem we are provided with the matrix a, that is 66, 33, 3, negative 17, negative 24, negative 12, 0, positive 7, 6, 3, 1 and negative 1.
00:25
And we are asked to write down this matrix a in the row echelon form by making use of elementary row operations.
00:36
So first we begin by using the elementary row operation r1 implies 1 over 66 times r1.
00:46
So this gives us the equivalent matrix 1, 1 over 2, 1 over 22, negative 17 over 66 and the other two rows remain as they are since we have not applied any row operation for r2 and r3.
01:06
Next, we make use of the row operations.
01:10
R2 implies r2 plus 24 times r1 and r3 implies r3 minus 6 times r1.
01:21
So this gives us 1, 1 over 2, 1 over 22, negative 17 over 66.
01:29
The second row becomes 0, 0, 12 over 11, 9 over 11, and the third row becomes 0, 0, 8 over 11, 6 over 11.
01:44
Next, we perform the row operation, r2 implies 11 over 12 multiplied with r2...