00:01
We are provided with the matrix minus 1, 0, 1, 2, 0 minus 1, 0, 0, 0, negative 3 over 4 minus 2, 1 over 2, negative 3 over 4 minus 2, 1 over 3.
00:28
And we have to reduce this matrix in row -reduced acronym 4, that is, r3 changes to r3 plus 1 over 2 times, of r4.
00:43
Apply this operation we get minus 1, 0, 1, 2, 0 minus 1, 0 minus 3 over 4 minus 3 over 2, 4 over 3.
01:03
Now again apply the row operation that is r3 changes to r3 minus 3 over 4 times of r3.
01:18
We get minus 1, 0, 1, 2, 0 minus 1, 2, 0 minus 1, 2, 0, 0 minus 3, 4 over 3.
01:36
Again, apply the row operation that is r3 changes through 3 times of r3.
01:43
We get minus 1, 0, 1, 2, 0, 0, minus 1.
01:52
To 0 0 0 minus 9 4.
02:00
So this is the required road reduce form of the given matrix.
02:16
In part e, we have to solve the matrix by using the matrix in part a.
02:24
That is, we can rewrite this matrix as minus x plus 0 times y plus 1 times z equals to 2, market is equation 1.
02:37
0 times x minus 1 times y plus 2 times z equals to 0 .0 times x plus 0 times y minus 9 times z equals to 4.
02:56
Market is equation 3...