00:01
So to solve this, these equation first we have to write these equation into matrix of form.
00:08
So we have here ax is equal to b.
00:11
So matrix a will be then we have here 1 1 1.
00:17
Second row is 2 minus 1 minus 1.
00:22
Third row is 3 2 3 and last row will be 1 minus 3 minus 2.
00:33
So multiply with a variable matrix those are x y and z and that is equal to our b matrix that is column matrix.
00:46
So we have here 4 minus 1 14 and 0.
00:54
So first we need to write this matrix at augmented matrix in a form of augmented matrix.
01:02
So this will be a ratio b.
01:06
So that is equal to we have here 1 1 1 2 minus 1 minus 1 3 2 3 1 minus 3 minus 2 and this will be augmented with 4 minus 1 14 and 0.
01:32
Now we have to use here the reduction method.
01:35
So we have to reduction here by row.
01:37
So first we have to apply here r2 is r2 minus 2 r1.
01:48
R3 is r3 minus 3 r1 and r4 is r4 minus r1.
02:03
So our matrix will be then 1 1 1 and next line will be then 0 minus 3 minus 3 and this will be the augmented matrix.
02:19
So ratio here will be 4 and here will be minus of 9.
02:25
Next line will be 0 minus 1 0 and last row will be 0 minus 4 minus 3 minus 4.
02:37
So now we are applying here that is we are interchanging r2 to r3.
02:44
So r2 interchanging with r3...