00:01
Hello everyone from the question we are given that matrix a is equal to 1201 0111 1201 and we have to find row equivalent of matrix a its range free variables and null space of a so first we convert into row equal in form this implies first row is as it is 1201 second row is also as it is 0 110 and replace r3 by r3 minus r1 third row is 0, 0, 0, 0.
00:33
From here we can see that they are pivots and they are free variables.
00:39
So if we consider x1, x2, x3, x4, then x3 and x4 are free variables and rank of matrix a is equals to 2.
00:49
Because there are two non -zero rows, then rank of matrix a is equals to 2.
00:54
And x3, x4 are free variables.
00:56
Now from here we have x1 plus 2x2.
00:59
Plus x4 is equal to 0 and x2 plus x3 is equal to 0 so from here x2 is equal to minus x3 now we put the value of x2 in this equation then x1 minus x3 minus 2x3 plus x4 is equal to 0 so from here x1 is equal to 2x3 minus x4 then we have x1 x2 x3 x4 is equals to 2x3 minus x4 minus x3 x3 x3 x4 x3 and x4 are free variables this is equals to 2 minus 110 x3 plus minus 1 001 1 x4 so the null space of matrix a is equals to 2 minus 1 110 transpose minus 1 0 01 transpose.
02:03
Also we have matrix b is equals to 1203 to 407...