00:01
In the given question we have a row transformation which is going from r square to r square and is given by t a of u is equal to product of a into u where for the first part a is equal to one zero minus one two and u1 is equal to one two and u2 is equal to minus one one one so now, t -au1 will be given by a into u1 which will be a1 0 minus 1 2 into 1 2.
00:53
Applying the matrix multiplication we will have minus 1 into 1 plus minus 1 into 2 for the second row we have 0 into 1 plus 2 plus 2.
01:08
So our required matrix will be minus 1 and 4.
01:18
Similarly, we will now calculate t a2 which will be t a into u2 which will be 1 0 minus 1 2 into minus 1 1 which will be 1 1 which will be given by 1 into minus 1 plus minus 1 1 1 1 1 1 plus minus 1 1 1 1 1 1 1 1 1.
01:47
On 0 into minus 1 plus 2 into 1 it will be minus 2 on 2.
01:58
Now we have to show that the value of t .a .u .1 is minus 1 4 and the value of t .a .u .2 which is minus 2 2 and their set that is t .a .u .1 and t .a .u .2 span.
02:29
R squared.
02:32
Now in order to prove this, let us consider ab which belongs to r squared.
02:44
Now can we write x1 into 1 minus 4 plus x2 minus 2 2 is equal to ab? so spanning r square means that any vector i that i choose ab as a weighted combination of t a u1 and that of t a u2 that is if any vector in r square can be written as a linear combination of t a u1 and t a u2 so this will give us minus 1 4 minus 2 2 into x1 x2 will be equals to ab so let us find the value of these unknowns so the unknowns are x1 and x2 to make this equation.
03:56
In order to do this we will apply the matrix operations so let us write the augmented matrix which will be minus 1 4 minus 2 2 a b now we will apply the new operation r1 which is given as r1 plus r2 so this will give us minus 1 plus 4 will give 3 minus 2 plus 2 will give 0 the letter rule will remain as it is and a plus b in the first and d in the second one now we will apply r1 implies 1 by 3 r1 so this will give us 1 0 a plus b upon 3 this is 3 4 2d so the second row remains as such for the next case we will apply r2 implies 1 by 2 of r2 this will give us 1 0 the first row as it is for the second row we have 2 1 and b by 2 the last operation that we will apply is r2 implies r2 minus 2 r1 so we will get 1 0 that is the first row as it is and in the later row 0 1 and b minus 2 a upon 6 so we can clear see that the value of x1 is a plus p by 3 and the value of x2 is b minus 2 a upon 6 so yes p a u1 comma t a u2 vectors spans r squared now for the second question we have the value of a equals to 1 minus 2 minus 1 2 following the same procedure we will now find the value of t .a .u .1 and t .a .u .2...