00:01
Hi, now we are going to find for what values of p and r if the system is consistent.
00:07
Now from the given matrix a and b, i write the argumented matrix ab is equal to 1 -1 to p 2 0 5 or 0 2 1 7 2 -2 4 8.
00:31
So, first we are going to do r2 as r2 -2 times of r1 and r4 as r4 -2 times of r1, then the matrix will be equivalent to 1 -1 to p 0 2 1 or –2p 0 2 1 7 0 0 0 8 -2p and then i am going to do r2 as r2 divide by 2, then this matrix will be equivalent to 1 -1 to p 0 1 1 divide by 2 or –2p divide by 2 0 2 1 7 0 0 0 8 -2p and then i am going to do r3 as r3 -2 times of r2, then the matrix will be equivalent to 1 -1 to p 0 1 1 divide by 2 or –2p divide by 2 0 0 0 7 -r plus 2p 0 0 0 8 -2p.
02:11
Now, the system is consistent if rank of a is equal to rank of the argument matrix ab.
02:30
So, here we should have 7 -r plus 2p will be equal to 0 and 8 -2p is equal to 0.
02:40
So, from this we get the value of p becomes 4.
02:44
Now, we have to substitute this value of p in this equation, then we get 7 -r plus 8 is equal to 0, then the value of r becomes 15.
02:54
Therefore, the answer for part a is the value of p is 4 and the value of r is 15 and next in part b we are going to find the general solution for the system.
03:07
So, now from the reduced matrix we can write x -y plus 2z is equal to 4 and y plus z divide by 2 is equal to 7 divide by 2...