 Hello, and welcome to the session. I am Deepika here. Let's discuss the question. Examine the consistency of the system of equations x plus y plus z is equal to 1, 2x plus 3y plus 2z is equal to 2, ax plus ay plus 2z is equal to 4. Let's start the solution. Solution? We know that the system of equations is said to be consistent if its solution exists and a system of equation is said to be inconsistent if its solution does not exist. So we will see whether its solution exists or not. The given system of equations is plus y plus z is equal to 1 and 2x plus 3y plus 2z is equal to 2, and ax plus ay plus 2z is equal to 4. We know that the system of equations can be written in the form to be where x plus z is equal to x, y, z, v is equal to 1, 2, 4. Find out determinant a. That is equal to a, 1, 2, a. That is the common factor from R3. So by taking out a as a common factor from R3, we get R determinant a is equal to 1, 3, 1, 1 by applying equals to R1 minus R3. We get determinant a is equal to 1 minus 1, 0, 1 minus 1, 0, 1 minus 2 is minus 1, rho2 is as it is 2, 3, 2 and rho3 is also as it is 1, 1, 2. Now we will expand along R1. We will get our determinant a is equal to a into 0 minus 0 minus 1 into 2 minus 3. A is equal to 1. That is equal to a, which is not equal to 0. This implies determinant a is not equal to 0. This implies the given system of equation is consistent. Since the answer for the above system of equations is consistent. I hope the question is clear to you. Bye and have a good day.