 Hello and welcome to the session. Let us understand the following problem today. If a is equal to 101, 012, 004, then show that determinant of 3a is equal to 27 into determinant of a. Before adding the solution, let us see the key idea. Consider the determinant of the square matrix a is equal to a i to a 3 into 3, that is determinant a is equal to a11, a12, a13, a21, a22, a23, a31, a32, a33. Therefore, the determinant of a is equal to a11 into a22, a33 minus a23 into a32 minus a12 multiplied by a21, a33 minus a31, a23 plus a13 multiplied by a21, a32 minus a22, a31. Now, let us write the solution. Given to us this determinant a is equal to 101, 012, 004. Now, solving this we get which is equal to 1 multiplied by 1 into 4 minus 2 into 0, then minus 0, 0 into 4 minus 0 into 2, then plus 1, 0 into 0 minus 1 into 0. So, we get which is equal to 1 into 4 minus 0 minus 0 plus 0 which is nothing but 4. Therefore, determinant of a is equal to 4. Now, determinant of 3a can be written as 303, 036, 0012. Now, let us find that which is equal to 3 into 3 into 12 minus 6 into 0 minus 0, 0 into 12 minus 0 into 6 plus 3, 0 into 0 minus 0 into 3 which is equal to 3 into 36 minus 0 plus 0 plus 0 which is equal to 3 into 36 which is equal to 108. We have 27 into determinant a is equal to 27 into 4 which is equal to 108. Hence, determinant of 3a is equal to 27 multiplied by determinant of a hence proved. I hope you understood the problem. Bye and have a nice day.