 Hello and welcome to the session. Let us understand the following problem today. If x, y, z are non-zero real numbers then the inverse of matrix A is equal to x00, 0, y0, 0, 0, z is a x inverse 0, 0, 0, y inverse 0, 0, 0, z inverse b x, y, z multiplied by x inverse 0, 0, 0, 0, y inverse 0, 0, 0, z inverse c 1 divided by x, y, z multiplied by x00, 0, y0, 0, 0, z d1 divided by x, y, z multiplied by 1, 0, 0, 0, 1, 0, 0, 0, 1. Now let us write the solution. Given to us is A which is equal to x00, 0, y0, 0, 0, z. Now taking x, y, z common from rows r1, r2, r3 respectively we get x, y, z multiplied by 1, 0, 0, 0, 1, 0, 0, 0, 1. Therefore a joint of A is equal to 1, 0, 0, 0, 1, 0, 0, 0, 1. Now let us first find determinant A which is equal to x, y, z. Now therefore A inverse is equal to 1 by determinant A multiplied by a joint of A which is equal to 1 by x, y, z multiplied by 1, 0, 0, 0, 1, 0, 0, 0, 1 which is equal to x inverse 00, 0, y inverse 0, 00, z inverse. This is the required answer. So the correct option is A. I hope you understood this problem by and have a nice day.