 Hello friends, welcome to the session i markup. We are going to discuss matrices. We are given with the question that find values of x, y, z if the matrix A equal to 0 to y, z x, y minus z x minus y, z satisfy the equation transpose of A into A equal to i. Now let us have to the solution. We are given A equal to matrix 0 to y, z x, y minus z x minus y z. Now since we are required to form transpose of A, therefore A transpose equal to 0 to y, z x, y minus z x minus y z. Now we are given in the question that A and A transpose satisfy the equation A transpose A equal to i. Now hence we have 2 y y minus y z minus z z into matrix A which is 2 0 2 y z y minus y minus z equal to identity matrix which is 1 0 0 0 1 0 0 0 1. Now this implies 0 into 0 plus x into x plus x into x 0 into 2 y plus x into y plus x into minus y then x 0 into 0 into z plus x into minus z plus x into z then 2 y into 0 plus y into x plus minus y into x then 2 y into 2 y plus y into y plus minus y into minus y then 2 y plus y into minus z plus minus y into z then z into 0 minus z into x plus z into x then z into 2 y minus z into y plus z into minus y then z into z minus z into minus z plus z into z is equal to identity matrix which is 1 0 0 0 1 0 0 0 1. Now we see that this implies 0 plus x square plus x square 0 plus x y minus x y 0 minus x z plus x z then 0 plus x y minus x y then 4 y square plus y square plus y square then 2 z y minus z y minus z y then 0 minus x z plus x z 2 y z minus y z minus y z then z square plus z square plus z square equal to identity matrix which is 1 0 0 0 1 0 0 1 this is equal this implies 2 x square 0 0 0 6 y square 0 0 0 and 3 z square equal to identity matrix which is 1 0 0 0 1 0 0 0 1. Now on equating the corresponding elements we have x square equal to 1 this is our first equation 6 y square equal to 1 this is our second equation and 3 z square equal to 1 this is our third equation. Now from equation first we get x square equal to 1 by 2 this implies x equal to plus minus 1 by root 2 from equation second we get y square equal to 1 by 6 this implies y equal to plus minus 1 by root 6 from equation third we have z square equal to 1 by 3 this implies z equal to plus minus 1 by root 3 therefore x equal to plus minus 1 by root 2 y equal to plus minus 1 by root 6 and z equal to plus minus 1 by root 3 is the required answer hope you understood this solution and enjoyed the session goodbye and take care.