NDA Mathematics Question (2025) — Solution

Step-by-step solution

For an orthogonal matrix A, the condition A A^T = I must be satisfied. Given A = pmatrix y & z & x \\ z & x & y \\ x & y & z pmatrix , we find A^T = pmatrix y & z & x \\ z & x & y \\ x & y & z pmatrix . Multiplying A and A^T gives: A A^T = pmatrix y & z & x \\ z & x & y \\ x & y & z pmatrix pmatrix y & z & x \\ z & x & y \\ x & y & z pmatrix = pmatrix x^2+y^2+z^2 & xy+yz+zx & xy+yz+zx \\ xy+yz+zx & x^2+y^2+z^2 & xy+yz+zx \\ xy+yz+zx & xy+yz+zx & x^2+y^2+z^2 pmatrix Equating A A^T to the identity matrix I = pmatrix 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 pmatrix , we compare the corresponding elements. From the diagonal elements, we get: x^2 + y^2 + z^2 = 1 From the non-diagonal elements, we get: xy + yz + zx = 0 Since x, y, and z are integers, the condition x^2 + y^2 + z^2 = 1 implies that one of the variables must be 1 and the other two must be 0. This also satisfies xy + yz + zx = 0. Therefore, the value of x^2 + y^2 + z^2 is 1. Answer: 1