UPSC NDA I 2025 Mathematics Paper

The matrix A is:$A=\begin{pmatrix} y & x & x \\ z & x & y \\ x & y & z \end{pmatrix}$Since A is an orthogonal matrix, we know that:$A^{T}=A^{-1} \;\; \Rightarrow \;\; A^{T} A=I$This property tells us that $A$ is orthogonal, and it implies that $A^{T} A$ (the product of $A^{'}$ s transpose and $A$ is equal to the identity matrix $I$, which is:$A^{T} A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$Now, let's calculate $A^{T} A$ step by step. The transpose of matrix $A$, denoted $A^{T}$ is:$A^{T}=\begin{pmatrix} y & z & x \\ x & x & y \\ x & y & z \end{pmatrix}$ Now, we perform matrix multiplication between $A^{T}$ and A :$A^{T} A=\begin{pmatrix} y & z & x \\ x & x & y \\ x & y & z \end{pmatrix}\begin{pmatrix} y & x & x \\ z & x & y \\ x & y & z \end{pmatrix}$Performing this multiplication, we get the following matrix:$A^{T} A=\begin{pmatrix} y^2+z^2+x^2 & xy+zx+xy & xz+yz+x^2 \\ xy+zx+xy & x^2+x^2+y^2 & xy+xz+yz \\ xz+yz+x^2 & xy+xz+yz & x^2+y^2+z^2 \end{pmatrix}$This matrix must be equal to the identity matrix I, which is:$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$By comparing the elements of the matrices, we get the following system of equations:$\;\text{ 1. }\; y^2+z^2+x^2=12 \cdot xy+zx+xy=03 \cdot xz+yz+x^2=1$Thus, the key result from the orthogonality condition is:$x^2+y^2+z^2=1$