UPSC NDA I 2025 Mathematics Paper

The matrix A is:
A=[

y	x	x
z	x	y
x	y	z

]
Since A is an orthogonal matrix, we know that:
AT=A−1‌‌⇒‌‌ATA=I
This property tells us that A is orthogonal, and it implies that ATA (the product of A′ s transpose and A is equal to the identity matrix I, which is:
ATA=[

1	0	0
0	1	0
0	0	1

]
Now, let's calculate ATA step by step. The transpose of matrix A, denoted AT is:
AT=[

y	z	x
x	x	y
x	y	z

]

Now, we perform matrix multiplication between AT and A :
ATA=[

y	z	x
x	x	y
x	y	z

][

y	x	x
z	x	y
x	y	z

]
Performing this multiplication, we get the following matrix:
ATA=[

y2+z2+x2	xy+zx+xy	xz+yz+x2
xy+zx+xy	x2+x2+y2	xy+xz+yz
xz+yz+x2	xy+xz+yz	x2+y2+z2

]
This matrix must be equal to the identity matrix I, which is:
[

1	0	0
0	1	0
0	0	1

]
By comparing the elements of the matrices, we get the following system of equations:
‌ 1. ‌y2+z2+x2=12⋅xy+zx+xy=03⋅xz+yz+x2=1
Thus, the key result from the orthogonality condition is:
x2+y2+z2=1