UPSC NDA I 2025 Mathematics Paper

Multiply Matrix 1 and Matrix 2:
‌[

x

1

1

]×[

1	2	3
4	5	6
7	8	9

]=[

x+4+7

2x+5+8

3x+6+9

]
‌⇒[

x+11

2x+13

3x+15

]
Multiply the resulting matrix with Matrix 3:
‌[

x+11

2x+13

3x+15

]×[

1

1

x

]
‌=(x+11)⋅1+(2x+13)⋅1+(3x+15)⋅x
‌⇒(x+11)+(2x+13)+(3x2+15x)
‌⇒(3x2+18x+24)
Equate the result to 45 :
(3x2+18x+24=45)

‌⇒(3x2+18x−21=0)
‌(x2+6x−7=0)
‌(x2+7x−x−7=0)
‌⇒(x=1‌ or ‌x=−7)
Step 5: Verify:
For X=1, substitute back:
‌(3(1)2+18(1)+24=45)
‌45=45
∴ The correct value of x is 1 .