If [\table 2,-1;1,1] [\table x;y] = [\table 4/5], then x^2+y^2 =

AP ICET 2019 Shift 2 Paper

Question : 133 of 200

Marks: +1, -0

[\table 2,-1;1,1] [\table x;y]

[\table 4/5]

, then

x^2+y^2

Solution:

2x-y = 4

→ (1)

x + y = 5

→(2)
Solving (1) & (2)

x = 3; y = 2

x^{2}+y^{2}=9+4=13

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