TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the system of equation.
[

2	8
3	7

][

a

b

]=k[

a

b

]
The above equation is solved as,
[

2a+8b

3a+7b

]=[

ka

kb

]
[

a(2−k)+8b

3a+(7−k)b

]=[

0

0

]
[

2−k	8
3	7−k

][

a

b

]=[

0

0

]
It has non trivial solution. Therefore,
[

2−k	8
3	7−k

]=0
(2−k)(7−k)−24=0
k2−9k−10=0
(k−10)(k+1)=0
since k>0
So, k=10 and a=b=1
Therefore, the solutions are 10,10 .