Let A= bmatrix1 & 2 & 7\ 4 & -2 & 8\ 3 & 8 & -7 bmatrix and (A- I)=0, where is a real number. If the largest possible value of is p, then the circle (x-p)2+(y-2p)2=320, intersects the co-ordinate axes at
- 1 point
- 2 points
- 3 points
- 4 points
Answer
(3) 3 points
Solution
(A- I)=0 3-(tr A) 2+(sum of principal minors) - A=0
tr A=1-2-7=-8
Principal minors: (14-64)+(-7-21)+(-2-8)=-50-28-10=-88
A=1(14-64)-2(-28-24)+7(32+6)=-50+104+266=320
3+8 2-88 -320=0
=8 satisfies it: 512+512-704-320=0
( -8) ( 2+16 +40 )=0 =8, -8 2 6
p=8 and the circle is (x-8)2+(y-16)2=320, centre C(8, 16), radius 8 5
[figure]
x-axis (y=0): (x-8)2=320-256=64 x=0, 16
y-axis (x=0): (y-16)2=320-64=256 y=0, 32
Distinct points: (0, 0), (16, 0), (0, 32) (the origin is common to both axes)
the circle meets the co-ordinate axes at 3 points.
Answer: (3)