JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Matrices

Let S be the set of all integer solutions,

(x, y, z)

, of the system of equation

x - 2 y + 5 z = 0

- 2 x + 4 y + z = 0

- 7 x + 14 y + 9 z = 0

such that

15 \leq x^{2} + y^{2} + z^{2} \leq 150

. Then, the number of elements in the set

S

is equal to ______.

Answer: 8

Solution

Combine the first two equations as twice the first plus the second:

(2 x - 4 y + 10 z) + (- 2 x + 4 y + z) = 11 z = 0

, so

z = 0

, and this value also satisfies the third equation. The first equation then gives

x = 2 y

. The bound

15 \leq x^{2} + y^{2} + z^{2} \leq 150

becomes

15 \leq 4 y^{2} + y^{2} \leq 150

, i.e.

3 \leq y^{2} \leq 30

. The integers with

y^{2}

in this range are

y = \pm 2, \pm 3, \pm 4, \pm 5

, eight values, each fixing

x

and

z

, so the set has

8

elements.

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Integer answer