Let a, b, c, x, y and z be real
numbers that satisfy the
three equations
13x+by+cz = 0
ax +23y +cz = 0
ax+by+42z = 0
Suppose that a +13 and x = 0
What is the absolute value of
13 23 42
a-13 b-23 c-42 ?
+
+
Answers
13/a−13 + 23/b−23 + 42/c−42 = -2
Step-by-step explanation:
Let a,b,c,x,y,z be real numbers that satisfy the three equations
13x + by + cz = 0
ax + 23y + cz = 0
ax + by + 42z = 0
Suppose that a≠13 and x≠0.
13/a−13 + 23/b−23 + 42/c−42
Solving the equations:
(13−a)x + (b−23)y = 0
(23−b)y + (c−42)z = 0
(13−a)x + (c−42)z = 0
We get:
(a−13)x = (b−23)y = (c−42)z
So 1/(a−13)x = 1/(b−23)y = 1/(c−42)z
13/a−13 + 23/b−23 + 42/c−42 = 13x/(a−13)x + 23y/(b−23)y + 42z/(c−42)z
= (13x + 23y + 42z) / (a-13)x due to the equalities.
Adding the three equations, we get:
13x + 23y + 42z = -2(ax + by + cz)
= -2(ax-13x)
= -2x (a-13)
From the first equation, we get:
13/a−13 + 23/b−23 + 42/c−42 = -2(a-13)x / (a-13)x
13/a−13 + 23/b−23 + 42/c−42 = -2
The values of 13, 23, 42 are arbitrary and this works the same for any set of three equations of non-zero integers.