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Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx, and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to
  • a)
    2
  • b)
    –1
  • c)
    0
  • d)
    1
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Let a, b, c be any real numbers. Suppose that there are real numbers x...
The given equations are
–x + cy + bz = 0
cx –y + az = 0
bx + ay – z = 0
∵ x, y, z are not all zero
∴ The above system should not have unique (zero) solution
⇒ –1(1– a2) – c(– c – ab) + b(ac + b) = 0
⇒–1 + a2 + b2 + c2 + 2abc = 0
⇒ a2 + b2 + c2 + 2abc = 1
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Most Upvoted Answer
Let a, b, c be any real numbers. Suppose that there are real numbers x...
Given Information:
a, b, c are real numbers.
x = cy + bz
y = az + cx
z = bx + ay

Solution:

Step 1: Finding a2 + b2 + c2 + 2abc
From the given equations, we can write:
x = cy + bz
=> x = c(az + cx) + b(bx + ay)
=> x = acz + c^2x + bbx + aby
=> (1 - ac)x = c^2x + b^2x
=> (1 - ac - c^2 - b^2)x = 0
Since x is not zero, we must have:
1 - ac - c^2 - b^2 = 0
=> a^2 + b^2 + c^2 + 2abc = 1
Therefore, a2 + b2 + c2 + 2abc = 1
Hence, the correct option is (D) 1.
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Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx, and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal toa)2b)–1c)0d)1Correct answer is option 'D'. Can you explain this answer?
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