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Question
If x, y, z are all different from zero and `|(1 + x, 1, 1),(1, 1 + y, 1),(1, 1, 1 + z)|` = 0, then value of x–1 + y–1 + z–1 is ______.
Options
x y z
x–1 y–1 z–1
– x – y – z
–1
Solution
If x, y, z are all different from zero and `|(1 + x, 1, 1),(1, 1 + y, 1),(1, 1, 1 + z)|` = 0, then value of x–1 + y–1 + z–1 is –1.
Explanation:
We have, `|(1 + x, 1, 1),(1, 1 + y, 1),(1, 1, 1 + z)|` = 0
Applying C1 → C1 – C3 and C2 → C2 – C3
⇒ `|(x, 0, 1),(0, y, 1),(-z, -z, 1 + z)|` = 0
Expanding along R1
x[y(1 + z) + z] – 0 + 1(yz) = 0
⇒ xy + xyz + xz + yz = 0
⇒ `1/x + 1/y + 1/z + 1` = 0 .....[Dividing by (xyz) on both sides]
⇒ `1/x + 1/y + 1/z` = –1
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