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By using properties of determinants, prove that |x+yy+zz+xzxy111| = 0. - Mathematics and Statistics

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Question

By using properties of determinants, prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0.

Sum
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Solution

L.H.S. = `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` 

Applying R1 → R1 + R2, we get

L.H.S. = `|(x + y + z, x + y + z, x + y + z),(z, x , y),(1, 1, 1)|`

Taking (x + y + z) common from R1, we get

L.H.S. = `(x + y + z)|(1, 1, 1),(z, x, y),(1, 1, 1)|`

= (x + y + z) (0)          …[∵ R1 and R3 are identical]
= 0
= R.H.S.

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Chapter 6: Determinants - MISCELLANEOUS EXERCISE - 6 [Page 95]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 11 Maharashtra State Board
Chapter 6 Determinants
MISCELLANEOUS EXERCISE - 6 | Q 3) | Page 95

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