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Prove that |x+yy+zz+xz+xx+yy+xy+zz+xx+y|=2|xyzzxyyzx| - Mathematics and Statistics

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प्रश्न

Prove that `|(x + y, y + z, z + x),(z + x, x + y, y + z),(y + z, z + x, x + y)| = 2|(x, y, z),(z, x, y),(y, z, x)|`

योग

उत्तर

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

Applying R1 → R1 + R2 + R3, we get

L.H.S. = `|(2(x + y + z), 2(x + y + z),2(x + y + z)),(z + x, x + y, y + z),(y + z, z + x, x + y)|`

Taking 2 common from R1, we get

L.H.S. = `2|(x + y + z, x + y + z, x + y + z),(z + x, x + y, y + z),(y + z, z + x, x + y)|`

Applying R1 → R1 – R3, we get

L.H.S. = `2|(x, y, z),(z + x, x + y, y + z),(y + z, z + x, x + y)|`

Applying R2 → R2 – R1, we get

L.H.S. = `2|(x, y, z),(z, x, y),(y + z, z + x, x + y)|`

Applying R3 → R3 – R2, we get

L.H.S. = `2|(x, y, z),(z, x, y),(y,  z, x)|`

= R.H.S.

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अध्याय 4: Determinants and Matrices - Exercise 4.2 [पृष्ठ ६८]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 4 Determinants and Matrices
Exercise 4.2 | Q 2 | पृष्ठ ६८

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