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Question
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)|`
Solution
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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