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Question: If \(\Delta = \left| \begin{matrix} 3 & 4 & 5 & x \\ 4 & 5 & 6 & y \\ 5 & 6 & 7 & z \\ x & y & z & 0...

If Δ=∣345x456y567zxyz0∣\Delta = \left| \begin{matrix} 3 & 4 & 5 & x \\ 4 & 5 & 6 & y \\ 5 & 6 & 7 & z \\ x & y & z & 0 \end{matrix} \right|, then ∆ equals-

A

(y – 2z + 3x)2

B

(x – 2y + z)2

C

(x + y + z)2

D

x2 + y2 + z2 – xy – yz –zx

Answer

(x – 2y + z)2

Explanation

Solution

R1 → R1 + R3 –2R2

Δ=∣000X+Z−2Y456Y567ZXYZ0∣\Delta = \left| \begin{matrix} 0 & 0 & 0 & X + Z - 2Y \\ 4 & 5 & 6 & Y \\ 5 & 6 & 7 & Z \\ X & Y & Z & 0 \end{matrix} \right|= –(x + z –2y)⇒ Δ=∣456567XYZ∣\Delta = \left| \begin{matrix} 4 & 5 & 6 \\ 5 & 6 & 7 \\ X & Y & Z \end{matrix} \right|

C1 → C 1 + C 3 –2 C 2, C 2 → C 2  – C 3

∆ = –(x + z – 2y) ∣0−160−17X−2y+Zy−zz∣\left| \begin{matrix} 0 & - 1 & 6 \\ 0 & - 1 & 7 \\ X - 2y + Z & y - z & z \end{matrix} \right|= – (x + z –2y)2