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Question

Show that ∣ ∣ ∣xx2yzyy2zxzz2xy∣ ∣ ∣=(xy)(yz)(zx)(xy+yz+zx)

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Solution

∣ ∣ ∣xx2yzyy2zxzz2xy∣ ∣ ∣

Divide and multiply respectively 1st row by x 2nd row by y 3rd row by z

we get

1xyz∣ ∣ ∣x2x3xyzy2y3xyzz2z3xyz∣ ∣ ∣

Δ=xyzxyz∣ ∣ ∣x2x31y2y31z2z31∣ ∣ ∣

Δ=∣ ∣ ∣x2x31y2y31z2z31∣ ∣ ∣

R2R2R1

we get

Δ=∣ ∣ ∣x2x31y2x2y3x30z2x2z3x30∣ ∣ ∣

Expand the determinant over 3rd column

We get
Δ=(y2x2)(z3x3)(y3x3)(z2x2)
Δ=(xy)(yz)(zx)(xy+yz+zx) [henceproved]

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