Answer :
Applying, C1 → C1 – 4 C3, we get,
Applying, R1 → R1 + R2 and R3→ R3 – R2, we get
Now, applying R2 → R2 + 3 R1, we get,
= 1[(109)(12) – (119)(11)] = 1308 – 1309
= – 1
So, Δ = – 1
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