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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