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1. Chapter 4 Class 12 Determinants
2. Concept wise
3. Solving by simplifying det.

Transcript

Example 11 Prove that |โ 8(๐&๐+๐&๐+๐+๐@2๐&3๐+2๐&4๐+3๐+2๐@3๐&6๐+3๐&10๐+6๐+3๐)| = a3 Let ฮ = |โ 8(๐&๐+๐&๐+๐+๐@2๐&3๐+2๐&4๐+3๐+2๐@3๐&6๐+3๐&10๐+6๐+3๐)| Applying R2 โ R2 โ 2R1 = |โ 8(๐&๐+๐&๐+๐+๐@๐๐โ๐(๐)&3๐+3๐โ2(๐+๐)&4๐+3๐+2๐โ2(๐+๐+๐)@3๐&6๐+3๐&10๐+6๐+3๐)| = |โ 8(๐&๐+๐&๐+๐+๐@๐&3๐+3๐โ2๐โ2๐&4๐+3๐+2๐โ2๐โ2๐โ2๐@3๐&6๐+3๐&10๐+6๐+3๐)| = |โ 8(๐&๐+๐&๐+๐+๐@๐&๐&2๐+๐@3๐&6๐+3๐&10๐+6๐+3๐)| Applying R3 โ R3 โ 3R1 = |โ 8(๐&๐+๐&๐+๐+๐@0&๐&2๐+๐@๐๐โ๐(๐)&6๐+3๐โ3(๐+๐)&10๐+6๐+3๐โ3๐+๐+๐)| = |โ 8(๐&๐+๐&๐+๐+๐@0&๐&2๐+๐@๐&6๐+3๐โ3๐โ3๐&10๐+6๐+3๐โ3๐+๐+๐)| = |โ 8(๐&๐+๐&๐+๐+๐@0&๐&2๐+๐@0&3๐&7๐+3๐)| Expanding along C1 = = a |โ 8(๐&2๐+๐@3๐&7๐+3๐)| โ 0 + 0 = a (a(7a + 3b) โ 3a (2a + b) = a (7a2 + 3ab โ 6a2 โ 3ab) = a(a2) = a3 = R.H.S Hence proved |โ 8(๐&2๐+๐@3๐&7๐+3๐)| |โ 8(๐+๐&๐+๐+๐@3๐&7๐+3๐)|

Solving by simplifying det.