Ex 4.5, 7 - Find inverse of matrix (if it exists) - Chapter 4 - Ex 4.5

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.5, 7 Find the inverse of each of the matrices (if it exists). 1﷮2﷮3﷮0﷮2﷮4﷮0﷮0﷮5﷯﷯ Let A = 1﷮2﷮3﷮0﷮2﷮4﷮0﷮0﷮5﷯﷯ We know that A–1 = 1﷮|A|﷯ (adj A) exists if |A|≠ 0 Step 1: Calculate |A| |A| = 1﷮2﷮3﷮0﷮2﷮4﷮0﷮0﷮5﷯﷯ = 1 2﷮4﷮0﷮5﷯﷯– 2 0﷮4﷮0﷮5﷯﷯ – 3 0﷮2﷮0﷮0﷯﷯ = 1(10 – 0) –2 (0 – 0) –3 (0 – 0) = 1 (10) – 0 – 0 = 10 Since |A| ≠ 0 ∴ A–1 exists Step 2: Calculating adj A adj A = = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ A = 1﷮2﷮3﷮0﷮5﷮4﷮0﷮0﷮5﷯﷯ M11 = 2﷮4﷮0﷮5﷯﷯ = 2(5) – 0(4) = 10 M12 = 0﷮4﷮0﷮5﷯﷯ = 0(5) – 0(4) = 0 M13 = 0﷮2﷮0﷮0﷯﷯ = 0 – 0 = 0 M21 = 2﷮3﷮0﷮5﷯﷯ = 10 – 0 = 10 A11 = ( – 1)1-1 M11 = ( – 1)2 (10) = 1(10) = 10 A12 = ( – 1)1+2 M12 = ( – 1)3 0 = 0 A13 = ( – 1)1+3 M13 = ( – 1)4 0 = 0 A21 = ( – 1)2+1 M21 = ( – 1)3 (10) = – 1 (10) = – 10 A22 = ( – 1)2+2 M22 = ( – 1)4 5 = 5 A23 = ( – 1)2+3 M23 = ( – 1)4 0 = 0 A31 = ( – 1)3+1 M31 = ( – 1)4 2 = 2 A32 = ( – 1)3+2 M32 = ( – 1)5 4= – 1 (4) = – 4 A33 = ( – 1)3+3 M33 = ( – 1)6 2 = 2 Thus adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯= 10﷮−10﷮2﷮0﷮5﷮−4﷮0﷮0﷮2﷯﷯ Step 3: Calculate inverse Now , A– 1 = 1﷮|A|﷯ ( adj (A)) = 𝟏﷮𝟏𝟎﷯ 𝟏𝟎﷮−𝟏𝟎﷮𝟐﷮𝟎﷮𝟓﷮−𝟒﷮𝟎﷮𝟎﷮𝟐﷯﷯

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