Ex 4.5, 11 - Class 12

Transcript

Ex 4.5, 11 Find the inverse of each of the matrices 1﷮0﷮0﷮0﷮ cos﷮𝛼﷯﷮ sin﷮𝛼﷯﷮0﷮ sin﷮𝛼﷯﷮− cos﷮𝛼﷯﷯﷯ Let A = 1﷮0﷮0﷮0﷮ cos﷮𝛼﷯﷮ sin﷮𝛼﷯﷮0﷮ sin﷮𝛼﷯﷮− cos﷮𝛼﷯﷯﷯ We know that A–1 = 1﷮|A|﷯ (adj A) exists if |A|≠ 0 Step 1: Calculate |A| |A| = 1﷮0﷮0﷮0﷮ cos﷮𝛼﷯﷮ sin﷮𝛼﷯﷮0﷮ sin﷮𝛼﷯﷮− cos﷮𝛼﷯﷯﷯ = 1 cos﷮𝛼﷯﷮ sin﷮𝛼﷯﷮ sin﷮𝛼﷯﷮− cos﷮𝛼﷯﷯﷯– 0 0﷮ sin﷮𝛼﷯﷮0﷮ − cos﷮𝛼﷯﷯﷯+ 0 0﷮ cos﷮𝛼﷯﷮0﷮𝑠𝑖𝑛 𝛼﷯﷯ = 1(– cos2α – sin2α ) – 0 + 0 = – ( cos2α + sin2α ) = – 1 Since 𝐴﷯≠ 0 Thus A-1 exits Step 2: Calculating adj A adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ A = 1﷮0﷮0﷮0﷮ cos﷮𝛼﷯﷮ sin﷮𝛼﷯﷮0﷮ sin﷮𝛼﷯﷮− cos﷮𝛼﷯﷯﷯ M11 = cos﷮α﷯﷮ sin﷮α﷯﷮ sin﷮α﷯﷮− cos﷮α﷯﷯﷯ = – cos2α – sin2α = –( cos﷮2﷯α + 𝑠𝑖𝑛﷮2﷯α )= –1 M12 = 0﷮sin 𝛼﷮0﷮− cos 𝛼﷯﷯ = 0 – 0 = 0 M13 = 0﷮ cos﷮𝛼﷯﷮0﷮sin 𝛼﷯﷯ = 0 – 0 = 0 M21 = 0﷮0﷮sin 𝛼﷮− cos﷮𝛼﷯﷯﷯ = 0 – 0 = 0 M22 = 1﷮0﷮0﷮−cos 𝛼﷯﷯ = –cos α – 0 = – cos α M23 = 1﷮0﷮0﷮ sin﷮𝛼﷯﷯﷯ = sin α = 0 = sin α M31 = 0﷮0﷮cos 𝛼﷮sin 𝛼﷯﷯ = 0 – 0 = – 0 M32 = 1﷮0﷮0﷮sin 𝛼﷯﷯ = sin α – 0 = sin α M33 = 1﷮0﷮0﷮cos 𝛼﷯﷯ = cos α + 0 = cos α A11 = (−1)﷮1+1﷯ M11 = (−2)﷮2﷯ ( – 1) = – 1 A12 = (−1)﷮1+2﷯ M12 = (−1)﷮3﷯ 0 = 0 A13 = (−1)﷮1+3﷯ M13 = (−4)﷮4﷯ 0 = 0 A21 = ( – 1)2+1 M21 = ( – 1)3 0 = 0 A22 = (−1)﷮2+2﷯ M22 = (−1)﷮4﷯(– cos α) = – cos α A23 = ( – 1)﷮2+3﷯ M23 = ( – 1)﷮5﷯s sin α = – sin α A31 = (−1)﷮3+1﷯. M31 = (−1)﷮4﷯( – 1)4 0 = 0 A32 = ( – 1)3+2 sin α = ( – 1)5 sin α = – sin α A33 = ( – 1)3+3 M33 = ( – 1)6 cos α = cos α So, adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ = −1﷮0﷮0﷮0﷮− cos﷮𝛼﷯﷮− sin﷮𝛼﷯﷮0﷮− sin﷮𝛼﷯﷮ cos﷮𝛼﷯﷯﷯ Step 3: Calculating inverse Now, A– 1 = 1﷮|A|﷯ ( adj (A)) = 1﷮−1﷯ −1﷮0﷮0﷮0﷮− cos﷮𝛼﷯﷮− sin﷮𝛼﷯﷮0﷮− sin﷮𝛼﷯﷮ cos﷮𝛼﷯﷯﷯ = – −1﷮0﷮0﷮0﷮− cos﷮𝛼﷯﷮− sin﷮𝛼﷯﷮0﷮− sin﷮𝛼﷯﷮ cos﷮𝛼﷯﷯﷯ = 𝟏﷮𝟎﷮𝟎﷮𝟎﷮ 𝒄𝒐𝒔﷮𝜶﷯﷮ 𝒔𝒊𝒏﷮𝜶﷯﷮𝟎﷮ 𝒔𝒊𝒏﷮𝜶﷯﷮ −𝒄𝒐𝒔﷮𝜶﷯﷯﷯