Aljabar Linear

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Home Page Title Page ◭◭ ◮◮ ◭ ◮ Page 124 of 132 Go Back Full Screen Close Quit ⎛ ⎝ ⎛ ⎝ 2 3 0 4 0 1 2 3 0 4 0 1 ⎞ ⎠ x = ⎞ ⎠ x = ⎛ ⎝ ⎛ ⎝ 0 16 17 4 17 0 4 17 1 17 ⎞ ⎠ ⇒ x = ⎞ ⎠ ⇒ x = Sehingga didapat matriks A − adalah: A − = 1 2 − 6 17 0 4 17 3 −34 1 17 . − 16 17 4 17 − 3 34 1 17 Penyelesaian pendekatan dari persamaan ⎛ ⎞ ⎛ ⎞ 2 3 5000 ⎝ 0 4 ⎠ p = ⎝ 4000 ⎠ q 0 1 2000 diberikan oleh p = A q − ⎛ ⎝ 5000 4000 2000 ⎞ ⎠ = 1 2 − 6 17 0 4 17 3 −34 1 17 ⎛ ⎝ 5000 4000 2000 ⎞ ⎠ = 911.76 1058.82 Hasil yang didapat sama dengan hasil perhitungan sebelumnya. .
Home Page Title Page ◭◭ ◮◮ ◭ ◮ Page 125 of 132 Go Back Full Screen Close Quit Soal UAS GASAL 2006/2007 1. 1.Selidiki apakah matriks berikut bisa didiagonalkan atau tidak bisa didiagonalkan? ⎡ Berikan alasan dari jawaban! ⎣ 0 1 3 0 1 2 1 −1 0 JawabPolinomial kharakteristik dari matriks diberikan oleh: λ p(λ) = 0 −1 −1 −3 λ − 1 −2 1 λ = (λ − 1)2 (λ + 1) Untuk λ = 1, didapat: ⎡ E1(R 3 ) = Ker( ⎣ ⎤ ⎦ 1 −1 −3 0 0 −2 −1 1 1 ⎤ ⎦) = ⎛ ⎝ 1 1 0 ⎞ Terlihat bahwa untuk λ = 1 multiplisitas geometri = 1 < 2 = multiplisitas aljabar. Jadi matriks tidak bisa didiagonalkan. ⎠ .
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Aljabar Linear

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