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תהי A מטריצה הפיכה המקיימת את המשוואה: \[ -5 A^{19}-10 A^{17}+20 A^{16}-50 A^{15}+10 I=0 \] רשמו את A \( A^{-1} \) כפולינום ב- A.

Ask by Allan Haynes. in Israel
Jan 22,2025

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\( A^{-1} = \frac{1}{2}A^{18} + A^{16} - 2A^{15} + 5A^{14} \)

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נתחיל מהמשוואה הנתונה: \[ -5 A^{19}-10 A^{17}+20 A^{16}-50 A^{15}+10 I=0 \] נמציא את הביטוי של \( A^{-1} \) כפולינום ב- \( A \). נבודד את \( I \): \[ 5 A^{19}+10 A^{17}-20 A^{16}+50 A^{15} = 10 I \] נחלק את שניהם ב-10: \[ \frac{1}{2} A^{19} + A^{17} - 2 A^{16} + 5 A^{15} = I \] כעת, אם נבצע שינוי סדר בלבול של ממדי \( A^{-1} \): \[ A^{-1} = \frac{1}{2} A^{19} + A^{17} - 2 A^{16} + 5 A^{15} \] ולכן, \( A^{-1} \) ניתן לכתיבה כפולינום ב- \( A \) בצורה של: \[ A^{-1} = \frac{1}{2} A^{19} + A^{17} - 2 A^{16} + 5 A^{15} \]

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