1 00:00:05,245 --> 00:00:10,490 å¤§å®¶å¥½ï¼Œæ¬¢è¿Žæ”¶çœ‹çº¿æ€§ä»£æ•°ä¹ é¢˜è¯¾ã€‚ 2 00:00:10,490 --> 00:00:16,010 今天这节课我们将ç€é‡ç»ƒä¹ å¦‚ä½•åœ¨è€ƒè¯•ä¸æ±‚解试题。 3 00:00:16,010 --> 00:00:25,740 ä½ ä¹Ÿè®¸æœ‰å¦‚ä¸‹ä½“ä¼šï¼šåœ¨ä½ åšä½œä¸šä¹ 题的时候, ä½ å¯ä»¥å¯¹æ¯é“题甚至æ¯ä¸€ä¸ªæ¥éª¤è¿›è¡Œéžå¸¸ä»”细的æ€è€ƒã€‚ 4 00:00:25,740 --> 00:00:29,570 ä½ è¿˜å¯ä»¥ç”¨ä¸åŒçš„æ–¹æ³•求解åŒä¸€é“题。 5 00:00:29,570 --> 00:00:35,340 è¿™æ ·ä¸€æ–¹é¢ä½ å¯ä»¥éªŒè¯ä½ çš„ç”æ¡ˆï¼Œå¦å¤–ä½ ä¹Ÿå¯ä»¥æ‰¾åˆ°æœ€ä½³è§£æ³•。 6 00:00:35,340 --> 00:00:39,730 但是这些在考试ä¸é€šå¸¸æ˜¯ä¸å¯è¡Œçš„。 7 00:00:39,730 --> 00:00:44,835 å› ä¸ºæ—¶é—´é€šå¸¸å¾ˆç´§è¿«ï¼Œé‚£ä¹ˆæˆ‘ä»¬éœ€è¦ç»ƒä¹ çš„ 8 00:00:44,835 --> 00:00:49,940 就是在考试ä¸å¿«é€Ÿå¹¶å‡†ç¡®æ±‚è§£ä¹ é¢˜çš„æ–¹æ³•ã€‚ 9 00:00:49,940 --> 00:00:58,540 䏋颿ˆ‘们æ¥çœ‹è¿™é“题: è¿™é“题出自一个线性代数50分钟的考试。 10 00:00:58,540 --> 00:01:04,380 ç›®å‰ä¸ºæ¢ä½ å·²ç»å¦äº†è¶³å¤Ÿçš„çŸ¥è¯†æ¥æ±‚解这é“题, 11 00:01:04,380 --> 00:01:12,930 既然这是一个50åˆ†é’Ÿçš„è€ƒè¯•ï¼Œé‚£ä¹ˆèŠ±åœ¨è¿™é“ é¢˜ä¸Šçš„æ—¶é—´æ¯”è¾ƒåˆç†çš„应该是ä¸è¶…过15分钟。 12 00:01:12,930 --> 00:01:22,840 也就是说在15åˆ†é’Ÿä¹‹å†…ï¼Œä½ åº”è¯¥å®Œæˆè¯»é¢˜ï¼Œ ç†è§£é¢˜æ„,并且完全求解这三个部分。 13 00:01:22,840 --> 00:01:30,871 çŽ°åœ¨è¯·ä½ æš‚åœè¿™ä¸ªè§†é¢‘,并å°è¯•独立求解, ç„¶åŽä½ å¯ä»¥è®¡æ—¶15分钟, 14 00:01:30,871 --> 00:01:36,450 15分钟之åŽï¼Œæˆ‘将回æ¥å¹¶ä¸€èµ·æ£€éªŒä½ çš„ç”æ¡ˆã€‚ 15 00:01:36,450 --> 00:01:41,830 åŒæ—¶å¦‚æžœä½ æå‰å®Œæˆçš„è¯ï¼Œä¹Ÿè¯·ç‹¬ç«‹æ£€æŸ¥ä½ çš„ç”æ¡ˆã€‚ 16 00:01:41,830 --> 00:01:46,250 å› ä¸ºåœ¨è€ƒè¯•ä¸ä½ 应该尽é‡å¾—到所有应得的分数。 17 00:01:46,250 --> 00:01:50,750 好,一会儿è§ã€‚ 18 00:01:56,490 --> 00:02:03,380 ä½ å®Œæˆäº†å—? 䏋颿ˆ‘们æ¥ä¸€èµ·è§£å†³è¿™é“问题。 19 00:02:03,380 --> 00:02:11,190 ä½ å¯ä»¥çœ‹å‡ºè¿™é“问题是关于 矩阵行列å¼å®šä¹‰å…¬å¼çš„ç»ƒä¹ é¢˜ã€‚ 20 00:02:11,190 --> 00:02:18,190 我们有一个4乘4的矩阵,那么第一个问题就是è¦è®¡ç®—这个矩阵A的行列å¼ã€‚ 21 00:02:18,190 --> 00:02:22,760 åªä½¿ç”¨è¡Œåˆ—å¼çš„å®šä¹‰å¼æ¥è®¡ç®—。 22 00:02:22,760 --> 00:02:32,540 好,我们å¯ä»¥å†™å‡ºï¼ŒçŸ©é˜µA也就是第一部分, 矩阵A的行列å¼ç‰äºŽ 23 00:02:32,540 --> 00:02:37,700 一些乘积的求和, 24 00:02:37,700 --> 00:02:45,160 è¿™äº›å…ƒç´ 25 00:02:45,160 --> 00:02:51,520 å–自Açš„ä¸åŒè¡Œï¼Œä¹Ÿå°±æ˜¯è¯´æˆ‘们在æ¯ä¸€è¡Œä¸å–Açš„ä¸€ä¸ªå…ƒç´ 26 00:02:51,520 --> 00:02:57,200 并且ä¿è¯å®ƒä»¬çš„åˆ—æŒ‡æ ‡éƒ½å®Œå…¨ä¸åŒï¼Œä¹Ÿå°±æ˜¯è¯´ 27 00:02:57,200 --> 00:03:03,450 è¿™äº›Î±ï¼ŒÎ²ï¼ŒÎ³ï¼Œå’ŒÎ´æ˜¯ç›¸åº”çš„åˆ—æŒ‡æ ‡ã€‚ 28 00:03:03,450 --> 00:03:05,730 我把它记在这里, 29 00:03:05,730 --> 00:03:13,585 我们需è¦è¿™äº›æˆä¸ºæ•°å—1,2,3,4的一个置æ¢ï¼Œ 30 00:03:13,585 --> 00:03:17,892 é‚£ä¹ˆå¦‚æžœä½ æŒ‰ç…§è¿™ä¸ªé¡ºåºæ±‚解的è¯ï¼Œ 31 00:03:17,892 --> 00:03:26,000 从第一行é€è¡Œå‘下找到所有å¯èƒ½çš„1,2,3,4çš„ç½®æ¢ 32 00:03:26,000 --> 00:03:33,348 最åŽå†åŽ»æŽ‰ç‰äºŽ0的那些项的è¯ï¼Œæœ‰å¤šå°‘é¡¹ä½ éœ€è¦è®¡ç®—呢? 33 00:03:33,348 --> 00:03:39,940 ä½ éœ€è¦è®¡ç®—4的阶乘那么多项,也就是24项。 34 00:03:39,940 --> 00:03:49,006 è¿™æ ·çš„è¯åœ¨è€ƒè¯•ä¸å°±æ¯”è¾ƒèŠ±æ—¶é—´ï¼Œå…¶å®žæœ‰ä¸€ç§æ›´å¿«çš„æ–¹æ³•æ¥æ‰¾ 到这个求和å¼ä¸çš„éž0项。 35 00:03:49,006 --> 00:03:53,770 æ—¢ç„¶è¦æ‰¾éž0项的è¯ï¼Œæˆ‘们æ¥è§‚察矩阵Aä¸çš„0å…ƒç´ åœ¨å“ªé‡Œã€‚ 36 00:03:53,770 --> 00:04:03,250 ä½ å¯ä»¥çœ‹åˆ°çŸ©é˜µAçš„0å…ƒç´ ï¼Œå‡ºçŽ°åœ¨è¿™å››ä¸ªä½ç½®ï¼Œ 它们分别在第三行和第四行 37 00:04:03,250 --> 00:04:09,850 ä¹Ÿå°±æ˜¯è¯´åœ¨ä½ å†™å‡ºé‚£ä¸ªå…¬å¼çš„æ—¶å€™ï¼ŒA3γ和A4δ, 38 00:04:09,850 --> 00:04:19,160 åªæœ‰å¯èƒ½åœ¨è¿™å››ä¸ªæ•°å—ä¸é€‰å–, 也就是我所画的这个红方å—ä¸é€‰å–, 39 00:04:19,160 --> 00:04:24,130 也就是说A3γ和A4δ 40 00:04:24,130 --> 00:04:33,560 åªå¯ä»¥æ˜¯9,12,或者是10å’Œ11, å¦‚æžœæ˜¯è¿™æ ·æƒ…å†µçš„è¯ï¼Œ 41 00:04:33,560 --> 00:04:42,680 我们åˆéœ€è¦è¿™äº›å…ƒç´ 都从ä¸åŒçš„列ä¸é€‰å–, ä¹Ÿå°±æ˜¯è¯´æ˜Žä½ åœ¨ç¬¬ä¸€è¡Œå’Œç¬¬äºŒè¡Œé€‰å–çš„å…ƒç´ ï¼Œ 42 00:04:42,680 --> 00:04:49,320 必须是这四个数å—。 43 00:04:49,320 --> 00:04:55,960 第一行和第二行åªå¯ä»¥ä»Žä¸Šé¢è¿™ä¸ªçº¢æ–¹å—ä¸é€‰å–, 44 00:04:55,960 --> 00:05:05,230 也就是说A1α,A2β, è¦ä¹ˆæ˜¯1,6,è¦ä¹ˆæ˜¯2å’Œ5, 45 00:05:05,230 --> 00:05:13,990 ä½ å¯ä»¥çœ‹åˆ°å…¶å®žæˆ‘们å¯ä»¥å¿½ç•¥è¿™ä¸ªä½ç½®çš„四个 æ•°å—ã€‚å› ä¸ºæ— è®ºé€‰å–哪一个数å—çš„è¯ 46 00:05:13,990 --> 00:05:20,700 我们将ä¸å¯é¿å…的在第三行或第四行ä¸é€‰åˆ°è‡³å°‘一个0å…ƒç´ ã€‚ 47 00:05:20,700 --> 00:05:28,540 å¥½ï¼Œä¸‹é¢æˆ‘们å¯ä»¥çœ‹åˆ° ç»è¿‡è¿™æ ·ç®€åŒ–çš„è¯ï¼Œè¿™ä¸ªå¼åä¸è¿˜æœ‰å¤šå°‘项éž0呢? 48 00:05:28,540 --> 00:05:36,804 这里有两ç§å–æ³•ï¼Œè¿™é‡Œåˆæœ‰ä¸¤ç§å–法, 那么最åŽä¹Ÿå°±æ˜¯è¯´éž0é¡¹ï¼Œåªæœ‰å››é¡¹ï¼Œ 49 00:05:36,804 --> 00:05:43,585 æˆ‘ä»¬æŠŠå®ƒä¾æ¬¡å†™å‡ºï¼Œ 在第一个方å—ä¸ï¼Œæˆ‘å¯ä»¥é€‰å–1乘以6, 50 00:05:43,585 --> 00:05:48,459 第二个方å—䏿ˆ‘选9å’Œ12,乘以9, 51 00:05:48,459 --> 00:05:56,300 乘以12ï¼Œé‚£ä¹ˆè¿™ä¸€é¡¹ç›¸åº”çš„åˆ—æŒ‡æ ‡å°±æ˜¯ï¼Œå› ä¸ºå®ƒä»¬å‡ä¸ºå¯¹è§’çº¿ä¸Šå…ƒç´ ï¼Œ 52 00:05:56,300 --> 00:06:02,868 也就是A11,A22,A33,A44, 53 00:06:02,868 --> 00:06:09,013 这就是第一项,那么这是1,2,3,4的一个 54 00:06:09,013 --> 00:06:13,680 æ£åºæŽ’列。所以这一项的符å·å°±æ˜¯ä¸€ä¸ªæ£å·ã€‚ 55 00:06:13,680 --> 00:06:21,170 好看下一项,下一项在第一个方å—䏿ˆ‘还选å–1å’Œ6, 56 00:06:21,170 --> 00:06:26,820 在第二个方å—䏿ˆ‘选å–10å’Œ11 57 00:06:26,820 --> 00:06:34,270 å¦‚æžœæˆ‘è¿™æ ·é€‰çš„è¯ï¼Œåˆ—æŒ‡æ ‡å°±å°†æˆä¸ºA11,A22 58 00:06:34,270 --> 00:06:43,230 10是Açš„ç¬¬ä¸‰è¡Œç¬¬å››åˆ—çš„å…ƒç´ ï¼Œ ä¹Ÿå°±æ˜¯è¯´è¿™é‡Œçš„åˆ—æŒ‡æ ‡åº”æ˜¯A34, 59 00:06:43,230 --> 00:06:48,180 11是第四行第三列ä¸çš„å…ƒç´ ï¼Œæ‰€ä»¥æœ€åŽåº”是3。 60 00:06:48,180 --> 00:06:51,310 è¿™æ˜¯è¿™é¡¹ç›¸åº”çš„åˆ—æŒ‡æ ‡ï¼Œ 61 00:06:51,310 --> 00:06:58,850 ä½†æ˜¯ä½ å¯ä»¥çœ‹åˆ°æˆ‘们需è¦å¯¹æœ€åŽä¸¤ä¸ªä½ç½®è¿›è¡Œå¯¹æ¢ï¼Œæ‰å¯ä»¥ä½¿è¿™ä¸ªç½®æ¢å›žåˆ°1,2,3,4 62 00:06:58,850 --> 00:07:02,650 也就是说这一项å‰åº”该有一个负å·ã€‚ 63 00:07:02,650 --> 00:07:09,880 好,我们在这里继ç»ï¼Œ 64 00:07:09,880 --> 00:07:17,190 下一项 我们在第一个方å—ä¸é€‰å–2å’Œ5, 65 00:07:17,190 --> 00:07:23,620 那么在第二个方å—ä¸ä»ç„¶é€‰å–9å’Œ12, 66 00:07:23,620 --> 00:07:32,800 è¿™ä¸€é¡¹çš„åˆ—æŒ‡æ ‡å°†æ˜¯2 是Aä¸ç¬¬ä¸€è¡Œç¬¬äºŒåˆ—å…ƒç´ ï¼Œæ‰€ä»¥æ˜¯A12, 67 00:07:32,800 --> 00:07:41,200 5æ˜¯ç¬¬äºŒè¡Œç¬¬ä¸€ä¸ªå…ƒç´ ï¼ŒA21, 那么下é¢è¿˜æ˜¯å¯¹è§’线上A33,A44, 68 00:07:41,200 --> 00:07:49,870 åŒæ ·å¯¹äºŽè¿™ä¸ªç½®æ¢æˆ‘ä»¬éœ€è¦ å¯¹æ¢è¿™ä¸¤ä¸ªä½ç½®ä½¿å¾—这个置æ¢å›žåˆ°1,2,3,4 69 00:07:49,870 --> 00:07:58,950 也就是说这项å‰é¢ä¹Ÿåº”该还有一个负å·ï¼Œ 所以把一个负å·åŠ åœ¨è¿™é‡Œã€‚ 70 00:07:58,950 --> 00:08:04,830 那么最åŽä¸€ä¸ªå°±æ˜¯ 71 00:08:04,830 --> 00:08:13,840 2,5,10å’Œ11, åŒç† 72 00:08:13,840 --> 00:08:19,260 è¿™ä¸€é¡¹çš„åˆ—æŒ‡æ ‡å°†æ˜¯2,1,4,3, 73 00:08:19,260 --> 00:08:24,960 那么这一次我们需è¦å¯¹è¿™ä¸¤ä¸ªä½ç½®è¿›è¡Œå¯¹æ¢ï¼Œ 74 00:08:24,960 --> 00:08:31,030 并且这两个ä½ç½®è¿›è¡Œå¯¹æ¢ï¼Œæ‰å¯ä»¥å›žåˆ°1,2,3,4,那么两次对æ¢ä½¿å¾— 75 00:08:31,030 --> 00:08:40,390 这一项的符å·ä¸ºæ£ã€‚好,这就是 行列å¼Aä¸ä¸ä¸º0的四项。 76 00:08:40,390 --> 00:08:46,950 é‚£ä¹ˆä½ å¯¹è¿™å››é¡¹è¿›è¡Œæ±‚å’Œçš„è¯ï¼Œæ£ç¡®ç”案应该ç‰äºŽ8。 77 00:08:46,950 --> 00:08:50,550 ä½ çš„ç”æ¡ˆæ£ç¡®å—? 78 00:08:50,550 --> 00:08:57,085 åœ¨è€ƒè¯•ä¸æ—¶é—´é€šå¸¸éžå¸¸ç´§è¿«ï¼Œé‚£ä¹ˆå®Œæˆç¬¬ä¸€éƒ¨åˆ†åŽæˆ‘ä»¬å°†ç«‹å³ 79 00:08:57,085 --> 00:09:03,620 æ¥è¿›è¡Œå¯¹ç¬¬äºŒéƒ¨åˆ†çš„æ±‚解。第二 80 00:09:03,620 --> 00:09:08,840 éƒ¨åˆ†æ˜¯è¦æ±‚ä½ è®¡ç®—çŸ©é˜µA的第一行的代数余åå¼ã€‚ 81 00:09:08,840 --> 00:09:16,350 C11,C12,C13å’ŒC14,那么A是一个4乘4矩阵。 82 00:09:16,350 --> 00:09:22,750 æ¯ä¸€ä¸ªä»£æ•°ä½™åå¼å°†æ¶‰åŠåˆ°è®¡ç®—一个3乘3矩阵的行列å¼ã€‚ 83 00:09:22,750 --> 00:09:29,930 也就是说看起æ¥å¯¹äºŽç¬¬äºŒéƒ¨åˆ†ï¼Œæˆ‘们需è¦è®¡ç®—四个3乘3矩阵的行列å¼ã€‚ 84 00:09:29,930 --> 00:09:38,965 æˆ‘ä»¬ä¸‹é¢æŠŠå®ƒä»¬åˆ†åˆ«å†™å‡ºæ¥ã€‚å…ˆæ¥çœ‹C11, C11是这个ä½ç½®ä»£æ•°ä½™åå¼ã€‚ 85 00:09:38,965 --> 00:09:43,150 那么我们è¦è®¡ç®—的就是这个剩余ä½ç½®çš„3乘3矩阵的行列å¼ã€‚ 86 00:09:43,150 --> 00:09:49,980 把它写在这里。C11ç‰äºŽ 87 00:09:49,980 --> 00:09:57,460 å¦‚ä¸‹çŸ©é˜µçš„è¡Œåˆ—å¼ 6,7,8, 88 00:09:57,460 --> 00:10:01,865 0,9,10, 0, 89 00:10:01,865 --> 00:10:10,050 11,12, ä½ è¦ç”¨ä»€ä¹ˆæ–¹æ³•æ¥æ±‚解这个3乘3 90 00:10:10,050 --> 00:10:15,310 矩阵的代数行列å¼å‘¢ï¼Ÿä½ 当然å¯ä»¥ç”¨è¡Œåˆ—å¼çš„定义å¼ï¼Œ 91 00:10:15,310 --> 00:10:21,390 也就是对于这个3乘3çŸ©é˜µå†™å‡ºç±»ä¼¼äºŽè¿™æ ·çš„ä¸€ä¸ªæ±‚å’Œå…¬å¼ã€‚ 92 00:10:21,390 --> 00:10:26,485 ä½†æ˜¯å¦‚æžœä½ ç”¨è¿™ç§æ–¹æ³•çš„è¯ï¼Œä½ 需è¦è®¡ç®—3çš„ 93 00:10:26,485 --> 00:10:31,580 阶乘,åƒä¹˜ç§¯ï¼Œä¹Ÿå°±æ˜¯è¯´ä½ 需è¦è®¡ç®—6个乘积。 94 00:10:31,580 --> 00:10:36,810 有没有更为快æ·çš„åŠžæ³•å‘¢ï¼Ÿä½ å¯ä»¥æ³¨æ„到 95 00:10:36,810 --> 00:10:42,280 这个3乘3çŸ©é˜µçš„ç¬¬ä¸€åˆ—åªæœ‰è¿™ä¸ªä½ç½®çš„å…ƒç´ éž0。 96 00:10:42,280 --> 00:10:49,760 那么也就是说我们应该利用代数余åå¼çš„æ–¹æ³•æ¥è®¡ç®—这个矩阵的行列å¼ã€‚ 97 00:10:49,760 --> 00:10:53,090 我们对第一列进行展开。 98 00:10:53,090 --> 00:10:59,400 那么结果应该是6乘以它的代数余åå¼ã€‚ 99 00:10:59,400 --> 00:11:04,830 那么它代数余åå¼å°±æ˜¯å‰©ä½™2乘2矩阵的行列å¼ã€‚è¿™æ · 100 00:11:04,830 --> 00:11:10,260 就很简å•,也就应该是9乘以12å‡åŽ» 101 00:11:10,260 --> 00:11:16,420 10乘以11,计算这个å¼åçš„è¯ 102 00:11:16,420 --> 00:11:25,480 108å‡åŽ»110å¾—è´Ÿ2,å†ä¹˜ä»¥6 ç‰äºŽè´Ÿ12,这就是C11。 103 00:11:25,480 --> 00:11:33,232 䏋颿¥çœ‹C12, åŒæ ·C12也是一个3乘3矩阵的 104 00:11:33,232 --> 00:11:41,759 行列å¼ï¼Œæˆ‘们把这个 矩阵写在这里,应该是5,7,8, 105 00:11:41,759 --> 00:11:47,958 0 ,9, 10 ,0, 11, 106 00:11:47,958 --> 00:11:54,159 12, ä¸è¦å¿˜è®°çŽ°åœ¨æˆ‘ä»¬çœ‹åˆ°çš„æ˜¯1,2 107 00:11:54,159 --> 00:12:01,660 ä½ç½®ä»£æ•°ä½™åå¼ï¼Œæ‰€ä»¥åœ¨è¿™ä¸ªè¡Œåˆ—å¼ä¹‹å‰æˆ‘还需è¦ä¸€ä¸ªè´Ÿå·ã€‚ 108 00:12:01,660 --> 00:12:06,890 这也就是负1çš„1åŠ 2次方ç‰äºŽè´Ÿ1。 109 00:12:06,890 --> 00:12:13,850 åŒæ ·çš„这个3乘3çŸ©é˜µçš„ç¬¬ä¸€åˆ—ä¹Ÿåªæœ‰ä¸€ä¸ªéž0å…ƒç´ ã€‚ 110 00:12:13,850 --> 00:12:18,640 对它进行展开就ç‰äºŽè´Ÿ5乘以 111 00:12:18,640 --> 00:12:24,710 åŒæ ·çš„这个2乘2矩阵的行列å¼ï¼Œä¹Ÿå°±æ˜¯9乘以12 112 00:12:24,710 --> 00:12:30,320 å‡åŽ»10乘以11,这一次应该ç‰äºŽ10, 113 00:12:30,320 --> 00:12:35,460 这就是C12ã€‚æˆ‘ä»¬ç»§ç» 114 00:12:35,460 --> 00:12:44,020 C13ç‰äºŽå¦‚下矩阵的行列å¼ï¼š 5, 115 00:12:44,020 --> 00:12:47,937 6 ,8, 0, 116 00:12:47,937 --> 00:12:52,840 0, 10, 0 ,0, 12, 117 00:12:52,840 --> 00:13:00,820 ä½ å½“ç„¶å¯ä»¥åŒæ ·çš„用对第一列展开的方法求这个3乘3矩阵的行列å¼ã€‚ 118 00:13:00,820 --> 00:13:04,080 å¦‚æžœä½ é‚£æ ·åšçš„è¯ï¼Œç»“果应为0。 119 00:13:04,080 --> 00:13:11,860 ä½†æ˜¯å…¶å®žä½ åº”è¯¥å¯ä»¥ç›´æŽ¥çœ‹å‡º 这个矩阵行列å¼åº”为0,并ä¸éœ€è¦ä»»ä½•计算。 120 00:13:11,860 --> 00:13:15,710 å› ä¸ºå¾ˆæ˜¾ç„¶è¿™ä¸¤åˆ—æ˜¯çº¿æ€§ç›¸å…³çš„ã€‚ 121 00:13:15,710 --> 00:13:22,790 第二列就ç‰äºŽç¬¬ä¸€åˆ—乘以5分之6, 也就是说这实际上是一个奇异的矩阵。 122 00:13:22,790 --> 00:13:29,320 那么奇异矩阵的行列å¼å‡ä¸º0,所以立å³ä½ å¯ä»¥å†™å‡ºC13ç‰äºŽ0。 123 00:13:29,320 --> 00:13:36,228 æœ€åŽæ¥çœ‹C14, 把剩余部分矩阵抄下æ¥ï¼Œ5, 124 00:13:36,228 --> 00:13:41,343 6, 7, 0 ,0 ,9, 125 00:13:41,343 --> 00:13:44,924 0, 0, 11, 126 00:13:44,924 --> 00:13:53,630 这也是一个奇异矩阵,这两列线性相关,那么这个结果也应该为0。 127 00:13:53,630 --> 00:14:00,890 当然通常情况下,我需è¦ä¸€ä¸ªè´Ÿ1在这里,但是这里负1乘以0还是ç‰äºŽ0。 128 00:14:00,890 --> 00:14:07,540 好,这就是A的第一行的代数余åå¼ï¼Œä½ å¯ä»¥çœ‹åˆ°ï¼Œå¦‚æžœä½ è§‚å¯Ÿåˆ° 129 00:14:07,540 --> 00:14:13,520 这些矩阵的特点的è¯ï¼Œè¿™æ ·çš„计算其实是并ä¸èŠ±æ—¶é—´çš„ã€‚ 130 00:14:13,520 --> 00:14:18,565 好,下é¢ä½ å¯ä»¥ç»§ç»åšç¬¬ä¸‰éƒ¨åˆ†ï¼Œä½†æ˜¯æˆ‘们注æ„到 131 00:14:18,565 --> 00:14:23,610 有了第二部分的结果,我们å¯ä»¥å¾ˆå®¹æ˜“åœ°æ£€éªŒç¬¬ä¸€éƒ¨åˆ†ç”æ¡ˆæ˜¯å¦æ£ç¡®ã€‚ 132 00:14:23,610 --> 00:14:29,000 é‚£ä¹ˆå¦‚ä½•æ¥æ£€éªŒå‘¢ï¼Ÿæˆ‘们现在有A的第一列的代数余åå¼ã€‚ 133 00:14:29,000 --> 00:14:36,360 åŒæ ·æˆ‘们还知é“A的行列å¼å¯ä»¥ç”±A的第一列 与其代数余åå¼çš„内积给出。 134 00:14:36,360 --> 00:14:42,780 我们就æ¥çœ‹ä¸€ä¸‹å®ƒçš„å†…ç§¯æ˜¯ä¸æ˜¯ç‰äºŽæˆ‘们在第一部分ä¸å¾—åˆ°çš„ç”æ¡ˆã€‚ 135 00:14:42,780 --> 00:14:45,920 䏋颿ˆ‘把它写在这里。 136 00:14:45,920 --> 00:14:48,717 矩阵A的行列å¼è¿˜ç‰äºŽAçš„ 137 00:14:48,717 --> 00:14:56,642 第一列与代数余åå¼å†…积,就是a11乘以C11åŠ ä¸Ša12 138 00:14:56,642 --> 00:15:01,770 乘以C12,那么åŽé¢ä¸¤é¡¹éƒ½ä¸º0了。 139 00:15:01,770 --> 00:15:07,610 实际上我们就是需è¦è®¡ç®—这两项的和,a11ç‰äºŽ1, 140 00:15:07,610 --> 00:15:15,880 C11ç‰äºŽè´Ÿ12, A12ç‰äºŽ2,C12ç‰äºŽ10, 141 00:15:15,880 --> 00:15:19,900 ä½ å¯ä»¥çœ‹åˆ°è¿™ä¸ªç»“果确实为8。 142 00:15:19,900 --> 00:15:24,830 æ‰€ä»¥è‡³å°‘ç¬¬äºŒéƒ¨åˆ†ä¸Žç¬¬ä¸€éƒ¨åˆ†ç”æ¡ˆæ˜¯ä¸€è‡´çš„。 143 00:15:24,830 --> 00:15:30,670 䏋颿ˆ‘们æ¥è¿›è¡Œå¯¹ç¬¬ä¸‰é¢˜çš„æ±‚解。 144 00:15:30,670 --> 00:15:36,070 ç¬¬ä¸‰é¢˜æ˜¯è¦æ±‚ä½ å†™å‡ºA的逆矩阵的第一列。 145 00:15:36,070 --> 00:15:42,000 çœ‹èµ·æ¥æˆ‘们åˆè¦æ±‚解很多3乘3矩阵的行列å¼ï¼Œ 146 00:15:42,000 --> 00:15:47,160 但是æ£å¦‚很多ç»è¿‡ä»”ç»†è®¾è®¡çš„è€ƒè¯•ä¹ é¢˜ä¸€æ ·ï¼Œ 147 00:15:47,160 --> 00:15:52,500 ç¬¬ä¸‰é—®çš„ç”æ¡ˆå…¶å®žå¯ä»¥ç”±ç¬¬ä¸€é—®ä¸Žç¬¬äºŒé—®ç”案直接给出。 148 00:15:52,500 --> 00:16:01,530 我们先æ¥å›žå¿†ä¸€ä¸‹Açš„é€†çŸ©é˜µçš„å…¬å¼æ˜¯ä»€ä¹ˆï¼Ÿ A的逆矩阵ç‰äºŽ1除以Açš„è¡Œåˆ—å¼ 149 00:16:01,530 --> 00:16:09,450 å†ä¹˜ä»¥ä¸€ä¸ªçŸ©é˜µC的转置,这个矩阵Cçš„å…ƒç´ å°±ç”±A的代数余åå¼ç»™å‡ºã€‚ 150 00:16:09,450 --> 00:16:16,160 å¦‚æžœæˆ‘ä»¬æƒ³è¦æ‰¾åˆ°Açš„é€†çŸ©é˜µçš„ç¬¬ä¸€åˆ—çš„è¯ 151 00:16:21,235 --> 00:16:26,310 这实际上就ç‰äºŽè¿™ä¸ªå¸¸æ•° 152 00:16:26,310 --> 00:16:32,560 乘以矩阵C的转置的第一列, 153 00:16:32,560 --> 00:16:41,950 那么也就应该是矩阵C的第一行 å†è½¬ç½®ï¼Œ 154 00:16:41,950 --> 00:16:50,030 矩阵A的行列å¼ç”±ç¬¬ä¸€ é—®ç”æ¡ˆå¾—出,也就是八分之一。 155 00:16:50,030 --> 00:16:54,370 Cçš„ç¬¬ä¸€è¡Œç”±ç¬¬äºŒé—®çš„ç”æ¡ˆç»™å‡ºï¼Œ 156 00:16:54,370 --> 00:17:01,810 咱们抄下æ¥ï¼Œå°±æ˜¯è´Ÿ12, 10 157 00:17:01,810 --> 00:17:09,320 ,0 ,0, è¿™æ ·ä½ å°±å¾—åˆ°äº†A的逆矩阵的第一列。 158 00:17:09,320 --> 00:17:13,570 好,我们已ç»å®Œæˆäº†å…¨éƒ¨ä¸‰ä¸ªéƒ¨åˆ†çš„æ±‚解。 159 00:17:13,570 --> 00:17:19,420 ä½ çš„ç”æ¡ˆæ£ç¡®å—?在结æŸä¹‹å‰ï¼Œæˆ‘希望æé†’ä½ ä¸¤ç‚¹ï¼š 160 00:17:19,420 --> 00:17:28,720 ç¬¬ä¸€ç‚¹æ˜¯ä½ å¯ä»¥çœ‹å‡ºï¼Œè¿™ä¸ªè€ƒè¯•ä¸»è¦æ˜¯ åœ¨æµ‹éªŒä½ å¯¹äºŽçŸ©é˜µè¡Œåˆ—å¼å…¬å¼ï¼Œå®šä¹‰å…¬å¼çš„应用。 161 00:17:28,720 --> 00:17:37,020 我们在之å‰çš„ä¹ é¢˜è¯¾ä¸ç»ƒä¹ 了利用消元 法与代数余å弿–¹æ³•çš„ç»“åˆæ¥æ±‚解矩阵的行列å¼ã€‚ 162 00:17:37,020 --> 00:17:40,220 但是我们ä¸åº”该忘记这个公å¼ã€‚ 163 00:17:40,220 --> 00:17:45,550 在很多情况下,这个公å¼å而å¯ä»¥ç»™å‡ºæ¯”较简便的解法。 164 00:17:45,550 --> 00:17:49,980 æ£å¦‚我们看到的这个矩阵A。 165 00:17:49,980 --> 00:17:54,670 第二点是我希望在求解过程ä¸ï¼Œä½ å¯ä»¥å°†ä½ çš„ 166 00:17:54,670 --> 00:18:01,593 æ±‚è§£è¿‡ç¨‹åˆ—å‡ºå¦‚ä¸‹ï¼Œå¦‚æžœä½ å¯ä»¥æ¯”较清晰有æ¡ç†åœ°åˆ—出求解过程, 167 00:18:01,593 --> 00:18:07,623 一方é¢è¿™æ ·ä¾¿äºŽæ£€éªŒä½ çš„ç”æ¡ˆï¼Œå¦å¤–一方é¢å³ä½¿ä½ 最åŽç”案 168 00:18:07,623 --> 00:18:13,430 䏿£ç¡®çš„è¯ï¼Œä½ 也å¯ä»¥é€šè¿‡è¿™äº›è¿‡ç¨‹å¾—到部分的分数。 169 00:18:13,430 --> 00:18:18,930 好,我今天就讲到这里,ç¥å¤§å®¶è€ƒè¯•有好è¿ï¼Œè°¢è°¢ã€‚ 170 00:18:18,930 --> 00:18:24,600 Funding for this video was provided by the Lord Foundation. 171 00:18:24,600 --> 00:18:29,154 To help OCW continue to provide free and open acceess to MIT courses, 172 00:18:29,154 --> 00:18:36,710 please make a donation at ocw.mit.edu/donate