化简行阶梯最简矩阵出现一个-1是不是可以直接在那一行乘以-1就可以了 矩阵化成行阶梯形矩阵前，必须把第一行变为一再转换，还是什么数...
\u7ebf\u6027\u4ee3\u6570\u4e2d\uff0c\u6c42\u7ebf\u6027\u65b9\u7a0b\u7ec4\u65f6\uff0c\u5316\u7b80\u884c\u6700\u7b80\u5f62\u77e9\u9635\u4e2d\u76841\u53ef\u4ee5\u662f-1\u4e48\uff1f\uff1f\u53ef\u4ee5\uff0c \u4e0d\u8fc7\u8981\u8bb0\u4f4f\u3002
\u4f8b\u5982\u65b9\u7a0b\u7ec4
x1+2x2 + x3 = 4
x1+ x2 +x3 = 2
\u589e\u5e7f\u77e9\u9635 (A, b) =
[1 2 1 4]
[1 1 1 2]
\u884c\u521d\u7b49\u53d8\u6362\u4e3a
[1 2 1 4]
[0 -1 0 -2]
\u884c\u521d\u7b49\u53d8\u6362\u4e3a
[1 0 1 0]
[0 -1 0 -2]
r(A,b) =r(A) = 2 < 3,
\u65b9\u7a0b\u7ec4\u6709\u65e0\u7a77\u591a\u89e3\u3002
\u65b9\u7a0b\u7ec4\u540c\u89e3\u53d8\u5f62\u4e3a
x1 = -x3
-x2 = -2
\u53d6 x3 = 0\uff0c \u5f97\u7279\u89e3 (0, 2, 0)^T,
\u5bfc\u51fa\u7ec4\u5373\u5bf9\u5e94\u7684\u9f50\u6b21\u65b9\u7a0b\u7ec4\u662f
x1 = -x3
-x2 = 0
\u53d6 x3 = 1\uff0c \u5f97\u57fa\u7840\u89e3\u7cfb (-1, 0, 1)^T,
\u5219\u539f\u65b9\u7a0b\u7ec4\u7684\u901a\u89e3\u662f
x = (0, 2, 0)^T+ k(-1, 0, 1)^T\uff0c
\u5176\u4e2d k \u4e3a\u4efb\u610f\u5e38\u6570\u3002
\u5316\u9636\u68af\u77e9\u9635\u65f6\u53ef\u4ee5\u76f4\u63a5\u9010\u5217\u5316\u7b80,\u8fd9\u9898\u4e2d\u5148\u5c06\u5404\u884c\u7b2c\u4e00\u5217\u5316\u4e3a0
\u5c06\u7b2c\u4e00\u884c\u7684-1\u500d\u52a0\u81f3\u7b2c\u4e8c\u884c,-2\u500d\u52a0\u81f3\u7b2c\u4e09\u884c,4\u500d\u52a0\u81f3\u7b2c\u56db\u884c\u5f97\uff1a
1,1,2,3
0,1,1,1
0,1,0,-5
0,8,9,14
\u7136\u540e\u518d\u5316\u7b2c\u4e8c\u5217,\u5c06\u7b2c\u4e8c\u884c\u7684-1\u500d\u52a0\u81f3\u7b2c\u4e09\u884c,-8\u500d\u52a0\u81f3\u7b2c\u56db\u884c\u5f97\uff1a
1,1,2,3
0,1,1,1
0,0,-1,-6
0,0,1,6
\u4e3a\u65b9\u4fbf,\u5148\u5c06\u7b2c\u4e09\u884c\u4e58\u4ee5-1\u5f97\uff1a
1,1,2,3
0,1,1,1
0,0,1,6
0,0,1,6
\u7136\u540e\u5c06\u7b2c\u4e09\u884c\u7684-1\u500d\u52a0\u81f3\u7b2c\u56db\u884c\u5373\u53ef\u5f97\uff1a
1,1,2,3
0,1,1,1
0,0,1,6
0,0,0,0
\u8fd9\u5c31\u662f\u6700\u7ec8\u7684\u9636\u68af\u77e9\u9635\u4e86,\u90fd\u53ef\u4ee5\u7528\u7c7b\u4f3c\u7684\u65b9\u6cd5\u53d8\u6362
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