A是一个3×3的矩阵（4,2,3.； 1,1,0； -1,2,3。）且AX=3A+X，求矩阵X 求矩阵A=（1,1,2,2,1,；0,2,1,5，-1；2,...
\u8bbeA=(3,1,-2;1,5,2),B=(4,1;2,3:;1,2) ,C= (0,3,1;3,-1,,2),\u52193\u00d72\u4e0b\u5217\u77e9\u9635\u8fd0\u7b97\u7ed3\u679c\u4e3a3\u00d72\u7684\u77e9\u9635\u662f\uff08 \uff09.A \u662f 2\u00d73 \u77e9\u9635\uff0cB \u662f 3\u00d72 \u77e9\u9635\uff0cC \u662f 2\u00d73 \u77e9\u9635\uff0c
\u56e0\u6b64 ABC \u662f 2\u00d73 \u77e9\u9635\uff0c
AC^TB^T \u662f 2\u00d73 \u77e9\u9635\uff0c
C^TB^TA^T=(ABC)^T \u662f 3\u00d72 \u77e9\u9635\uff0c
CBA \u662f 2\u00d73 \u77e9\u9635\u3002
\u9009 C
\u5177\u4f53\u56de\u7b54\u5982\u56fe\uff1a
\u5728\u7ebf\u6027\u4ee3\u6570\u4e2d\uff0c\u4e00\u4e2a\u77e9\u9635A\u7684\u5217\u79e9\u662fA\u7684\u7ebf\u6027\u72ec\u7acb\u7684\u7eb5\u5217\u7684\u6781\u5927\u6570\u76ee\u3002\u7c7b\u4f3c\u5730\uff0c\u884c\u79e9\u662fA\u7684\u7ebf\u6027\u65e0\u5173\u7684\u6a2a\u884c\u7684\u6781\u5927\u6570\u76ee\u3002\u901a\u4fd7\u4e00\u70b9\u8bf4\uff0c\u5982\u679c\u628a\u77e9\u9635\u770b\u6210\u4e00\u4e2a\u4e2a\u884c\u5411\u91cf\u6216\u8005\u5217\u5411\u91cf\uff0c\u79e9\u5c31\u662f\u8fd9\u4e9b\u884c\u5411\u91cf\u6216\u8005\u5217\u5411\u91cf\u7684\u79e9\uff0c\u4e5f\u5c31\u662f\u6781\u5927\u65e0\u5173\u7ec4\u4e2d\u6240\u542b\u5411\u91cf\u7684\u4e2a\u6570\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u5728\u9636\u68af\u5f62\u77e9\u9635\u4e2d\uff0c\u9009\u5b9a1\uff0c3\u884c\u548c3\uff0c4\u5217\uff0c\u5b83\u4eec\u4ea4\u53c9\u70b9\u4e0a\u7684\u5143\u7d20\u6240\u7ec4\u6210\u76842\u9636\u5b50\u77e9\u9635\u7684\u884c\u5217\u5f0f \u5c31\u662f\u77e9\u9635A\u7684\u4e00\u4e2a2\u9636\u5b50\u5f0f\u3002
\u5728m*n\u77e9\u9635A\u4e2d\uff0c\u4efb\u610f\u51b3\u5b9ak\u884c\u548ck\u5217\u4ea4\u53c9\u70b9\u4e0a\u7684\u5143\u7d20\u6784\u6210A\u7684\u4e00\u4e2ak\u9636\u5b50\u77e9\u9635\uff0c\u6b64\u5b50\u77e9\u9635\u7684\u884c\u5217\u5f0f\uff0c\u79f0\u4e3aA\u7684\u4e00\u4e2ak\u9636\u5b50\u5f0f\u3002
\u5f53r(A)<=n-2\u65f6\uff0c\u6700\u9ad8\u9636\u975e\u96f6\u5b50\u5f0f\u7684\u9636\u6570<=n-2\uff0c\u4efb\u4f55n-1\u9636\u5b50\u5f0f\u5747\u4e3a\u96f6\uff0c\u800c\u4f34\u968f\u9635\u4e2d\u7684\u5404\u5143\u7d20\u5c31\u662fn-1\u9636\u5b50\u5f0f\u518d\u52a0\u4e0a\u4e2a\u6b63\u8d1f\u53f7\uff0c\u6240\u4ee5\u4f34\u968f\u9635\u4e3a0\u77e9\u9635\u3002
\u5f53r(A)<=n-1\u65f6\uff0c\u6700\u9ad8\u9636\u975e\u96f6\u5b50\u5f0f\u7684\u9636\u6570<=n-1\uff0c\u6240\u4ee5n-1\u9636\u5b50\u5f0f\u6709\u53ef\u80fd\u4e0d\u4e3a\u96f6\uff0c\u6240\u4ee5\u4f34\u968f\u9635\u6709\u53ef\u80fd\u975e\u96f6\uff08\u7b49\u53f7\u6210\u7acb\u65f6\u4f34\u968f\u9635\u5fc5\u4e3a\u975e\u96f6\uff09\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1\u2014\u2014\u77e9\u9635\u7684\u79e9
然后用初等变换的方法解AX=B类型的矩阵方程
解: 由AX=3A+X
得 (A-E)X = 3A (整理后得矩阵方程的类型)
(A-E, A) = (构造矩阵, 求(A-E)^-1A, 常数3不必考虑, 最后乘它就行了)
3 2 3 4 2 3
1 0 0 1 1 0
-1 2 2 -1 2 3
r1-3r2,r3+r2
0 2 3 1 -1 3
1 0 0 1 1 0
0 2 2 0 3 3
r1-r3, r3*(1/2)
0 0 1 1 -4 0
1 0 0 1 1 0
0 1 1 0 3/2 3/2
r3-r1
0 0 1 1 -4 0
1 0 0 1 1 0
0 1 0 -1 11/2 3/2
交换行得
1 0 0 1 1 0
0 1 0 -1 11/2 3/2
0 0 1 1 -4 0
所以 (A-E)^-1A =
1 1 0
-1 11/2 3/2
1 -4 0
X = 3(A-E)^-1A =
3 3 0
-3 33/2 9/2
3 -12 0
满意请采纳^_^
见图片
x=(A-E)的逆矩阵乘以3A
绛旓細瑙: 鐢盇X=3A+X 寰 (A-E)X = 3A (鏁寸悊鍚庡緱鐭╅樀鏂圭▼鐨勭被鍨)(A-E, A) = (鏋勯鐭╅樀, 姹(A-E)^-1A, 甯告暟3涓嶅繀鑰冭檻, 鏈鍚庝箻瀹冨氨琛屼簡)3 2 3 4 2 3 1 0 0 1 1 0 -1 2 2 -1 2 3 r1-3r2,r3+r2 0 2 3 1 -1 3 1 0 0 1 ...
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绛旓細R(AB)=2鍝 鍥犱负A鏄鍙嗙殑 鎵浠鍙互琛ㄧず鎴怤涓垵绛夋柟闃电殑涔樼Н 鐒跺悗鍒濈瓑鍙樻崲涓嶄細鏀瑰彉鐭╅樀鐨勭З 浠ヤ笂閮芥槸涔︿笂鐨勫熀鏈畾涔 鎵浠(AB)=R锛圔锛=2 婊℃剰璇烽噰绾
