[理工] [工數] 矩陣

[IMBA5566 (許孟哲)]

Write the general solutions to

AX=[1  2  2][X1]=[1]
   [2  4  5][X2] [4]
            [X3]

as the sum of a particular solution to AX=b

and the general solution to AX=0


請問是要分成AX=b 和 AX=0 求出 Xp 和 Xh再相加嗎?

AX=[1  2  2][X1]=[0] ===> [1 2 2 0] ====>X3=0     >>>[X1]=[-2*X2]=Xh
   [2  4  5][X2] [0]      [0 0 1 0]      X1=-2*X2    [X2] [   X2]
            [X3]                                     [X3] [    0]

AX=[1  2  2][X1]=[1] ===> [1 2 2 1] ====>X3=2       >>> [-2*X2 -3]=Xp
   [2  4  5][X2] [4]      [0 0 1 2]      X1=-2*X2 -3    [   X2   ]
            [X3]                                        [       2]

嗯... 以上是我自己亂解的東西... 由於礙於沒有解答

而且我解的東東看起來一點都沒有正確答案的感覺 @////@"

因此上來求助 感謝解答~~~

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◆ From: 180.177.95.158
→ mp8113f:直接解Ax=b就好了吧  會包含xh在內 會有常數項
→ IMBA5566:喔~~ 感謝 那請問一下我的AX=b這樣OK嗎??
→ mp8113f:你x1 x2沒解出來 假設x2是個常數 則x1=-2c-3
→ mp8113f:則[x1 x2 x3]^T = c[-2 1 0]^T + [-3 0 2]^T
→ IMBA5566:了解~~ 感謝 謝謝~~!!!