5x +2y +z =-8
x -2y -3z =0
-x +y +2z =3
Solved this problem by using Gauess Gordan method.
{5x+2y+z=−8x−2y−3z=0−x+y+2z=3\begin{cases} 5x+2y+z=-8 \\ x-2y-3z=0 \\ -x+y+2z=3 \end{cases}⎩⎨⎧5x+2y+z=−8x−2y−3z=0−x+y+2z=3
[521∣−81−2−3∣0−112∣3]→R3+R2[521∣−81−2−3∣00−1−1∣3]\begin{bmatrix} 5&2&1&|&-8 \\ 1&-2&-3&|&0 \\ -1&1&2&|&3 \end{bmatrix} \overset{R_3+R_2}{\rightarrow} \begin{bmatrix} 5&2&1&|&-8 \\ 1&-2&-3&|&0 \\ 0&-1&-1&|&3 \end{bmatrix}⎣⎡51−12−211−32∣∣∣−803⎦⎤→R3+R2⎣⎡5102−2−11−3−1∣∣∣−803⎦⎤
→R1−5R2[01216∣−81−2−3∣00−1−1∣3]\qquad \qquad \qquad \qquad \qquad \quad\overset{R_1-5R_2}{\rightarrow} \begin{bmatrix} 0&12&16&|&-8 \\ 1&-2&-3&|&0 \\ 0&-1&-1&|&3 \end{bmatrix}→R1−5R2⎣⎡01012−2−116−3−1∣∣∣−803⎦⎤
→R1+12R3[004∣281−2−3∣00−1−1∣3]\qquad \qquad \qquad \qquad \qquad \quad\overset{R_1+12R_3}{\rightarrow} \begin{bmatrix} 0&0&4&|&28 \\ 1&-2&-3&|&0 \\ 0&-1&-1&|&3 \end{bmatrix}→R1+12R3⎣⎡0100−2−14−3−1∣∣∣2803⎦⎤
→R2−2R3[004∣2810−1∣−60−1−1∣3]\qquad \qquad \qquad \qquad \qquad \quad\overset{R_2-2R_3}{\rightarrow} \begin{bmatrix} 0&0&4&|&28 \\ 1&0&-1&|&-6 \\ 0&-1&-1&|&3 \end{bmatrix}→R2−2R3⎣⎡01000−14−1−1∣∣∣28−63⎦⎤
→R1/4[001∣710−1∣−60−1−1∣3]\qquad \qquad \qquad \qquad \qquad \quad\overset{R_1/4}{\rightarrow} \begin{bmatrix} 0&0&1&|&7 \\ 1&0&-1&|&-6 \\ 0&-1&-1&|&3 \end{bmatrix}→R1/4⎣⎡01000−11−1−1∣∣∣7−63⎦⎤
→R2+R1[001∣7100∣10−1−1∣3]\qquad \qquad \qquad \qquad \qquad \quad\overset{R_2+R_1}{\rightarrow} \begin{bmatrix} 0&0&1&|&7 \\ 1&0&0&|&1 \\ 0&-1&-1&|&3 \end{bmatrix}→R2+R1⎣⎡01000−110−1∣∣∣713⎦⎤
→R3+R1[001∣7100∣10−10∣10]\qquad \qquad \qquad \qquad \qquad \quad\overset{R_3+R_1}{\rightarrow} \begin{bmatrix} 0&0&1&|&7 \\ 1&0&0&|&1 \\ 0&-1&0&|&10 \end{bmatrix}→R3+R1⎣⎡01000−1100∣∣∣7110⎦⎤
→−R3[001∣7100∣1010∣−10]\qquad \qquad \qquad \qquad \qquad \quad\overset{-R_3}{\rightarrow} \begin{bmatrix} 0&0&1&|&7 \\ 1&0&0&|&1 \\ 0&1&0&|&-10 \end{bmatrix}→−R3⎣⎡010001100∣∣∣71−10⎦⎤
Answer: x=1, y=−10, z=7.x=1,\ \ y=-10,\ \ z=7.x=1, y=−10, z=7.
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