Question #272505

5x +2y +z =-8


x -2y -3z =0


-x +y +2z =3


Solved this problem by using Gauess Gordan method.


1
Expert's answer
2021-11-30T13:37:06-0500

{5x+2y+z=8x2y3z=0x+y+2z=3\begin{cases} 5x+2y+z=-8 \\ x-2y-3z=0 \\ -x+y+2z=3 \end{cases}


[521812301123]R3+R2[521812300113]\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}


R15R2[01216812300113]\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+12R3[0042812300113]\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}


R22R3[0042810160113]\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}


R1/4[001710160113]\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}


R2+R1[001710010113]\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}


R3+R1[0017100101010]\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[0017100101010]\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}


Answer: x=1,  y=10,  z=7.x=1,\ \ y=-10,\ \ z=7.


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