Question #205547

Solve by gaussian method

X+2Y-3Y=11

3X+2Y+Z=1

2X+Y-5Z=11


1
Expert's answer
2021-06-11T03:02:55-0400

Augmented matrix


(12311321121511)\begin{pmatrix} 1 & 2 & -3 & & 11 \\ 3 & 2 & 1 & & 1 \\ 2 & 1 & -5 & & 11 \\ \end{pmatrix}



R2=R23R1R_2=R_2-3R_1


(1231104103221511)\begin{pmatrix} 1 & 2 & -3 & & 11 \\ 0 & -4 & 10 & & -32 \\ 2 & 1 & -5 & & 11 \\ \end{pmatrix}


R3=R32R1R_3=R_3-2R_1


(1231104103203111)\begin{pmatrix} 1 & 2 & -3 & & 11 \\ 0 & -4 & 10 & & -32 \\ 0 & -3 & 1 & & -11 \\ \end{pmatrix}



R2=R2/4R_2=-R_2/4


(12311015/2803111)\begin{pmatrix} 1 & 2 & -3 & & 11 \\ 0 & 1 & -5/2 & & 8 \\ 0 & -3 & 1 & & -11 \\ \end{pmatrix}

R1=R12R2R_1=R_1-2R_2


(1025015/2803111)\begin{pmatrix} 1 & 0 & 2 & & -5 \\ 0 & 1 & -5/2 & & 8 \\ 0 & -3 & 1 & & -11 \\ \end{pmatrix}

R3=R3+3R2R_3=R_3+3R_2


(1025015/280013/213)\begin{pmatrix} 1 & 0 & 2 & & -5 \\ 0 & 1 & -5/2 & & 8 \\ 0 & 0 & -13/2 & & 13 \\ \end{pmatrix}

R3=(2/13)R3R_3=-(2/13)R_3


(1025015/280012)\begin{pmatrix} 1 & 0 & 2 & & -5 \\ 0 & 1 & -5/2 & & 8 \\ 0 & 0 & 1 & & -2\\ \end{pmatrix}

R1=R12R3R_1=R_1-2R_3


(1001015/280012)\begin{pmatrix} 1 & 0 & 0 & & -1 \\ 0 & 1 & -5/2 & & 8 \\ 0 & 0 & 1 & & -2\\ \end{pmatrix}

R2=R2+(5/2)R3R_2=R_2+(5/2)R_3


(100101030012)\begin{pmatrix} 1 & 0 & 0 & & -1 \\ 0 & 1 & 0 & & 3 \\ 0 & 0 & 1 & & -2\\ \end{pmatrix}

Then


x=1,y=3,z=2x=-1, y=3, z=-2

(1,3,2)(-1, 3, -2)



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