Question #258113

Solve the following system of linear equations for ๐‘ฅ, ๐‘ฆ, ๐‘ง, ๐‘ก: โˆ’๐‘ฆ+3๐‘ง+2๐‘ฅ+4๐‘ก = 9

๐‘ฅโˆ’2๐‘ง+7๐‘ก =11 ๐‘งโˆ’3๐‘ฆ+3๐‘ฅ+5๐‘ก = 8 ๐‘ฆ + 2๐‘ฅ + 4๐‘ง + 4๐‘ก = 10


1
Expert's answer
2021-10-29T03:01:54-0400
2xโˆ’y+3z+4t=9xโˆ’2z+7t=113xโˆ’3y+z+5t=82x+y+4z+4t=10\begin{matrix} 2x-y+3z+4t=9 \\ x-2z+7t=11\\ 3x-3y+z+5t=8\\ 2x+y+4z+4t=10 \end{matrix}

Augmented matrix


(2โˆ’134910โˆ’27113โˆ’3158214410)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 1 & 0 & -2 & 7 & & 11 \\ 3 & -3 & 1 & 5 & & 8 \\ 2 & 1 & 4 & 4 & & 10 \\ \end{pmatrix}

R2=R2โˆ’R1/2R_2=R_2-R_1/2


(2โˆ’134901/2โˆ’7/2513/23โˆ’3158214410)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 3 & -3 & 1 & 5 & & 8 \\ 2 & 1 & 4 & 4 & & 10 \\ \end{pmatrix}

R3=R3โˆ’3R1/2R_3=R_3-3R_1/2


(2โˆ’134901/2โˆ’7/2513/20โˆ’3/2โˆ’7/2โˆ’1โˆ’11/2214410)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & -3/2 & -7/2 & -1 & & -11/2 \\ 2 & 1 & 4 & 4 & & 10 \\ \end{pmatrix}

R4=R4โˆ’R1R_4=R_4-R_1


(2โˆ’134901/2โˆ’7/2513/20โˆ’3/2โˆ’7/2โˆ’1โˆ’11/202101)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & -3/2 & -7/2 & -1 & & -11/2 \\ 0 & 2 & 1 & 0 & & 1 \\ \end{pmatrix}

R3=R3+3R2R_3=R_3+3R_2


(2โˆ’134901/2โˆ’7/2513/200โˆ’14141402101)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 14 & & 14 \\ 0 & 2 & 1 & 0 & & 1 \\ \end{pmatrix}

R4=R4โˆ’4R2R_4=R_4-4R_2


(2โˆ’134901/2โˆ’7/2513/200โˆ’1414140015โˆ’20โˆ’25)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 14 & & 14 \\ 0 & 0 & 15 & -20 & & -25 \\ \end{pmatrix}

R4=R4+15R3/14R_4=R_4+15R_3/14


(2โˆ’134901/2โˆ’7/2513/200โˆ’141414000โˆ’5โˆ’10)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 14 & & 14 \\ 0 & 0 & 0 & -5 & & -10 \\ \end{pmatrix}

R4=R4/(โˆ’5)R_4=R_4/(-5)


(2โˆ’134901/2โˆ’7/2513/200โˆ’14141400012)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 14 & & 14 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R3=R3โˆ’14R4R_3=R_3-14R_4


(2โˆ’134901/2โˆ’7/2513/200โˆ’140โˆ’1400012)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 0 & & -14 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R3=R3/(โˆ’14)R_3=R_3/(-14)


(2โˆ’134901/2โˆ’7/2513/20010100012)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R2=R2โˆ’5R4R_2=R_2-5R_4


(2โˆ’134901/2โˆ’7/20โˆ’7/20010100012)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 0 & & -7/2 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R2=R2+7R3/2R_2=R_2+7R_3/2


(2โˆ’134901/20000010100012)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R2=2R2R_2=2R_2


(2โˆ’1349010000010100012)\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R1=R1โˆ’4R4R_1=R_1-4R_4


(2โˆ’1301010000010100012)\begin{pmatrix} 2 & -1 & 3 & 0 & & 1 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R1=R1โˆ’3R3R_1=R_1-3R_3


(2โˆ’100โˆ’2010000010100012)\begin{pmatrix} 2 & -1 & 0 & 0 & & -2 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R1=R1+R2R_1=R_1+R_2


(2000โˆ’2010000010100012)\begin{pmatrix} 2 & 0 & 0 & 0 & & -2 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

R1=R1/2R_1=R_1/2


(1000โˆ’1010000010100012)\begin{pmatrix} 1 & 0 & 0 & 0 & & -1 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}

The solution is


x=โˆ’1,y=0,z=1,t=2x=-1, y=0, z=1, t=2

(x,y,z,t)=(โˆ’1,0,1,2)(x,y,z,t)=(-1,0,1,2)


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