use gaussian elimination to find the complete solution to the system of equations, or show that none…

use gaussian elimination to find the complete solution to the system of equations, or show that none exists.\n\\( \\left\\{ \\begin{array} { r } { x + 3 y - 2 z - w = 7 } \\\\ { 4 x + y + 2 z + 4 w = 3 } \\\\ { - 3 x - y + z - 2 w = - 8 } \\\\ { x - y - 3 z - 5 w = - 2 } \\end{array} \\right. \\)\nselect the correct choice below and, if necessary, fill in the answer boxes to complete your choice.\na. there is one solution. the solution set is \\( \\{ ( \\square, \\square, \\square, \\square ) \\} \\).\n(simplify your answers.)\nb. there are infinitely many solutions. the solution set is \\( \\{ ( \\square, \\square, \\square, w ) \\} \\), where \\( w \\) is any real number.\n(simplify your answers. type expressions using \\( w \\) as the variable. use integers or fractions for any numbers in the expression.)\nc. there is no solution. the solution set is \\( \\varnothing \\).

use gaussian elimination to find the complete solution to the system of equations, or show that none exists.\n\\( \\left\\{ \\begin{array} { r } { x + 3 y - 2 z - w = 7 } \\\\ { 4 x + y + 2 z + 4 w = 3 } \\\\ { - 3 x - y + z - 2 w = - 8 } \\\\ { x - y - 3 z - 5 w = - 2 } \\end{array} \\right. \\)\nselect the correct choice below and, if necessary, fill in the answer boxes to complete your choice.\na. there is one solution. the solution set is \\( \\{ ( \\square, \\square, \\square, \\square ) \\} \\).\n(simplify your answers.)\nb. there are infinitely many solutions. the solution set is \\( \\{ ( \\square, \\square, \\square, w ) \\} \\), where \\( w \\) is any real number.\n(simplify your answers. type expressions using \\( w \\) as the variable. use integers or fractions for any numbers in the expression.)\nc. there is no solution. the solution set is \\( \\varnothing \\).

Answer

Explanation:

Step1: Write the augmented matrix

$$ \left[\begin{array}{cccc|c} 1&3&- 2&-1&7\ 4&1&2&4&3\ -3&-1&1&-2&-8\ 1&-1&-3&-5&-2 \end{array}\right] $$

Step2: Eliminate (x) from the second, third and fourth rows

  • (R_2=R_2 - 4R_1): $$ \left[\begin{array}{cccc|c} 1&3&-2&-1&7\ 0&-11&10&8&-25\ -3&-1&1&-2&-8\ 1&-1&-3&-5&-2 \end{array}\right] $$
  • (R_3=R_3 + 3R_1): $$ \left[\begin{array}{cccc|c} 1&3&-2&-1&7\ 0&-11&10&8&-25\ 0&8&-5&-5&13\ 1&-1&-3&-5&-2 \end{array}\right] $$
  • (R_4=R_4 - R_1): $$ \left[\begin{array}{cccc|c} 1&3&-2&-1&7\ 0&-11&10&8&-25\ 0&8&-5&-5&13\ 0&-4&-1&-4&-9 \end{array}\right] $$

Step3: Eliminate (y) from the third and fourth rows

  • (R_3=\frac{8}{11}R_2+R_3): $$ R_2=\left[\begin{array}{cccc|c} 1&3&-2&-1&7\ 0&-11&10&8&-25\ 0&0&\frac{25}{11}&\frac{9}{11}&-\frac{61}{11}\ 0&-4&-1&-4&-9 \end{array}\right] $$
  • (R_4 =-\frac{4}{11}R_2+R_4): $$ \left[\begin{array}{cccc|c} 1&3&-2&-1&7\ 0&-11&10&8&-25\ 0&0&\frac{25}{11}&\frac{9}{11}&-\frac{61}{11}\ 0&0&-\frac{51}{11}&-\frac{12}{11}&-\frac{139}{11} \end{array}\right] $$

Step4: Eliminate (z) from the fourth row

  • (R_4=\frac{51}{25}R_3+R_4): $$ \left[\begin{array}{cccc|c} 1&3&-2&-1&7\ 0&-11&10&8&-25\ 0&0&\frac{25}{11}&\frac{9}{11}&-\frac{61}{11}\ 0&0&0&\frac{3}{25}&-\frac{3}{25} \end{array}\right] $$

Step5: Back - substitute

From ( \frac{3}{25}w=-\frac{3}{25}), we get (w=- 1) From (\frac{25}{11}z+\frac{9}{11}w=-\frac{61}{11}), substitute (w =-1): (\frac{25}{11}z-\frac{9}{11}=-\frac{61}{11}), then (z=- \frac{52}{25}) From (-11y + 10z+8w=-25), substitute (z =-\frac{52}{25}) and (w=-1): (-11y+10\times(-\frac{52}{25})+8\times(-1)=-25) (-11y-\frac{104}{5}-8=-25) (-11y=-\frac{21}{5}) (y=\frac{21}{55}) From (x + 3y-2z-w=7), substitute (y=\frac{21}{55}), (z =-\frac{52}{25}) and (w=-1): (x+3\times\frac{21}{55}-2\times(-\frac{52}{25})-(-1)=7) (x+\frac{63}{55}+\frac{104}{25}+1 = 7) (x=7 - 1-\frac{63}{55}-\frac{104}{25}) (x=\frac{8750 - 875- 315 - 2288}{1375}=\frac{5272}{1375})

Answer:

A. There is one solution. The solution set is (\left{\left(\frac{5272}{1375},\frac{21}{55},-\frac{52}{25},-1\right)\right})