Math  /  Algebra

QuestionUse Gaussian elimination to find the complete solution to the system of equations, or show that none exists. {5x+6y+4z=112x5y4z=8x+yz=10012\left.\begin{array}{c} \left\{\begin{array}{rr} 5 x+6 y+4 z= & -11 \\ 2 x-5 y-4 z= & -8 \\ x+y-z= & 1 \end{array}\right. \\ \left|\begin{array}{lll} 0 & 0 & 1 \end{array}\right|-2 \end{array} \right\rvert\,
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. There is one solution. The solution set is {(,,D)}\{(\square, \square, D)\}. (Simplify your answers. Use integers or fractions for any numbers in the expressions.) B. There are infinitely many solutions. The solution set is {(,,z)}\{(,, z)\}, where zz is any real number. (Type expressions using zz as the variable. Use integers or fractions for any numbers in the expressions.) C. There is no solution. The solution set is \varnothing.

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables: xx, yy, and zz.
2. We need to use Gaussian elimination to find the solution.
3. Gaussian elimination involves transforming the system into an upper triangular matrix and then performing back substitution.
4. The system may have a unique solution, infinitely many solutions, or no solution.

STEP 2

1. Write the augmented matrix for the system of equations.
2. Use row operations to transform the matrix into row-echelon form.
3. Determine if the system has a unique solution, infinitely many solutions, or no solution.
4. If a solution exists, perform back substitution to find the complete solution.

STEP 3

Write the augmented matrix for the system of equations:
[5641125481111]\begin{bmatrix} 5 & 6 & 4 & | & -11 \\ 2 & -5 & -4 & | & -8 \\ 1 & 1 & -1 & | & 1 \end{bmatrix}

STEP 4

Use row operations to transform the matrix into row-echelon form. Start by making the first element of the second row zero using the first row:
Subtract 25\frac{2}{5} times the first row from the second row:
R2=R225R1R_2 = R_2 - \frac{2}{5}R_1
New matrix:
[5641103752852751111]\begin{bmatrix} 5 & 6 & 4 & | & -11 \\ 0 & -\frac{37}{5} & -\frac{28}{5} & | & -\frac{27}{5} \\ 1 & 1 & -1 & | & 1 \end{bmatrix}

STEP 5

Make the first element of the third row zero using the first row:
Subtract 15\frac{1}{5} times the first row from the third row:
R3=R315R1R_3 = R_3 - \frac{1}{5}R_1
New matrix:
[56411037528527501595165]\begin{bmatrix} 5 & 6 & 4 & | & -11 \\ 0 & -\frac{37}{5} & -\frac{28}{5} & | & -\frac{27}{5} \\ 0 & \frac{1}{5} & -\frac{9}{5} & | & \frac{16}{5} \end{bmatrix}

STEP 6

Make the second element of the third row zero using the second row:
Multiply the second row by 137-\frac{1}{37} and add to the third row:
R3=R3+137R2R_3 = R_3 + \frac{1}{37}R_2
New matrix:
[56411037528527500613713337]\begin{bmatrix} 5 & 6 & 4 & | & -11 \\ 0 & -\frac{37}{5} & -\frac{28}{5} & | & -\frac{27}{5} \\ 0 & 0 & -\frac{61}{37} & | & \frac{133}{37} \end{bmatrix}

STEP 7

Examine the row-echelon form of the matrix to determine the nature of the solutions:
Since the last row corresponds to the equation:
0x+0y6137z=133370x + 0y - \frac{61}{37}z = \frac{133}{37}
This simplifies to:
z=13361z = -\frac{133}{61}
The system has a unique solution.

STEP 8

Perform back substitution to find the values of xx and yy:
From the second row:
375y285(13361)=275-\frac{37}{5}y - \frac{28}{5}\left(-\frac{133}{61}\right) = -\frac{27}{5}
Solve for yy:
y=137(275+28×1335×61)y = \frac{1}{37} \left( -\frac{27}{5} + \frac{28 \times 133}{5 \times 61} \right)
From the first row:
5x+6y+4(13361)=115x + 6y + 4\left(-\frac{133}{61}\right) = -11
Solve for xx:
x=15(116y4(13361))x = \frac{1}{5} \left( -11 - 6y - 4\left(-\frac{133}{61}\right) \right)
The solution set is:
A. There is one solution. The solution set is {(x,y,13361)}\{(x, y, -\frac{133}{61})\}.

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