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Math  /  Algebra

Question12x−12y+16z=446x+2y−6z=−1075x−5y−20z=0\begin{aligned} 12 x-12 y+16 z & =44 \\ 6 x+2 y-6 z & =-10 \\ 75 x-5 y-20 z & =0 \end{aligned}
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. There is one solution. The solution is x=x= □\square □\square , y=y= (Simplify your answer.) and z=z= □\square . B. The system is dependent. There are infinitely many solutions. C. The system is inconsistent. There is no solution.

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables xx, yy, and zz.
2. We will use methods such as substitution, elimination, or matrix operations to solve the system.
3. The system can have one solution, infinitely many solutions, or no solution.

STEP 2

1. Write the system of equations in matrix form.
2. Use Gaussian elimination to simplify the matrix to row-echelon form.
3. Determine the type of solution: unique, infinite, or none.
4. Solve for the variables if there is a unique solution.

STEP 3

Write the system of equations in matrix form. The augmented matrix for the system is:
[12−1216∣4462−6∣−1075−5−20∣0]\begin{bmatrix} 12 & -12 & 16 & | & 44 \\ 6 & 2 & -6 & | & -10 \\ 75 & -5 & -20 & | & 0 \end{bmatrix}

STEP 4

Use Gaussian elimination to simplify the matrix to row-echelon form. We start by making the first entry of the second row zero by replacing the second row with the second row minus half of the first row:
R2→R2−12R1R_2 \rightarrow R_2 - \frac{1}{2}R_1
[12−1216∣4408−14∣−3275−5−20∣0]\begin{bmatrix} 12 & -12 & 16 & | & 44 \\ 0 & 8 & -14 & | & -32 \\ 75 & -5 & -20 & | & 0 \end{bmatrix}
Next, make the first entry of the third row zero by replacing the third row with the third row minus 7512\frac{75}{12} times the first row:
R3→R3−7512R1R_3 \rightarrow R_3 - \frac{75}{12}R_1
[12−1216∣4408−14∣−32070−120∣−275]\begin{bmatrix} 12 & -12 & 16 & | & 44 \\ 0 & 8 & -14 & | & -32 \\ 0 & 70 & -120 & | & -275 \end{bmatrix}

STEP 5

Continue with Gaussian elimination by making the second entry of the third row zero by replacing the third row with the third row minus 708\frac{70}{8} times the second row:
R3→R3−708R2R_3 \rightarrow R_3 - \frac{70}{8}R_2
[12−1216∣4408−14∣−32005∣−5]\begin{bmatrix} 12 & -12 & 16 & | & 44 \\ 0 & 8 & -14 & | & -32 \\ 0 & 0 & 5 & | & -5 \end{bmatrix}

STEP 6

Determine the type of solution. The matrix is in row-echelon form and does not have any rows that are all zeros with a non-zero entry in the augmented column, indicating that the system is consistent and has a unique solution.

STEP 7

Solve for the variables using back substitution:
From the third row: 5z=−5 5z = -5 z=−1 z = -1
Substitute z=−1 z = -1 into the second row: 8y−14(−1)=−32 8y - 14(-1) = -32 8y+14=−32 8y + 14 = -32 8y=−46 8y = -46 y=−468=−234 y = -\frac{46}{8} = -\frac{23}{4}
Substitute y=−234 y = -\frac{23}{4} and z=−1 z = -1 into the first row: 12x−12(−234)+16(−1)=44 12x - 12\left(-\frac{23}{4}\right) + 16(-1) = 44 12x+69−16=44 12x + 69 - 16 = 44 12x+53=44 12x + 53 = 44 12x=−9 12x = -9 x=−912=−34 x = -\frac{9}{12} = -\frac{3}{4}
The solution is: x=−34,y=−234,z=−1 x = -\frac{3}{4}, \quad y = -\frac{23}{4}, \quad z = -1
The correct choice is A. There is one solution. The solution is x=−34 x = -\frac{3}{4} , y=−234 y = -\frac{23}{4} , z=−1 z = -1 .

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