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Math  /  Algebra

QuestionX incorrect.
Solve the system of linear equations. 2y+z−x=622x+3z−2y=0−2z+y−4x=44\begin{array}{l} 2 y+z-x=62 \\ 2 x+3 z-2 y=0 \\ -2 z+y-4 x=44 \end{array}
If there are an infinite number of solutions, enter IS in each answer field. If there are no solutions, enter NS in each answer huchit. x=−105.272y=4z=42545z=4.818\begin{array}{l} x=-105.272 \\ y=4 \\ z=42545 \\ z=4.818 \end{array}

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables xx, yy, and zz.
2. The goal is to find the values of xx, yy, and zz that satisfy all three equations simultaneously.
3. We will use methods such as substitution, elimination, or matrix operations (e.g., Gaussian elimination) to solve the system of equations.

STEP 2

1. Write the system of equations in matrix form.
2. Convert the matrix to row echelon form using Gaussian elimination.
3. Solve for the variables xx, yy, and zz by back-substitution if necessary.
4. Check for consistency and uniqueness of solutions.
5. Interpret the results to determine if there are unique solutions, infinite solutions (IS), or no solutions (NS).

STEP 3

Write the system of equations in matrix form.
(−1212−23−41−2)(xyz)=(62044) \begin{pmatrix} -1 & 2 & 1 \\ 2 & -2 & 3 \\ -4 & 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 62 \\ 0 \\ 44 \end{pmatrix}

STEP 4

Start Gaussian elimination by transforming the matrix to row echelon form. Begin by swapping rows if necessary to have a nonzero pivot in the first row.
R1↔R2 R_1 \leftrightarrow R_2
(2−23−121−41−2)(xyz)=(06244) \begin{pmatrix} 2 & -2 & 3 \\ -1 & 2 & 1 \\ -4 & 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 62 \\ 44 \end{pmatrix}

STEP 5

Eliminate xx from R2R_2 and R3R_3 by adding appropriate multiples of R1R_1 to R2R_2 and R3R_3.
R2→R2+12R1 R_2 \rightarrow R_2 + \frac{1}{2}R_1
R3→R3+2R1 R_3 \rightarrow R_3 + 2R_1
(2−23012.50−34)(xyz)=(06244) \begin{pmatrix} 2 & -2 & 3 \\ 0 & 1 & 2.5 \\ 0 & -3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 62 \\ 44 \end{pmatrix}

STEP 6

Eliminate yy from R3R_3 by adding 3R23R_2 to R3R_3.
R3→R3+3R2 R_3 \rightarrow R_3 + 3R_2
(2−23012.50011.5)(xyz)=(062230) \begin{pmatrix} 2 & -2 & 3 \\ 0 & 1 & 2.5 \\ 0 & 0 & 11.5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 62 \\ 230 \end{pmatrix}

STEP 7

Solve for zz from R3R_3.
11.5z=230  ⟹  z=23011.5=20 11.5z = 230 \implies z = \frac{230}{11.5} = 20

STEP 8

Substitute z=20z = 20 into R2R_2 to solve for yy.
y+2.5×20=62  ⟹  y+50=62  ⟹  y=12 y + 2.5 \times 20 = 62 \implies y + 50 = 62 \implies y = 12

STEP 9

Substitute y=12y = 12 and z=20z = 20 into R1R_1 to solve for xx.
2x−2×12+3×20=0  ⟹  2x−24+60=0  ⟹  2x+36=0  ⟹  x=−18 2x - 2 \times 12 + 3 \times 20 = 0 \implies 2x - 24 + 60 = 0 \implies 2x + 36 = 0 \implies x = -18

STEP 10

Verify the solution by substituting x=−18x=-18, y=12y=12, and z=20z=20 back into the original equations.
1. 2y+z−x=2×12+20−(−18)=24+20+18=622y + z - x = 2 \times 12 + 20 - (-18) = 24 + 20 + 18 = 62 (satisfied)
2. 2x+3z−2y=2×(−18)+3×20−2×12=−36+60−24=02x + 3z - 2y = 2 \times (-18) + 3 \times 20 - 2 \times 12 = -36 + 60 - 24 = 0 (satisfied)
3. −2z+y−4x=−2×20+12−4×(−18)=−40+12+72=44-2z + y - 4x = -2 \times 20 + 12 - 4 \times (-18) = -40 + 12 + 72 = 44 (satisfied)
All three equations are satisfied.

STEP 11

Since all equations are satisfied, the solution is unique.
x=−18,y=12,z=20 \boxed{x = -18, y = 12, z = 20}

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