Math Problem Statement

Solve the system of linear equations using elementary row operations on an augmented matrix. This problem will have you enter intermediate steps of the solution.

⎧⎪⎨⎪⎩4x+5y−2z=−1y+2z=−2x+y−z=1{4x+5y-2z=-1y+2z=-2x+y-z=1

(a) Fill in the corresponding augmented matrix. The rightmost column represents the constant values from the linear equations.    

(b) Use elementary row operations on the augmented matrix in part (a) to make a matrix of the form

⎡⎢⎣1abc01de001f⎤⎥⎦[1abc01de001f], ⎡⎢⎣1abc01de0001⎤⎥⎦[1abc01de0001], ⎡⎢⎣1abc001d0000⎤⎥⎦[1abc001d0000], or ⎡⎢⎣1abc000d0000⎤⎥⎦[1abc000d0000].*

     

(c) Use elementary row operations on the augmented matrix in part (b) to make a matrix of the form

⎡⎢⎣100a010b001c⎤⎥⎦[100a010b001c], ⎡⎢⎣10a001c00001⎤⎥⎦[10a001c00001], or ⎡⎢⎣1a0b001c0000⎤⎥⎦[1a0b001c0000], or ⎡⎢⎣1ab000010000⎤⎥⎦[1ab000010000].      

(d) Using the matrix from in part (c), determine the solution of the corresponding system of linear equations. If the system has no solution or is inconsistent, enter DNE for all answers. If there are infinite number of solutions, enter zz for zz and find xx and yy in terms of zz. Enter only integer numbers or reduced fractions.

x=x=   

y=y=   

z=z=     

Solution

Let's solve this system step-by-step using elementary row operations on the augmented matrix. The given system of equations is:

1) \ & 4x + 5y - 2z = -1 \\ 2) \ & y + 2z = -2 \\ 3) \ & x + y - z = 1 \end{aligned}$$ ### (a) Augmented Matrix Representation We can express the system of equations in augmented matrix form, where the rightmost column contains the constants: $$\begin{bmatrix} 4 & 5 & -2 & | & -1 \\ 0 & 1 & 2 & | & -2 \\ 1 & 1 & -1 & | & 1 \end{bmatrix}$$ ### (b) Row Reduction to Row-Echelon Form We will use elementary row operations to reduce the matrix to an upper triangular form. #### Step 1: Make the first pivot element equal to 1. We can divide Row 3 (since it has a leading 1) by 1 and swap Row 1 with Row 3: $$R_1 \leftrightarrow R_3$$ This gives: $$\begin{bmatrix} 1 & 1 & -1 & | & 1 \\ 0 & 1 & 2 & | & -2 \\ 4 & 5 & -2 & | & -1 \end{bmatrix}$$ #### Step 2: Eliminate the first column below the pivot. We'll now eliminate the 4 in the third row's first column by performing: $$R_3 \leftarrow R_3 - 4R_1$$ This gives: $$\begin{bmatrix} 1 & 1 & -1 & | & 1 \\ 0 & 1 & 2 & | & -2 \\ 0 & 1 & 2 & | & -5 \end{bmatrix}$$ #### Step 3: Eliminate the second row's first column. Perform: $$R_2 \leftarrow R_2 - R_1$$ This gives: $$\begin{bmatrix} 1 & 1 & -1 & | & 1 \\ 0 & 0 & 3 & | & -3 \\ 0 & 1 & 2 & | & -5 \end{bmatrix}$$ We now have the matrix in an upper triangular form. ### (c) Further Row Reduction to Reduced Row-Echelon Form Now, we continue the row operations to get the matrix in reduced row-echelon form. #### Step 4: Make the second pivot element equal to 1. We'll divide Row 3 by 1: $$R_3 \leftarrow \frac{1}{1} R_3$$ This gives: $$\begin{bmatrix} 1 & 1 & -1 & | & 1 \\ 0 & 0 & 3 & | & -3 \\ 0 & 1 & 2 & | & -5 \end{bmatrix}$$ #### Step 5: Eliminate the third column above the pivot. We'll eliminate the entries above the pivot in Row 2 by: $$R_1 \leftarrow R_1 - 1R_3$$ This gives: \[ \begin{bmatrix} 1 &

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Linear Equations
Elementary Row Operations
Matrix Representation

Formulas

Row operations to perform Gaussian elimination
Matrix form of a system of equations

Theorems

Gaussian Elimination
Row-Echelon Form
Reduced Row-Echelon Form

Suitable Grade Level

Grades 11-12 or College Level

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