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.

⎧ ⎪ ⎨ ⎪ ⎩ 3 x + 4 y − 7 z

10 x + y − 3 z

3 x + 2 y − 2 z

3

(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

⎡ ⎢ ⎣ 1 a b c 0 1 d e 0 0 1 f ⎤ ⎥ ⎦ ,
⎡ ⎢ ⎣ 1 a b c 0 1 d e 0 0 0 1 ⎤ ⎥ ⎦ ,
⎡ ⎢ ⎣ 1 a b c 0 0 1 d 0 0 0 0 ⎤ ⎥ ⎦ , or
⎡ ⎢ ⎣ 1 a b c 0 0 0 d 0 0 0 0 ⎤ ⎥ ⎦ .*

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

⎡ ⎢ ⎣ 1 0 0 a 0 1 0 b 0 0 1 c ⎤ ⎥ ⎦ ,
⎡ ⎢ ⎣ 1 0 a 0 0 1 c 0 0 0 0 1 ⎤ ⎥ ⎦ , or
⎡ ⎢ ⎣ 1 a 0 b 0 0 1 c 0 0 0 0 ⎤ ⎥ ⎦ , or
⎡ ⎢ ⎣ 1 a b 0 0 0 0 1 0 0 0 0 ⎤ ⎥ ⎦ .

(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
z for
z and find
x and
y in terms of
z . Enter only integer numbers or reduced fractions.

x

y

z

Solution

Let's walk through this problem step by step:

(a) Write the augmented matrix

We start with the system of linear equations:

3x + 4y - 7z &= 10 \\ x + y - 3z &= 3 \\ x + 2y - 2z &= 3 \end{aligned}$$ The augmented matrix corresponding to this system is: $$\left[ \begin{array}{ccc|c} 3 & 4 & -7 & 10 \\ 1 & 1 & -3 & 3 \\ 1 & 2 & -2 & 3 \\ \end{array} \right]$$ --- ### **(b) Perform row operations to get an upper triangular matrix** We aim to perform row operations to get the matrix in upper triangular form (preferably with leading 1s). #### Step 1: Make the pivot in the first column equal to 1. We can do this by swapping Row 1 with Row 2: $$\left[ \begin{array}{ccc|c} 1 & 1 & -3 & 3 \\ 3 & 4 & -7 & 10 \\ 1 & 2 & -2 & 3 \\ \end{array} \right]$$ #### Step 2: Eliminate the first entry in Row 2 and Row 3. We can subtract $$3 \times \text{Row 1}$$ from Row 2 and subtract Row 1 from Row 3. - Row 2: $$\text{Row 2} - 3 \times \text{Row 1}$$ $$(3, 4, -7, 10) - 3 \times (1, 1, -3, 3) = (0, 1, 2, 1)$$ - Row 3: $$\text{Row 3} - \text{Row 1}$$ $$(1, 2, -2, 3) - (1, 1, -3, 3) = (0, 1, 1, 0)$$ So the updated matrix is: $$\left[ \begin{array}{ccc|c} 1 & 1 & -3 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 1 & 1 & 0 \\ \end{array} \right]$$ #### Step 3: Eliminate the second entry in Row 3. We subtract Row 2 from Row 3: - Row 3: $$\text{Row 3} - \text{Row 2}$$ $$(0, 1, 1, 0) - (0, 1, 2, 1) = (0, 0, -1, -1)$$ The matrix now looks like: $$\left[ \begin{array}{ccc|c} 1 & 1 & -3 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & -1 & -1 \\ \end{array} \right]$$ Now we have the matrix in upper triangular form. --- ### **(c) Use row operations to make the matrix diagonal** #### Step 1: Make the pivot in the third row a positive 1 by multiplying Row 3 by $$-1$$: $$\left[ \begin{array}{ccc|c} 1 & 1 & -3 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \\ \end{array} \right]$$ #### Step 2: Eliminate the third entry in Row 2. We subtract $$2 \times \text{Row 3}$$ from Row 2: - Row 2: $$\text{Row 2} - 2 \times \text{Row 3}$$ $$(0, 1, 2, 1) - 2 \times (0, 0, 1, 1) = (0, 1, 0, -1)$$ #### Step 3: Eliminate the third entry in Row 1. We add $$3 \times \text{Row 3}$$ to Row 1: - Row 1: $$\text{Row 1} + 3 \times \text{Row 3}$$ $$(1, 1, -3, 3) + 3 \times (0, 0, 1, 1) = (1, 1, 0, 6)$$ Now the matrix looks like this: $$\left[ \begin{array}{ccc|c} 1 & 1 & 0 & 6 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ \end{array} \right]$$ #### Step 4: Eliminate the second entry in Row 1. We subtract Row 2 from Row 1: - Row 1: $$\text{Row 1} - \text{Row 2}$$ $$(1, 1, 0, 6) - (0, 1, 0, -1) = (1, 0, 0, 7)$$ The final matrix is: $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 7 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ \end{array} \right]$$ --- ### **(d) Solution of the system** From the final matrix, we can directly read off the solutions: $$x = 7, \quad y = -1, \quad z = 1$$ --- ### Summary of Solutions: $$x = 7, \quad y = -1, \quad z = 1$$ --- Would you like more details on any part of the solution? Here are 5 related questions you could explore: 1. How can we verify the solution to a system of equations? 2. How do we solve a system of linear equations using Cramer's rule? 3. What are the differences between Gaussian elimination and Gauss-Jordan elimination? 4. How do row operations affect the determinant of a matrix? 5. Can a system of equations have no solution or infinitely many solutions? How do we detect this? **Tip:** Always check your solution by substituting the values back into the original system of equations!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Equations
Elementary Row Operations

Formulas

Row operations: swapping rows, multiplying a row by a constant, adding/subtracting rows

Theorems

Gaussian Elimination
Gauss-Jordan Elimination

Suitable Grade Level

Grades 10-12 or College-level