Solved on Nov 29, 2023

Find the intersection point of 3 one-way streets using Gaussian elimination to solve the system of $x+9=y+13$ , $z+15=x+10$ , $y+23=z+24$ .

STEP 1

Assumptions
1. The system of equations is given by: $\begin{array}{l} x+9=y+13 \\ z+15=x+10 \\ y+23=z+24 \end{array}$
2. We are to solve this system of equations using Gaussian elimination.

STEP 2

First, we need to rewrite the system of equations in matrix form. This will make it easier to apply Gaussian elimination.
$\begin{bmatrix} 1 & -1 & 0 & 4 \\ -1 & 0 & 1 & 5 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 3

The next step is to apply Gaussian elimination. We want to transform the matrix into row echelon form. This means we want to have a leading 1 (also known as a pivot) in each row, and for each pivot to be to the right of the pivot in the row above it.
First, we will swap the first and second rows to get a pivot in the first row.
$\begin{bmatrix} -1 & 0 & 1 & 5 \\ 1 & -1 & 0 & 4 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 4

Next, we will multiply the first row by -1 to get a positive pivot.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 1 & -1 & 0 & 4 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 5

Now, we will subtract the first row from the second row to get a zero in the second row, first column.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & -1 & 1 & 9 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 6

We will multiply the second row by -1 to get a positive pivot.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & 1 & -1 & -9 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 7

Finally, we will subtract the second row from the third row to get a zero in the third row, second column.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & 1 & -1 & -9 \\ 0 & 0 & 0 & 8 \end{bmatrix}$

STEP 8

The matrix is now in row echelon form. The last row tells us that 0 = 8, which is a contradiction. Therefore, the system of equations has no solution.

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Find the intersection point of 3 one-way streets using Gaussian elimination to solve the system of $x+9=y+13$ , $z+15=x+10$ , $y+23=z+24$ .

STEP 1

Assumptions
1. The system of equations is given by: $\begin{array}{l} x+9=y+13 \\ z+15=x+10 \\ y+23=z+24 \end{array}$
2. We are to solve this system of equations using Gaussian elimination.

STEP 2

First, we need to rewrite the system of equations in matrix form. This will make it easier to apply Gaussian elimination.
$\begin{bmatrix} 1 & -1 & 0 & 4 \\ -1 & 0 & 1 & 5 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 3

STEP 4

Next, we will multiply the first row by -1 to get a positive pivot.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 1 & -1 & 0 & 4 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 5

Now, we will subtract the first row from the second row to get a zero in the second row, first column.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & -1 & 1 & 9 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 6

We will multiply the second row by -1 to get a positive pivot.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & 1 & -1 & -9 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

STEP 7

Finally, we will subtract the second row from the third row to get a zero in the third row, second column.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & 1 & -1 & -9 \\ 0 & 0 & 0 & 8 \end{bmatrix}$

STEP 8

The matrix is now in row echelon form. The last row tells us that 0 = 8, which is a contradiction. Therefore, the system of equations has no solution.

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Find the intersection point of 3 one-way streets using Gaussian elimination to solve the system of x+9=y+13x+9=y+13x+9=y+13, z+15=x+10z+15=x+10z+15=x+10, y+23=z+24y+23=z+24y+23=z+24.

STEP 1

Assumptions1. The system of equations is given by: x+9=y+13z+15=x+10y+23=z+24 \begin{array}{l} x+9=y+13 \\ z+15=x+10 \\ y+23=z+24 \end{array} x+9=y+13z+15=x+10y+23=z+24​2. We are to solve this system of equations using Gaussian elimination.

STEP 2

First, we need to rewrite the system of equations in matrix form. This will make it easier to apply Gaussian elimination.[1−104−101501−1−1] \begin{bmatrix} 1 & -1 & 0 & 4 \\ -1 & 0 & 1 & 5 \\ 0 & 1 & -1 & -1 \end{bmatrix} ⎣⎡​1−10​−101​01−1​45−1​⎦⎤​

STEP 3

STEP 4

Next, we will multiply the first row by -1 to get a positive pivot.[10−1−51−10401−1−1] \begin{bmatrix} 1 & 0 & -1 & -5 \\ 1 & -1 & 0 & 4 \\ 0 & 1 & -1 & -1 \end{bmatrix} ⎣⎡​110​0−11​−10−1​−54−1​⎦⎤​

STEP 5

Now, we will subtract the first row from the second row to get a zero in the second row, first column.[10−1−50−11901−1−1] \begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & -1 & 1 & 9 \\ 0 & 1 & -1 & -1 \end{bmatrix} ⎣⎡​100​0−11​−11−1​−59−1​⎦⎤​

STEP 6

We will multiply the second row by -1 to get a positive pivot.[10−1−501−1−901−1−1] \begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & 1 & -1 & -9 \\ 0 & 1 & -1 & -1 \end{bmatrix} ⎣⎡​100​011​−1−1−1​−5−9−1​⎦⎤​

STEP 7

Finally, we will subtract the second row from the third row to get a zero in the third row, second column.[10−1−501−1−90008] \begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & 1 & -1 & -9 \\ 0 & 0 & 0 & 8 \end{bmatrix} ⎣⎡​100​010​−1−10​−5−98​⎦⎤​

STEP 8

The matrix is now in row echelon form. The last row tells us that 0 = 8, which is a contradiction. Therefore, the system of equations has no solution.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the intersection point of 3 one-way streets using Gaussian elimination to solve the system of $x+9=y+13$ , $z+15=x+10$ , $y+23=z+24$ .

Assumptions
1. The system of equations is given by: $\begin{array}{l} x+9=y+13 \\ z+15=x+10 \\ y+23=z+24 \end{array}$
2. We are to solve this system of equations using Gaussian elimination.

First, we need to rewrite the system of equations in matrix form. This will make it easier to apply Gaussian elimination.
$\begin{bmatrix} 1 & -1 & 0 & 4 \\ -1 & 0 & 1 & 5 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

Next, we will multiply the first row by -1 to get a positive pivot.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 1 & -1 & 0 & 4 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

Now, we will subtract the first row from the second row to get a zero in the second row, first column.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & -1 & 1 & 9 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

We will multiply the second row by -1 to get a positive pivot.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & 1 & -1 & -9 \\ 0 & 1 & -1 & -1 \end{bmatrix}$

Finally, we will subtract the second row from the third row to get a zero in the third row, second column.
$\begin{bmatrix} 1 & 0 & -1 & -5 \\ 0 & 1 & -1 & -9 \\ 0 & 0 & 0 & 8 \end{bmatrix}$