Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Certainly! Let's use Gaussian elimination to solve the given system of equations:
[tex]\[ \begin{cases} -x + y + z = -2 \\ -x + 4y - 5z = -14 \\ 4x - 2y - 8z = 0 \end{cases} \][/tex]
We'll start by writing the augmented matrix of the system:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ -1 & 4 & -5 & | & -14 \\ 4 & -2 & -8 & | & 0 \end{pmatrix} \][/tex]
### Step 1: Forming Pivot in the First Column
We'll transform the matrix to get a leading 1 in the first row of the first column. For simplicity, let the first row remain the same, and we'll perform row operations to eliminate the variable [tex]\(x\)[/tex] from the second and third rows.
[tex]\[ R1: \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ -1 & 4 & -5 & | & -14 \\ 4 & -2 & -8 & | & 0 \end{pmatrix} \][/tex]
[tex]\[ R2 \gets R2 - R1 \][/tex]
Second row operation:
[tex]\[ -1 + 1 = 0, \quad 4 - 1 = 3, \quad -5 - 1 = -6, \quad -14 + 2 = -12 \][/tex]
Resulting in:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 3 & -6 & | & -12 \\ 4 & -2 & -8 & | & 0 \end{pmatrix} \][/tex]
[tex]\[ R3 \gets R3 + 4 \cdot R1 \][/tex]
Third row operation:
[tex]\[ 4 + 4(-1) = 0, \quad -2 + 4(1) = 2, \quad -8 + 4(1) = 2, \quad 0 + 4(-2) = -8 \][/tex]
Resulting in:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 3 & -6 & | & -12 \\ 0 & 2 & -8 & | & -8 \end{pmatrix} \][/tex]
### Step 2: Forming Pivot in the Second Column
Next, we'll make sure the second row has a leading one in the second column.
[tex]\[ R2 \gets \frac{1}{3} R2 \][/tex]
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 1 & -2 & | & -4 \\ 0 & 2 & -8 & | & -8 \end{pmatrix} \][/tex]
To eliminate the variable [tex]\(y\)[/tex] from the third row, we perform:
[tex]\[ R3 \gets R3 - 2 \times R2 \][/tex]
Third row operation:
[tex]\[ 0, \quad 2 - 2(1) = 0, \quad -8 - 2(-2) = -8 + 4 = -4, \quad -8 - 2(-4) = -8 + 8 = 0 \][/tex]
Resulting in:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 1 & -2 & | & -4 \\ 0 & 0 & -4 & | & 0 \end{pmatrix} \][/tex]
### Step 3: Forming Pivot in the Third Column
Finally, for the third row, we make the leading coefficient 1.
[tex]\[ R3 \gets -\frac{1}{4} R3 \][/tex]
Resulting in:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 1 & -2 & | & -4 \\ 0 & 0 & 1 & | & 0 \end{pmatrix} \][/tex]
### Step 4: Back Substitution for Solution
With the matrix in row echelon form, we use back substitution to find the values of [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex].
[tex]\[ \begin{cases} z = 0 \\ y - 2z = -4 \implies y - 0 = -4 \implies y = -4 \\ -x + y + z = -2 \implies -x - 4 + 0 = -2 \implies -x - 4 = -2 \implies -x = 2 \implies x = -2 \end{cases} \][/tex]
So, the solutions to the system of equations are:
[tex]\[ x = -2, \quad y = -4, \quad z = 0 \][/tex]
However, note that during the Gaussian Elimination, we observed a problematic row:
[tex]\[ [nan, nan, nan, nan] \][/tex]
This indicates an inconsistency or an element where division is undefined. If we applied our steps correctly, it implies the system might be inconsistent or degenerate. Hence, this requires a deeper look or alternative methods for confirmation.
[tex]\[ \begin{cases} -x + y + z = -2 \\ -x + 4y - 5z = -14 \\ 4x - 2y - 8z = 0 \end{cases} \][/tex]
We'll start by writing the augmented matrix of the system:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ -1 & 4 & -5 & | & -14 \\ 4 & -2 & -8 & | & 0 \end{pmatrix} \][/tex]
### Step 1: Forming Pivot in the First Column
We'll transform the matrix to get a leading 1 in the first row of the first column. For simplicity, let the first row remain the same, and we'll perform row operations to eliminate the variable [tex]\(x\)[/tex] from the second and third rows.
[tex]\[ R1: \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ -1 & 4 & -5 & | & -14 \\ 4 & -2 & -8 & | & 0 \end{pmatrix} \][/tex]
[tex]\[ R2 \gets R2 - R1 \][/tex]
Second row operation:
[tex]\[ -1 + 1 = 0, \quad 4 - 1 = 3, \quad -5 - 1 = -6, \quad -14 + 2 = -12 \][/tex]
Resulting in:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 3 & -6 & | & -12 \\ 4 & -2 & -8 & | & 0 \end{pmatrix} \][/tex]
[tex]\[ R3 \gets R3 + 4 \cdot R1 \][/tex]
Third row operation:
[tex]\[ 4 + 4(-1) = 0, \quad -2 + 4(1) = 2, \quad -8 + 4(1) = 2, \quad 0 + 4(-2) = -8 \][/tex]
Resulting in:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 3 & -6 & | & -12 \\ 0 & 2 & -8 & | & -8 \end{pmatrix} \][/tex]
### Step 2: Forming Pivot in the Second Column
Next, we'll make sure the second row has a leading one in the second column.
[tex]\[ R2 \gets \frac{1}{3} R2 \][/tex]
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 1 & -2 & | & -4 \\ 0 & 2 & -8 & | & -8 \end{pmatrix} \][/tex]
To eliminate the variable [tex]\(y\)[/tex] from the third row, we perform:
[tex]\[ R3 \gets R3 - 2 \times R2 \][/tex]
Third row operation:
[tex]\[ 0, \quad 2 - 2(1) = 0, \quad -8 - 2(-2) = -8 + 4 = -4, \quad -8 - 2(-4) = -8 + 8 = 0 \][/tex]
Resulting in:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 1 & -2 & | & -4 \\ 0 & 0 & -4 & | & 0 \end{pmatrix} \][/tex]
### Step 3: Forming Pivot in the Third Column
Finally, for the third row, we make the leading coefficient 1.
[tex]\[ R3 \gets -\frac{1}{4} R3 \][/tex]
Resulting in:
[tex]\[ \begin{pmatrix} -1 & 1 & 1 & | & -2 \\ 0 & 1 & -2 & | & -4 \\ 0 & 0 & 1 & | & 0 \end{pmatrix} \][/tex]
### Step 4: Back Substitution for Solution
With the matrix in row echelon form, we use back substitution to find the values of [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex].
[tex]\[ \begin{cases} z = 0 \\ y - 2z = -4 \implies y - 0 = -4 \implies y = -4 \\ -x + y + z = -2 \implies -x - 4 + 0 = -2 \implies -x - 4 = -2 \implies -x = 2 \implies x = -2 \end{cases} \][/tex]
So, the solutions to the system of equations are:
[tex]\[ x = -2, \quad y = -4, \quad z = 0 \][/tex]
However, note that during the Gaussian Elimination, we observed a problematic row:
[tex]\[ [nan, nan, nan, nan] \][/tex]
This indicates an inconsistency or an element where division is undefined. If we applied our steps correctly, it implies the system might be inconsistent or degenerate. Hence, this requires a deeper look or alternative methods for confirmation.
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
