Math  /  Algebra

Question6. Solve (if possible) each of the following systems of three equations in three variables using the method of GaussJordan elimination. (c) x1+2x2x3=32x1+4x22x3=63x1+6x2+2x3=1\begin{array}{rr} x_{1}+2 x_{2}-x_{3}= & 3 \\ 2 x_{1}+4 x_{2}-2 x_{3}= & 6 \\ 3 x_{1}+6 x_{2}+2 x_{3}= & -1 \end{array}

Studdy Solution

STEP 1

1. The system of equations is given in three variables: x1x_1, x2x_2, and x3x_3.
2. We will use the Gauss-Jordan elimination method to solve the system.
3. The system may have a unique solution, infinitely many solutions, or no solution.

STEP 2

1. Write the augmented matrix for the system of equations.
2. Use row operations to transform the matrix into reduced row-echelon form (RREF).
3. Interpret the RREF to find the solution(s) to the system.

STEP 3

Write the augmented matrix for the system:
[121324263621]\begin{bmatrix} 1 & 2 & -1 & | & 3 \\ 2 & 4 & -2 & | & 6 \\ 3 & 6 & 2 & | & -1 \end{bmatrix}

STEP 4

Use row operations to transform the matrix into reduced row-echelon form (RREF).
First, eliminate the first column entries below the pivot in the first row:
Subtract 2 times the first row from the second row:
R2=R22R1R_2 = R_2 - 2R_1
Subtract 3 times the first row from the third row:
R3=R33R1R_3 = R_3 - 3R_1
The matrix becomes:
[1213000000510]\begin{bmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 5 & | & -10 \end{bmatrix}

STEP 5

Next, make the pivot in the third row a 1 by dividing the third row by 5:
R3=15R3R_3 = \frac{1}{5}R_3
The matrix becomes:
[121300000012]\begin{bmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 1 & | & -2 \end{bmatrix}

STEP 6

Eliminate the third column entry in the first row by adding the third row to the first row:
R1=R1+R3R_1 = R_1 + R_3
The matrix becomes:
[120100000012]\begin{bmatrix} 1 & 2 & 0 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 1 & | & -2 \end{bmatrix}

STEP 7

Interpret the RREF to find the solution(s) to the system.
The matrix corresponds to the following system of equations:
\begin{align*} x_1 + 2x_2 &= 1 \\ x_3 &= -2 \end{align*}
Since the second row is all zeros, it does not provide any new information. We have a free variable, x2x_2.
Express x1x_1 in terms of x2x_2:
x1=12x2x_1 = 1 - 2x_2
The solution is:
\begin{align*} x_1 &= 1 - 2t \\ x_2 &= t \\ x_3 &= -2 \end{align*}
where tt is a parameter representing any real number.

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