Solved on Jan 18, 2024

Solve the system of linear equations represented by the given augmented matrix using Gauss-Jordan method. Perform the row operations and find the values of $x$ , $y$ , and $z$ .

STEP 1

Assumptions
1. We are given an augmented matrix representing a system of linear equations.
2. We will use the Gauss-Jordan elimination method to solve the system.
3. The row operations are given step by step, and we need to apply them to the matrix.
4. The final matrix will be in reduced-row echelon form, from which we can read off the solutions for $x$ , $y$ , and $z$ .

STEP 2

The given augmented matrix is: $\left[\begin{array}{cccc} 2 & 1 & 1 & 6 \\ 1 & -1 & -1 & -3 \\ 3 & 1 & 2 & 7 \end{array}\right]$ The first row operation is $R_{1} \longleftrightarrow R_{2}$ , which means we swap the first and second rows.

STEP 3

After swapping $R_{1}$ and $R_{2}$ , the matrix becomes: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 2 & 1 & 1 & 6 \\ 3 & 1 & 2 & 7 \end{array}\right]$

STEP 4

The next row operations are $R_{2}=-2 R_{1}+R_{2}$ and $R_{3}=-3 R_{1}+R_{3}$ . Apply these operations to eliminate the coefficients below the leading 1 in the first column.

STEP 5

First, apply $R_{2}=-2 R_{1}+R_{2}$ : $R_{2} = -2 \cdot \left[\begin{array}{c} 1 \\ -1 \\ -1 \\ -3 \end{array}\right] + \left[\begin{array}{c} 2 \\ 1 \\ 1 \\ 6 \end{array}\right] = \left[\begin{array}{c} 0 \\ 3 \\ 3 \\ 12 \end{array}\right]$

STEP 6

Now, apply $R_{3}=-3 R_{1}+R_{3}$ : $R_{3} = -3 \cdot \left[\begin{array}{c} 1 \\ -1 \\ -1 \\ -3 \end{array}\right] + \left[\begin{array}{c} 3 \\ 1 \\ 2 \\ 7 \end{array}\right] = \left[\begin{array}{c} 0 \\ 4 \\ 5 \\ 14 \end{array}\right]$

STEP 7

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 0 & 3 & 3 & 12 \\ 0 & 4 & 5 & 14 \end{array}\right]$

STEP 8

The next operation is $R_{2}=\frac{1}{3} R_{2}$ , which scales the second row to create a leading 1 in the second column.

STEP 9

Apply the operation to $R_{2}$ : $R_{2} = \frac{1}{3} \cdot \left[\begin{array}{c} 0 \\ 3 \\ 3 \\ 12 \end{array}\right] = \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right]$

STEP 10

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 0 & 1 & 1 & 4 \\ 0 & 4 & 5 & 14 \end{array}\right]$

STEP 11

The next operations are $R_{1}=R_{2}+R_{1}$ and $R_{3}=-4 R_{2}+R_{3}$ to create zeros below and above the leading 1 in the second column.

STEP 12

First, apply $R_{1}=R_{2}+R_{1}$ : $R_{1} = \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right] + \left[\begin{array}{c} 1 \\ -1 \\ -1 \\ -3 \end{array}\right] = \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \end{array}\right]$

STEP 13

Now, apply $R_{3}=-4 R_{2}+R_{3}$ : $R_{3} = -4 \cdot \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right] + \left[\begin{array}{c} 0 \\ 4 \\ 5 \\ 14 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 2 \end{array}\right]$

STEP 14

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right]$

STEP 15

The last operation is $R_{2}=-R_{3}+R_{2}$ to create a zero above the leading 1 in the third column.

STEP 16

Apply the operation to $R_{2}$ : $R_{2} = - \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 2 \end{array}\right] + \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right] = \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 2 \end{array}\right]$

STEP 17

The final matrix in reduced-row echelon form is: $\left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]$

STEP 18

From the final matrix, we can read off the solutions for $x$ , $y$ , and $z$ directly as the values in the last column.

STEP 19

The solution for $x$ is: $x = 1$

STEP 20

The solution for $y$ is: $y = 2$

STEP 21

The solution for $z$ is: $z = 2$
The solutions for the variables are $x = 1$ , $y = 2$ , and $z = 2$ .

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Solve the system of linear equations represented by the given augmented matrix using Gauss-Jordan method. Perform the row operations and find the values of xxx, yyy, and zzz.

STEP 1

STEP 2

STEP 3

After swapping R1R_{1}R1​ and R2R_{2}R2​, the matrix becomes: [1−1−1−321163127] \left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 2 & 1 & 1 & 6 \\ 3 & 1 & 2 & 7 \end{array}\right] ⎣⎡​123​−111​−112​−367​⎦⎤​

STEP 4

The next row operations are R2=−2R1+R2R_{2}=-2 R_{1}+R_{2}R2​=−2R1​+R2​ and R3=−3R1+R3R_{3}=-3 R_{1}+R_{3}R3​=−3R1​+R3​. Apply these operations to eliminate the coefficients below the leading 1 in the first column.

STEP 5

STEP 6

STEP 7

The matrix now looks like this: [1−1−1−30331204514] \left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 0 & 3 & 3 & 12 \\ 0 & 4 & 5 & 14 \end{array}\right] ⎣⎡​100​−134​−135​−31214​⎦⎤​

STEP 8

The next operation is R2=13R2R_{2}=\frac{1}{3} R_{2}R2​=31​R2​, which scales the second row to create a leading 1 in the second column.

STEP 9

Apply the operation to R2R_{2}R2​: R2=13⋅[03312]=[0114] R_{2} = \frac{1}{3} \cdot \left[\begin{array}{c} 0 \\ 3 \\ 3 \\ 12 \end{array}\right] = \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right] R2​=31​⋅⎣⎡​03312​⎦⎤​=⎣⎡​0114​⎦⎤​

STEP 10

The matrix now looks like this: [1−1−1−3011404514] \left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 0 & 1 & 1 & 4 \\ 0 & 4 & 5 & 14 \end{array}\right] ⎣⎡​100​−114​−115​−3414​⎦⎤​

STEP 11

The next operations are R1=R2+R1R_{1}=R_{2}+R_{1}R1​=R2​+R1​ and R3=−4R2+R3R_{3}=-4 R_{2}+R_{3}R3​=−4R2​+R3​ to create zeros below and above the leading 1 in the second column.

STEP 12

STEP 13

STEP 14

The matrix now looks like this: [100101140012] \left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right] ⎣⎡​100​010​011​142​⎦⎤​

STEP 15

The last operation is R2=−R3+R2R_{2}=-R_{3}+R_{2}R2​=−R3​+R2​ to create a zero above the leading 1 in the third column.

STEP 16

STEP 17

The final matrix in reduced-row echelon form is: [100101020012] \left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right] ⎣⎡​100​010​001​122​⎦⎤​

STEP 18

From the final matrix, we can read off the solutions for xxx, yyy, and zzz directly as the values in the last column.

STEP 19

The solution for xxx is: x=1 x = 1 x=1

STEP 20

The solution for yyy is: y=2 y = 2 y=2

STEP 21

The solution for zzz is: z=2 z = 2 z=2The solutions for the variables are x=1x = 1x=1, y=2y = 2y=2, and z=2z = 2z=2.

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Solve the system of linear equations represented by the given augmented matrix using Gauss-Jordan method. Perform the row operations and find the values of $x$ , $y$ , and $z$ .

After swapping $R_{1}$ and $R_{2}$ , the matrix becomes: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 2 & 1 & 1 & 6 \\ 3 & 1 & 2 & 7 \end{array}\right]$

The next row operations are $R_{2}=-2 R_{1}+R_{2}$ and $R_{3}=-3 R_{1}+R_{3}$ . Apply these operations to eliminate the coefficients below the leading 1 in the first column.

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 0 & 3 & 3 & 12 \\ 0 & 4 & 5 & 14 \end{array}\right]$

The next operation is $R_{2}=\frac{1}{3} R_{2}$ , which scales the second row to create a leading 1 in the second column.

Apply the operation to $R_{2}$ : $R_{2} = \frac{1}{3} \cdot \left[\begin{array}{c} 0 \\ 3 \\ 3 \\ 12 \end{array}\right] = \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right]$

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 0 & 1 & 1 & 4 \\ 0 & 4 & 5 & 14 \end{array}\right]$

The next operations are $R_{1}=R_{2}+R_{1}$ and $R_{3}=-4 R_{2}+R_{3}$ to create zeros below and above the leading 1 in the second column.

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right]$

The last operation is $R_{2}=-R_{3}+R_{2}$ to create a zero above the leading 1 in the third column.

The final matrix in reduced-row echelon form is: $\left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]$

From the final matrix, we can read off the solutions for $x$ , $y$ , and $z$ directly as the values in the last column.

The solution for $x$ is: $x = 1$

The solution for $y$ is: $y = 2$

The solution for $z$ is: $z = 2$
The solutions for the variables are $x = 1$ , $y = 2$ , and $z = 2$ .