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Solve the following system of equations using Gaussian elimination or​ Gauss-Jordan elimination.

4x−
y+
5z
=25
6x+
y−
z
=−4
2x+
y+
3z
=10
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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer boxes to complete your choice.
A.
The solution is enter your response here,enter your response here,enter your response here.
​(Type integers or simplified​ fractions.)
B.
There are infinitely many solutions of the form enter your response here,enter your response here,z.
​(Type expressions using z as the​ variable.)
C.
There is no solution.

User TGV
by
8.6k points

1 Answer

5 votes

Answer:

So, the correct choice is:

A. The solution is \( x \approx -1.57 \), \( y \approx -0.83 \), \( z \approx 0.33 \).

Explanation:

To solve the system of equations using Gaussian elimination or Gauss-Jordan elimination, we first write the system in augmented matrix form:

4 & -1 & 5 & \vert & 25 \\

6 & 1 & -1 & \vert & -4 \\

2 & 1 & 3 & \vert & 10 \\

\end{bmatrix} \]

Now, we'll perform row operations to transform the matrix into reduced row-echelon form (RREF).

Step 1: To make the entry in the first column of the second row (6) become zero, we'll subtract 1.5 times the first row from the second row:

\[ R_2 \rightarrow R_2 - 1.5R_1 \]

\[ \begin{bmatrix}

4 & -1 & 5 & \vert & 25 \\

0 & 2.5 & -8.5 & \vert & -41.5 \\

2 & 1 & 3 & \vert & 10 \\

\end{bmatrix} \]

Step 2: To make the entry in the first column of the third row (2) become zero, we'll subtract 0.5 times the first row from the third row:

\[ R_3 \rightarrow R_3 - 0.5R_1 \]

\[ \begin{bmatrix}

4 & -1 & 5 & \vert & 25 \\

0 & 2.5 & -8.5 & \vert & -41.5 \\

0 & 1.5 & 0.5 & \vert & -2.5 \\

\end{bmatrix} \]

Step 3: To make the entry in the second column of the third row (1.5) become zero, we'll subtract 0.6 times the second row from the third row:

\[ R_3 \rightarrow R_3 - 0.6R_2 \]

\[ \begin{bmatrix}

4 & -1 & 5 & \vert & 25 \\

0 & 2.5 & -8.5 & \vert & -41.5 \\

0 & 0 & 6 & \vert & 2 \\

\end{bmatrix} \]

Step 4: To make the entry in the third column of the second row (8.5) become zero, we'll divide the second row by 2.5:

\[ R_2 \rightarrow \frac{1}{2.5}R_2 \]

\[ \begin{bmatrix}

4 & -1 & 5 & \vert & 25 \\

0 & 1 & -3.4 & \vert & -16.6 \\

0 & 0 & 6 & \vert & 2 \\

\end{bmatrix} \]

Step 5: To make the entry in the third column of the first row (5) become zero, we'll subtract 5 times the third row from the first row, and then divide by 4:

\[ R_1 \rightarrow \frac{1}{4}(R_1 - 5R_3) \]

\[ \begin{bmatrix}

1 & -1 & 0 & \vert & 5 \\

0 & 1 & -3.4 & \vert & -16.6 \\

0 & 0 & 6 & \vert & 2 \\

\end{bmatrix} \]

Step 6: To make the entry in the second column of the first row (1) become zero, we'll add 1 times the second row to the first row:

\[ R_1 \rightarrow R_1 + R_2 \]

\[ \begin{bmatrix}

1 & 0 & -3.4 & \vert & -11.6 \\

0 & 1 & -3.4 & \vert & -16.6 \\

0 & 0 & 6 & \vert & 2 \\

\end{bmatrix} \]

Step 7: To make the entry in the third column of the second row (-3.4) become zero, we'll add 0.567 times the third row to the second row:

\[ R_2 \rightarrow R_2 + 0.567R_3 \]

\[ \begin{bmatrix}

1 & 0 & -3.4 & \vert & -11.6 \\

0 & 1 & 0 & \vert & -5 \\

0 & 0 & 6 & \vert & 2 \\

\end{bmatrix} \]

Step 8: To make the entry in the third column of the first row (-3.4) become zero, we'll add 0.567 times the third row to the first row:

\[ R_1 \rightarrow R_1 + 0.567R_3 \]

\[ \begin{bmatrix}

1 & 0 & 0 & \vert & -9.4 \\

0 & 1 & 0 & \vert & -5 \\

0 & 0 & 6 & \vert & 2 \\

\end{bmatrix} \]

Step 9: Finally, we can solve for the variables:

\[ x = -9.4/6 = -1.5666666 \approx -1.57 \]

\[ y = -5/6 \approx -0.83 \]

\[ z = 2/6 = 1/3 \approx 0.33 \]

The solution to the system of equations is:

\[ x \approx -1.57 \]

\[ y \approx -0.83 \]

\[ z \approx 0.33 \]

So, the correct choice is:

A. The solution is \( x \approx -1.57 \), \( y \approx -0.83 \), \( z \approx 0.33 \).

User Bsa
by
9.2k points