Question

The solution of the system of equations
\[ x_1 - 3x_2 + 4x_3 = 3 \]
\[ 2x_1 - 5x_2 + 7x_3 = 6 \]
\[ 3x_1 - 8x_2 + 11x_3 = 11 \]
is given by:

• A. \(x_1 = 3, x_2 = 0, x_3 = 2\)
• B. \(x_1 = 0, x_2 = 0, x_3 = 0\)
• C. \(x_1 = 1, x_2 = 0, x_3 = 2\)
• D. No solution exists

Hint:

You can also solve this by testing the given options.
Substitute the values from Option (A) into the first equation:
\[ 3 - 3(0) + 4(2) = 11 \neq 3 \]
This shows Option (A) is incorrect. Repeating this for the other options confirms that no solution exists.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A system of linear equations can have a unique solution, infinitely many solutions, or no solution.
We can determine the nature of the solutions by converting the system's augmented matrix into row-echelon form.

Step 2: Key Formula or Approach:
We write the augmented matrix \([A \mid B]\) and apply Gaussian elimination:
\[ [A \mid B] = \begin{bmatrix} 1 & -3 & 4 & \big| & 3 2 & -5 & 7 & \big| & 6 3 & -8 & 11 & \big| & 11 \end{bmatrix} \]

Step 3: Detailed Explanation:
Let us perform row operations to simplify the augmented matrix:
1. Eliminate the \(x_1\) term from the second row using the operation \(R_2 \to R_2 - 2R_1\):
\[ R_2 \text{ becomes: } [2 - 2(1), \ -5 - 2(-3), \ 7 - 2(4), \ \big| \ 6 - 2(3)] \]
\[ R_2 \text{ becomes: } [0, \ 1, \ -1, \ \big| \ 0] \]
2. Eliminate the \(x_1\) term from the third row using the operation \(R_3 \to R_3 - 3R_1\):
\[ R_3 \text{ becomes: } [3 - 3(1), \ -8 - 3(-3), \ 11 - 3(4), \ \big| \ 11 - 3(3)] \]
\[ R_3 \text{ becomes: } [0, \ 1, \ -1, \ \big| \ 2] \]
Now, our updated augmented matrix is:
\[ \begin{bmatrix} 1 & -3 & 4 & \big| & 3 0 & 1 & -1 & \big| & 0 0 & 1 & -1 & \big| & 2 \end{bmatrix} \]
3. Now, eliminate the \(x_2\) term from the third row using the operation \(R_3 \to R_3 - R_2\):
\[ R_3 \text{ becomes: } [0 - 0, \ 1 - 1, \ -1 - (-1), \ \big| \ 2 - 0] \]
\[ R_3 \text{ becomes: } [0, \ 0, \ 0, \ \big| \ 2] \]
This gives the final row-echelon form:
\[ \begin{bmatrix} 1 & -3 & 4 & \big| & 3 0 & 1 & -1 & \big| & 0 0 & 0 & 0 & \big| & 2 \end{bmatrix} \]
The third row corresponds to the equation:
\[ 0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 = 2 \implies 0 = 2 \]
This is a mathematical contradiction.
Therefore, the system of equations is inconsistent and has no solution.

Step 4: Final Answer:
Therefore, the correct option is (D).