Question

Let \( A \) be a \( 3 \times 5 \) matrix defined by
\[A = \begin{pmatrix}0 & 1 & 3 & 1 & 2 \\1 & 6 & 2 & 3 & 4 \\1 & 8 & 8 & 5 & 8\end{pmatrix}.\]
Consider the system of linear equations given by
\[A \begin{pmatrix}x_1 \\x_2 \\x_3 \\x_4 \\x_5\end{pmatrix} = \begin{pmatrix}3 \\1 \\10\end{pmatrix},\]
where \( x_1, x_2, x_3, x_4, x_5 \) are real variables. Then

• A. the rank of 𝐴 is 2 and the given system has a solution
• B. the rank of 𝐴 is 2 and the given system does NOT have a solution
• C. the rank of 𝐴 is 3 and the given system has a solution
• D. the rank of 𝐴 is 3 and the given system does NOT have a solution

Answer 1

The Correct Option is B

1. Compute the Rank of \( A \): - Perform row reduction to bring \( A \) to its row echelon form: \[ A = \begin{pmatrix} 0 & 1 & 3 & 1 & 2 \\ 1 & 6 & 2 & 3 & 4 \\ 1 & 8 & 8 & 5 & 8 \end{pmatrix}. \] - Row reduce: \[ \text{Subtract Row 2 from Row 3: } R_3 \to R_3 - R_2. \] \[ A = \begin{pmatrix} 0 & 1 & 3 & 1 & 2 \\ 1 & 6 & 2 & 3 & 4 \\ 0 & 2 & 6 & 2 & 4 \end{pmatrix}. \] - Further row reduction shows that two rows are linearly independent, confirming: \[ \text{Rank}(A) = 2. \] 2. Augmented Matrix Analysis: - Augment \( A \) with the column vector: \[ \left[A \, | \, \begin{pmatrix} 3 \\ 1 \\ 10 \end{pmatrix}\right]. \] - After row reduction, the last row of the augmented matrix leads to a contradiction, implying that the system is inconsistent. 3. Conclusion: - The rank of \( A \) is 2, and the system has no solution.