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.