Determine the rank of the matrix $B=\left(\begin{array}{ccc}1& 2& 3\\ 2& 4& 6\\ 3& 5& 9\end{array}\right)$ and state whether the system $B\mathbf{x}=\left(\begin{array}{c}7\\ 14\\ 21\end{array}\right)$ is consistent. If consistent, describe the solution set.

Step-by-step solution

1. Form the augmented matrix $[B\mid \mathbf{b}]$ with $\mathbf{b}=\left(\begin{array}{c}7\\ 14\\ 21\end{array}\right)$: $\left(\begin{array}{ccccc}1& 2& 3& |& 7\\ 2& 4& 6& |& 14\\ 3& 5& 9& |& 21\end{array}\right).$
2. Apply $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3-3R_1$: $\left(\begin{array}{ccccc}1& 2& 3& |& 7\\ 0& 0& 0& |& 0\\ 0& -1& 0& |& 0\end{array}\right).$
3. Swap $R_2$ and $R_3$ to get a non‑zero row in the second position: $\left(\begin{array}{ccccc}1& 2& 3& |& 7\\ 0& -1& 0& |& 0\\ 0& 0& 0& |& 0\end{array}\right).$
4. The coefficient matrix now has two non‑zero rows, so $\mathrm{rank}(B)=2$. The augmented matrix also has two non‑zero rows, so $\mathrm{rank}([B|\mathbf{b}])=2$.
5. Since the ranks are equal and less than the number of variables (3), the system is consistent and has infinitely many solutions.
6. Express the solution: from the second row $-y=0\Rightarrow y=0$. Substitute into the first row: $x+2(0)+3z=7\Rightarrow x+3z=7$, so $x=7-3z$ with $z$ free. Hence the solution set is $\{(7-3t,0,t{)}^T\mid t\in \mathbb{R}\}$.

Concept tested: Rank and consistency of a linear system

Published 19 September 2026 · How this solution was prepared and checked