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

Step-by-step solution

1. Convert $R$ to row‑echelon form by performing $R_2\to R_2-2R_1$ and $R_3\to R_3-3R_1$.
2. After the operations, the matrix becomes $\left(\begin{array}{ccc}1& 2& 3\\ 0& 0& -1\\ 0& 0& 0\end{array}\right)$.
3. The third column has a pivot in the second row, so the number of non‑zero rows is 2. Hence $\mathrm{rank}(R)=2$.
4. Apply the same row operations to the augmented part: the RHS vector $\left(\begin{array}{c}7\\ 14\\ 21\end{array}\right)$ transforms to $\left(\begin{array}{c}7\\ 0\\ 0\end{array}\right)$.
5. Since no row becomes $[000|\text{non‑zero}]$, the system is consistent.
6. With $\mathrm{rank}(R)=\mathrm{rank}$ of augmented matrix = 2 < number of variables (3), the solution set is infinite.
7. Thus the system is consistent and has infinitely many solutions.

Concept tested: Rank of matrix and consistency of linear system

Published 7 September 2026 · How this solution was prepared and checked