1. **State the problem:** We are given the system of linear equations: $$\begin{cases} x_1 + 3x_2 - 2x_3 = 6 \\ 4x_1 + 5x_2 + 2x_3 = 3 \\ x_1 + 3x_2 + 4x_3 = 3 \end{cases}$$ We need to find the rank of the coefficient matrix $A$ and the augmented matrix $A_b$, then determine the nature of the solution. 2. **Write the coefficient matrix $A$ and augmented matrix $A_b$:** $$A = \begin{bmatrix} 1 & 3 & -2 \\ 4 & 5 & 2 \\ 1 & 3 & 4 \end{bmatrix}, \quad A_b = \begin{bmatrix} 1 & 3 & -2 & 6 \\ 4 & 5 & 2 & 3 \\ 1 & 3 & 4 & 3 \end{bmatrix}$$ 3. **Recall the rank and solution rules:** - The rank of a matrix is the maximum number of linearly independent rows (or columns). - If $\text{rank}(A) = \text{rank}(A_b) = n$ (number of variables), the system has a unique solution. - If $\text{rank}(A) = \text{rank}(A_b) < n$, there are infinitely many solutions. - If $\text{rank}(A) \neq \text{rank}(A_b)$, the system is inconsistent (no solution). 4. **Find $\text{rank}(A)$ by row reducing $A$:** Start with $$\begin{bmatrix} 1 & 3 & -2 \\ 4 & 5 & 2 \\ 1 & 3 & 4 \end{bmatrix}$$ - Replace row 2 by row 2 minus 4 times row 1: $$R_2 = R_2 - 4R_1 = [4 - 4(1), 5 - 4(3), 2 - 4(-2)] = [0, 5 - 12, 2 + 8] = [0, -7, 10]$$ - Replace row 3 by row 3 minus row 1: $$R_3 = R_3 - R_1 = [1 - 1, 3 - 3, 4 - (-2)] = [0, 0, 6]$$ Now matrix is $$\begin{bmatrix} 1 & 3 & -2 \\ 0 & -7 & 10 \\ 0 & 0 & 6 \end{bmatrix}$$ All three rows are nonzero, so $\text{rank}(A) = 3$. 5. **Find $\text{rank}(A_b)$ by row reducing $A_b$ similarly:** Start with $$\begin{bmatrix} 1 & 3 & -2 & 6 \\ 4 & 5 & 2 & 3 \\ 1 & 3 & 4 & 3 \end{bmatrix}$$ - Replace row 2 by $R_2 - 4R_1$: $$[0, -7, 10, 3 - 4(6)] = [0, -7, 10, 3 - 24] = [0, -7, 10, -21]$$ - Replace row 3 by $R_3 - R_1$: $$[0, 0, 6, 3 - 6] = [0, 0, 6, -3]$$ Matrix becomes $$\begin{bmatrix} 1 & 3 & -2 & 6 \\ 0 & -7 & 10 & -21 \\ 0 & 0 & 6 & -3 \end{bmatrix}$$ All three rows are nonzero, so $\text{rank}(A_b) = 3$. 6. **Compare ranks and number of variables:** - Number of variables $n = 3$ - $\text{rank}(A) = 3$ - $\text{rank}(A_b) = 3$ Since $\text{rank}(A) = \text{rank}(A_b) = n$, the system has a unique solution. **Final answer:** - $\text{rank}(A) = 3$ - $\text{rank}(A_b) = 3$ - The system has a unique solution.