1. **Problem Statement:** Find the rank of each matrix (a, b, c) using the echelon method. 2. **Formula and Rules:** The rank of a matrix is the number of nonzero rows in its row echelon form (REF). 3. **Matrix a:** \[ \begin{bmatrix} 1 & 4 & 5 \\ 2 & 6 & 8 \\ 3 & 7 & 22 \end{bmatrix} \] - Step 1: Use row 1 to eliminate below: - $R_2 \to R_2 - 2R_1 = [0, 6 - 8, 8 - 10] = [0, -2, -2]$ - $R_3 \to R_3 - 3R_1 = [0, 7 - 12, 22 - 15] = [0, -5, 7]$ - Step 2: Use $R_2$ to eliminate below: - $R_3 \to R_3 - \frac{5}{2} R_2 = [0, -5 - (-5), 7 - (-5)] = [0, 0, 19.5]$ - REF matrix: \[ \begin{bmatrix} 1 & 4 & 5 \\ 0 & -2 & -2 \\ 0 & 0 & 19.5 \end{bmatrix} \] - All 3 rows are nonzero, so rank(a) = 3. 4. **Matrix b:** \[ \begin{bmatrix} 1 & 3 & 5 \\ 2 & -1 & 4 \\ -2 & 8 & 2 \end{bmatrix} \] - Step 1: Eliminate below row 1: - $R_2 \to R_2 - 2R_1 = [0, -1 - 6, 4 - 10] = [0, -7, -6]$ - $R_3 \to R_3 + 2R_1 = [0, 8 + 6, 2 + 10] = [0, 14, 12]$ - Step 2: Eliminate below $R_2$: - $R_3 \to R_3 - 2R_2 = [0, 14 - (-14), 12 - (-12)] = [0, 28, 24]$ (recalculate carefully) Actually, $R_3 \to R_3 - 2R_2 = [0, 14 - 2(-7), 12 - 2(-6)] = [0, 14 + 14, 12 + 12] = [0, 28, 24]$ - Step 3: Simplify $R_3$ by subtracting $4R_2$: - $R_3 \to R_3 - 4R_2 = [0, 28 - 4(-7), 24 - 4(-6)] = [0, 28 + 28, 24 + 24] = [0, 56, 48]$ This suggests a miscalculation; better to use $R_3 \to R_3 - 2R_2$ only. - Instead, check linear dependence: - $R_3 = -2 R_1 + 2 R_2$? No. - Let's proceed with $R_3 \to R_3 - 2R_2 = [0, 14 - 2(-7), 12 - 2(-6)] = [0, 14 + 14, 12 + 12] = [0, 28, 24]$ - Step 4: Use $R_2$ to eliminate $R_3$: - $R_3 \to R_3 - 4 R_2 = [0, 28 - 4(-7), 24 - 4(-6)] = [0, 28 + 28, 24 + 24] = [0, 56, 48]$ - This shows $R_3$ is a multiple of $R_2$, so $R_3$ is not independent. - REF matrix: \[ \begin{bmatrix} 1 & 3 & 5 \\ 0 & -7 & -6 \\ 0 & 0 & 0 \end{bmatrix} \] - Rank(b) = 2. 5. **Matrix c:** \[ \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 2 & 0 & 1 & 3 \\ -1 & 3 & 0 & 0 \end{bmatrix} \] - Step 1: Use $R_1$ to eliminate below: - $R_3 \to R_3 - 2R_1 = [0, 0 - 4, 1 - 0, 3 - 2] = [0, -4, 1, 1]$ - $R_4 \to R_4 + R_1 = [0, 3 + 2, 0 + 0, 0 + 1] = [0, 5, 0, 1]$ - Step 2: Use $R_2$ to eliminate below: - $R_3 \to R_3 + 4 R_2 = [0, -4 + 4, 1 + 4, 1 + 8] = [0, 0, 5, 9]$ - $R_4 \to R_4 - 5 R_2 = [0, 5 - 5, 0 - 5, 1 - 10] = [0, 0, -5, -9]$ - Step 3: Use $R_3$ to eliminate $R_4$: - $R_4 \to R_4 + R_3 = [0, 0, 0, 0]$ - REF matrix: \[ \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 5 & 9 \\ 0 & 0 & 0 & 0 \end{bmatrix} \] - Rank(c) = 3. **Final answers:** - Rank(a) = 3 - Rank(b) = 2 - Rank(c) = 3