1. We are given augmented matrices in row echelon form and need to identify pivot rows and columns and solve the system for each part. 2. Part (a): Matrix: $$\begin{bmatrix} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 7 \end{bmatrix}$$ - Pivot rows: 1, 2, 3 (all rows) - Pivot columns: 1, 2, 3 (leading 1s in each row) - Corresponding system: $$x_1 = -3$$ $$x_2 = 0$$ $$x_3 = 7$$ 3. Part (b): Matrix: $$\begin{bmatrix} 1 & 0 & 0 & -7 & 8 \\ 0 & 1 & 0 & 3 & 2 \\ 0 & 0 & 1 & 1 & -5 \end{bmatrix}$$ - Pivot rows: 1, 2, 3 - Pivot columns: 1, 2, 3 (leading 1s) - Variables: Let $x_1, x_2, x_3$ correspond to pivots; extra variables $x_4, x_5$ are free variables - The system can be written as: $$ x_1 - 7x_4 + 8x_5 = 0 $$ $$ x_2 + 3x_4 + 2x_5 = 0 $$ $$ x_3 + x_4 - 5x_5 = 0 $$ - Expressing pivot variables in terms of free variables: $$ x_1 = 7x_4 - 8x_5 $$ $$ x_2 = -3x_4 - 2x_5 $$ $$ x_3 = -x_4 + 5x_5 $$ - $x_4$ and $x_5$ are free parameters. 4. Part (c): Matrix: $$\begin{bmatrix} 1 & -6 & 0 & 0 & 3 & -2 \\ 0 & 0 & 1 & 0 & 4 & 7 \\ 0 & 0 & 0 & 1 & 5 & 8 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ - Pivot rows: 1, 2, 3 - Pivot columns: 1, 3, 4 (columns with leading 1s) - Variables: Assign $x_1$ to col 1, $x_2$ col 2, $x_3$ col 3, $x_4$ col 4, $x_5$ col 5 - Write equations: Row 1: $$ x_1 - 6x_2 + 3x_5 = -2 $$ Row 2: $$ x_3 + 4x_5 = 7 $$ Row 3: $$ x_4 + 5x_5 = 8 $$ - Solve for pivot variables: $$ x_1 = 6x_2 - 3x_5 - 2 $$ $$ x_3 = 7 - 4x_5 $$ $$ x_4 = 8 - 5x_5 $$ - Free variables: $x_2, x_5$. 5. Part (d): Matrix: $$\begin{bmatrix} 1 & -3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ - Pivot rows: 1, 2, 3 - Pivot columns: 1, 3, 4 - Variables: $x_1, x_2, x_3, x_4$ - Equations: $$ x_1 - 3x_2 = 0 $$ $$ x_3 = 0 $$ $$ x_4 = 0 $$ - Solve for pivot variables: $$ x_1 = 3x_2 $$ - Free variable: $x_2$ Final summary: (a) $x_1=-3, x_2=0, x_3=7$ (b) $x_1=7x_4-8x_5, x_2=-3x_4-2x_5, x_3=-x_4+5x_5, x_4,x_5$ free (c) $x_1=6x_2-3x_5-2, x_3=7-4x_5, x_4=8-5x_5, x_2,x_5$ free (d) $x_1=3x_2, x_3=0, x_4=0, x_2$ free