1. **Problem Statement:** Given the matrix equation $A\vec{x} = \vec{0}$, where $A$ is a $4 \times 4$ matrix, find the Reduced Row Echelon Form (RREF) of $A$, determine its rank, and analyze the solution to the homogeneous system. 2. **Matrix $A$:** $$ A = \begin{pmatrix} 0 & 4 & 0 & -3 \\ 3 & 0 & -4 & -3 \\ 1 & 4 & 3 & 2 \\ 2 & -1 & 0 & -3 \end{pmatrix} $$ 3. **Goal:** Find the RREF of $A$, then determine the rank (number of leading 1s in RREF), and conclude about the solution to $A\vec{x} = \vec{0}$. 4. **Step-by-step RREF calculation:** - Start with $A$: $$ \begin{pmatrix} 0 & 4 & 0 & -3 \\ 3 & 0 & -4 & -3 \\ 1 & 4 & 3 & 2 \\ 2 & -1 & 0 & -3 \end{pmatrix} $$ - Swap Row 1 and Row 2 to get a leading nonzero pivot in the first row: $$ \begin{pmatrix} 3 & 0 & -4 & -3 \\ 0 & 4 & 0 & -3 \\ 1 & 4 & 3 & 2 \\ 2 & -1 & 0 & -3 \end{pmatrix} $$ - Make leading 1 in Row 1 by dividing by 3: $$ \begin{pmatrix} \frac{3}{\cancel{3}} & 0 & \frac{-4}{\cancel{3}} & \frac{-3}{\cancel{3}} \\ 0 & 4 & 0 & -3 \\ 1 & 4 & 3 & 2 \\ 2 & -1 & 0 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -\frac{4}{3} & -1 \\ 0 & 4 & 0 & -3 \\ 1 & 4 & 3 & 2 \\ 2 & -1 & 0 & -3 \end{pmatrix} $$ - Eliminate first column entries below pivot: Row 3 = Row 3 - Row 1: $$ \begin{pmatrix} 1 & 0 & -\frac{4}{3} & -1 \\ 0 & 4 & 0 & -3 \\ \cancel{1} - \cancel{1} & 4 - 0 & 3 - \left(-\frac{4}{3}\right) & 2 - (-1) \\ 2 & -1 & 0 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -\frac{4}{3} & -1 \\ 0 & 4 & 0 & -3 \\ 0 & 4 & \frac{13}{3} & 3 \\ 2 & -1 & 0 & -3 \end{pmatrix} $$ Row 4 = Row 4 - 2*Row 1: $$ \begin{pmatrix} 1 & 0 & -\frac{4}{3} & -1 \\ 0 & 4 & 0 & -3 \\ 0 & 4 & \frac{13}{3} & 3 \\ 2 - 2*1 & -1 - 2*0 & 0 - 2*(-\frac{4}{3}) & -3 - 2*(-1) \end{pmatrix} = \begin{pmatrix} 1 & 0 & -\frac{4}{3} & -1 \\ 0 & 4 & 0 & -3 \\ 0 & 4 & \frac{13}{3} & 3 \\ 0 & -1 & \frac{8}{3} & -1 \end{pmatrix} $$ - Make leading 1 in Row 2 by dividing by 4: $$ \begin{pmatrix} 1 & 0 & -\frac{4}{3} & -1 \\ 0 & \frac{4}{\cancel{4}} & 0 & -\frac{3}{\cancel{4}} \\ 0 & 4 & \frac{13}{3} & 3 \\ 0 & -1 & \frac{8}{3} & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -\frac{4}{3} & -1 \\ 0 & 1 & 0 & -\frac{3}{4} \\ 0 & 4 & \frac{13}{3} & 3 \\ 0 & -1 & \frac{8}{3} & -1 \end{pmatrix} $$ - Eliminate second column entries in Rows 3 and 4: Row 3 = Row 3 - 4*Row 2: $$ \begin{pmatrix} 0 & 4 - 4*1 & \frac{13}{3} - 4*0 & 3 - 4*(-\frac{3}{4}) \end{pmatrix} = \begin{pmatrix} 0 & 0 & \frac{13}{3} & 6 \end{pmatrix} $$ Row 4 = Row 4 + Row 2: $$ \begin{pmatrix} 0 & -1 + 1 & \frac{8}{3} + 0 & -1 + (-\frac{3}{4}) \end{pmatrix} = \begin{pmatrix} 0 & 0 & \frac{8}{3} & -\frac{7}{4} \end{pmatrix} $$ - Now matrix is: $$ \begin{pmatrix} 1 & 0 & -\frac{4}{3} & -1 \\ 0 & 1 & 0 & -\frac{3}{4} \\ 0 & 0 & \frac{13}{3} & 6 \\ 0 & 0 & \frac{8}{3} & -\frac{7}{4} \end{pmatrix} $$ - Make leading 1 in Row 3 by multiplying by $\frac{3}{13}$: $$ \begin{pmatrix} 0 & 0 & \frac{13}{3} * \frac{3}{13} & 6 * \frac{3}{13} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & \frac{18}{13} \end{pmatrix} $$ - Eliminate third column entry in Row 4: Row 4 = Row 4 - $\frac{8}{3}$ * Row 3: $$ \begin{pmatrix} 0 & 0 & \frac{8}{3} - \frac{8}{3}*1 & -\frac{7}{4} - \frac{8}{3}*\frac{18}{13} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -\frac{7}{4} - \frac{144}{39} \end{pmatrix} $$ Calculate: $$ -\frac{7}{4} - \frac{144}{39} = -\frac{7}{4} - \frac{144}{39} = -\frac{273}{156} - \frac{576}{156} = -\frac{849}{156} \neq 0 $$ - Since last row is $[0\ 0\ 0\ c]$ with $c \neq 0$, the system is inconsistent if augmented, but here we only consider $A$ for rank. - The last row has a pivot in the 4th column. - Eliminate third column entry in Row 1: Row 1 = Row 1 + $\frac{4}{3}$ * Row 3: $$ \begin{pmatrix} 1 & 0 & -\frac{4}{3} + \frac{4}{3}*1 & -1 + \frac{4}{3}*\frac{18}{13} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & -1 + \frac{24}{13} = -\frac{13}{13} + \frac{24}{13} = \frac{11}{13} \end{pmatrix} $$ - Final RREF matrix: $$ \begin{pmatrix} 1 & 0 & 0 & \frac{11}{13} \\ 0 & 1 & 0 & -\frac{3}{4} \\ 0 & 0 & 1 & \frac{18}{13} \\ 0 & 0 & 0 & -\frac{849}{156} \end{pmatrix} $$ 5. **Rank:** There are 4 pivots (leading 1s) in columns 1, 2, 3, and 4. Therefore, $\text{rank}(A) = 4$. 6. **Solution to $A\vec{x} = \vec{0}$:** Since $A$ is full rank (rank = number of variables), the only solution is the trivial solution: $$ \vec{x} = \vec{0} $$ **Summary:** - RREF of $A$ is as above. - Rank of $A$ is 4. - The homogeneous system $A\vec{x} = \vec{0}$ has only the trivial solution.