1. Problem statement: Find the rank of the first 4x4 matrix and determine the type of solutions of the homogeneous system $Ax=0$ for that matrix. 2. Formula and rules: The rank of a matrix is the number of pivot (nonzero) rows in its RREF. If the rank equals the number of unknowns $n$ there is only the trivial solution. If the rank is less than $n$ there are infinitely many solutions. 3. Start with the given matrix: $$A=\begin{bmatrix}0 & 4 & 0 & -3\\ 3 & 0 & -4 & -3\\ 1 & 4 & 3 & 2\\ 2 & -1 & 0 & -3\end{bmatrix}$$ 4. Swap $R_1$ and $R_2$ to get a nonzero pivot in row 1: $$\begin{bmatrix}3 & 0 & -4 & -3\\ 0 & 4 & 0 & -3\\ 1 & 4 & 3 & 2\\ 2 & -1 & 0 & -3\end{bmatrix}$$ 5. Scale $R_1$ by $1/3$ to make the leading entry $1$. $$\frac{1}{3}\cdot 3 = \frac{\cancel{3}}{\cancel{3}} = 1$$ $$R_1:=\frac{1}{3}R_1=\begin{bmatrix}1 & 0 & -\frac{4}{3} & -1\\ 0 & 4 & 0 & -3\\ 1 & 4 & 3 & 2\\ 2 & -1 & 0 & -3\end{bmatrix}$$ 6. Eliminate entries below the pivot in column 1: $R_3\leftarrow R_3-R_1$, $R_4\leftarrow R_4-2R_1$. $$R_3\leftarrow R_3-R_1=\begin{bmatrix}0 & 4 & \frac{13}{3} & 3\end{bmatrix}$$ $$R_4\leftarrow R_4-2R_1=\begin{bmatrix}0 & -1 & \frac{8}{3} & -1\end{bmatrix}$$ Matrix becomes: $$\begin{bmatrix}1 & 0 & -\frac{4}{3} & -1\\ 0 & 4 & 0 & -3\\ 0 & 4 & \frac{13}{3} & 3\\ 0 & -1 & \frac{8}{3} & -1\end{bmatrix}$$ 7. Scale $R_2$ by $1/4$ to get a pivot $1$ in column 2. $$\frac{1}{4}\cdot 4 = \frac{\cancel{4}}{\cancel{4}} = 1$$ $$R_2:=\frac{1}{4}R_2=\begin{bmatrix}0 & 1 & 0 & -\frac{3}{4}\end{bmatrix}$$ 8. Eliminate column 2 entries: $R_3\leftarrow R_3-4R_2$, $R_4\leftarrow R_4+R_2$. $$R_3\leftarrow R_3-4R_2=\begin{bmatrix}0 & 0 & \frac{13}{3} & 6\end{bmatrix}$$ $$R_4\leftarrow R_4+R_2=\begin{bmatrix}0 & 0 & \frac{8}{3} & -\frac{7}{4}\end{bmatrix}$$ Matrix now is: $$\begin{bmatrix}1 & 0 & -\frac{4}{3} & -1\\ 0 & 1 & 0 & -\frac{3}{4}\\ 0 & 0 & \frac{13}{3} & 6\\ 0 & 0 & \frac{8}{3} & -\frac{7}{4}\end{bmatrix}$$ 9. Scale $R_3$ by $3/13$ to make the pivot in column 3 equal to $1$. $$\frac{3}{13}\cdot \frac{13}{3} = \frac{\cancel{13}}{\cancel{13}}\cdot\frac{\cancel{3}}{\cancel{3}} = 1$$ $$R_3:=\frac{3}{13}R_3=\begin{bmatrix}0 & 0 & 1 & \frac{18}{13}\end{bmatrix}$$ 10. Eliminate the 3rd entry of $R_4$. First compute the factor: $$\frac{\frac{8}{3}}{\frac{13}{3}}=\frac{8}{\cancel{3}}\cdot\frac{\cancel{3}}{13}=\frac{8}{13}$$ Then $R_4\leftarrow R_4-\frac{8}{13}R_3$, giving $$R_4=\begin{bmatrix}0 & 0 & 0 & -\frac{283}{52}\end{bmatrix}$$ 11. Now there are nonzero pivots in columns 1, 2, 3 and 4, so there are four pivots. Therefore the rank of the first matrix is 4. 12. Conclusion: Rank $=4$ and because rank equals the number of unknowns $4$, the homogeneous system $Ax=0$ has only the trivial solution $x=0$.