Consider the two $3\times 4$ matrices below \begin{align*} B&= \begin{bmatrix} 1 & 3 & -2 & 2 \\ -1 & -2 & -1 & -1 \\ -1 & -5 & 8 & -3 \end{bmatrix} & C&= \begin{bmatrix} 1 & 2 & 1 & 2 \\ 1 & 1 & 4 & 0 \\ -1 & -1 & -4 & 1 \end{bmatrix} \end{align*}

1. Row-reduce each matrix and determine that the reduced row-echelon forms of $B$ and $C$ are identical. From this argue that $B$ and $C$ are row-equivalent.
2. In the proof of Theorem RREFU, we begin by arguing that entries of row-equivalent matrices are related by way of certain scalars and sums. In this example, we would write that entries of $B$ from row $i$ that are in column $j$ are linearly related to the entries of $C$ in column $j$ from all three rows \begin{align*} \matrixentry{B}{ij} &= \delta_{i1}\matrixentry{C}{1j}+ \delta_{i2}\matrixentry{C}{2j}+ \delta_{i3}\matrixentry{C}{3j} & 1&\leq j\leq 4 \end{align*} For each $1\leq i\leq 3$ find the corresponding three scalars in this relationship. So your answer will be nine scalars, determined three at a time.