A First Course in Linear Algebra · Problem M40

Systems of Linear Equations — Robert Beezer

Mathematics — Linear Algebra · row-equivalent-matrices-rrefu

Problem

Consider the two 3×43 \times 4 matrices below B=[13−22−1−2−1−1−1−58−3]C=[12121140−1−1−41]B = \begin{bmatrix} 1 & 3 & -2 & 2 \\ -1 & -2 & -1 & -1 \\ -1 & -5 & 8 & -3 \end{bmatrix} \qquad C = \begin{bmatrix} 1 & 2 & 1 & 2 \\ 1 & 1 & 4 & 0 \\ -1 & -1 & -4 & 1 \end{bmatrix} 1. Row-reduce each matrix and determine that the reduced row-echelon forms of BB and CC are identical. From this argue that BB and CC 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 BB from row ii that are in column jj are linearly related to the entries of CC in column jj from all three rows [B]ij=δi1[C]1j+δi2[C]2j+δi3[C]3j1≤j≤4[B]_{ij} = \delta_{i1}[C]_{1j} + \delta_{i2}[C]_{2j} + \delta_{i3}[C]_{3j} \qquad 1 \leq j \leq 4 For each 1≤i≤31 \leq i \leq 3 find the corresponding three scalars in this relationship. So your answer will be nine scalars, determined three at a time.

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