Question

1. Which of the following elementary matrix corresponds to the elementary row operation of adding 2 times row 2 to row 4 applied to a system with 4 equations?

   1. \(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\)

   2. \(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ \end{bmatrix}\)

   3. \(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}\)

   The answer is b.

   Recall that you can obtain the elementary matrix for adding 2 times row 2 to row 4 by starting with the identity matrix (of size 4 in this case) and add 2 times row to 2 row 4.

2. Let \(A\) be a \(2\times 2\) matrix. Suppose that \(B\) is obtained from \(A\) by first multiplying the first row by \(2\) and then adding \(3\) times the first row of the resulting matrix to the second row. That is, \( A \stackrel{R_1 \leftarrow 2R_1}{\longrightarrow} A' \stackrel{R_2 \leftarrow R_2 + 3R_1}{\longrightarrow} B .\) Let \(P\) be the elementary matrix for the operation \(R_1 \leftarrow 2R_1\) and let \(Q\) be the elementary matrix for the operation \(R_2 \leftarrow R_2 + 3R_1\). Which of the following products is equal to \(B\)?

   1. \(PQA\)

   2. \(QPA\)

   The answer is b.

   Note that applying \(R_1 \leftarrow 2R_1\) to \(A\) is the same as forming the product \(PA\). Hence, \(A' = PA\). Now applying \(R_2 \leftarrow R_2 + 3R_1\) to \(A'\) is the same as \(QA'\). But \(B = QA' \). Hence, \(B = QA' = QPA\).