In the process of finding the inverse of the following matrix, explain the indicated row operatic 3 -3 6 1 00 1 -3 10 0 1 0 -1 3 -5 0 0 1 Step 1: Switch row 1 and row 2. 1 3 Answer to Step 1: -1 Answer to Step 2: Step 2: Explain the row operation to change the answer from Step 1 to the result below. 1 -3 10 0 1 0 0 6 -24 1 -3 0 -5 0 0 1 -3 10 0 1 0 -3 6 1 0 0 3 -50 0 1 -1 3

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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what is step 2 doing

In the process of finding the inverse of the following matrix, explain the indicated row operation.

\[
\begin{bmatrix}
3 & -3 & 6 & \vline & 1 & 0 & 0 \\
1 & -3 & 10 & \vline & 0 & 1 & 0 \\
-1 & 3 & -5 & \vline & 0 & 0 & 1 \\
\end{bmatrix}
\]

**Step 1:** Switch row 1 and row 2.

**Answer to Step 1:**

\[
\begin{bmatrix}
1 & -3 & 10 & \vline & 0 & 1 & 0 \\
3 & -3 & 6 & \vline & 1 & 0 & 0 \\
-1 & 3 & -5 & \vline & 0 & 0 & 1 \\
\end{bmatrix}
\]

**Step 2:** Explain the row operation to change the answer from Step 1 to the result below.

\[
\begin{bmatrix}
1 & -3 & 10 & \vline & 0 & 1 & 0 \\
0 & 6 & -24 & \vline & 1 & -3 & 0 \\
-1 & 3 & -5 & \vline & 0 & 0 & 1 \\
\end{bmatrix}
\]

Here, a row operation is performed on row 2 (subtracting 3 times row 1 from row 2) to get a zero in the first column of the second row. This is part of the process of using Gaussian elimination to find the inverse of the matrix.
Transcribed Image Text:In the process of finding the inverse of the following matrix, explain the indicated row operation. \[ \begin{bmatrix} 3 & -3 & 6 & \vline & 1 & 0 & 0 \\ 1 & -3 & 10 & \vline & 0 & 1 & 0 \\ -1 & 3 & -5 & \vline & 0 & 0 & 1 \\ \end{bmatrix} \] **Step 1:** Switch row 1 and row 2. **Answer to Step 1:** \[ \begin{bmatrix} 1 & -3 & 10 & \vline & 0 & 1 & 0 \\ 3 & -3 & 6 & \vline & 1 & 0 & 0 \\ -1 & 3 & -5 & \vline & 0 & 0 & 1 \\ \end{bmatrix} \] **Step 2:** Explain the row operation to change the answer from Step 1 to the result below. \[ \begin{bmatrix} 1 & -3 & 10 & \vline & 0 & 1 & 0 \\ 0 & 6 & -24 & \vline & 1 & -3 & 0 \\ -1 & 3 & -5 & \vline & 0 & 0 & 1 \\ \end{bmatrix} \] Here, a row operation is performed on row 2 (subtracting 3 times row 1 from row 2) to get a zero in the first column of the second row. This is part of the process of using Gaussian elimination to find the inverse of the matrix.
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