https://file.aiquickdraw.com/mathbot/file/1729227636432sabomiqm.jpg

The problem in the image asks to reduce the given matrix to its normal form. Here's the matrix from the image:

\[
\begin{pmatrix}
1 & 2 & 3 & -1 \\
-1 & -1 & -3 & -1 \\
0 & 1 & 1 & 1 \\
0 & 1 & -1 & 1
\end{pmatrix}
\]

To reduce this matrix to its normal form (likely referring to Row Echelon Form or Reduced Row Echelon Form), we can apply Gaussian or Gauss-Jordan elimination. Let’s proceed with the steps.

### Step 1: Use the first row to eliminate below it
- The pivot element is in the first row, first column (1).
- Add row 1 to row 2 to eliminate the -1 in row 2, column 1.

\[
R_2 \rightarrow R_2 + R_1 = (-1 + 1), (-1 + 2), (-3 + 3), (-1 + (-1)) = (0, 1, 0, -2)
\]

The matrix becomes:

\[
\begin{pmatrix}
1 & 2 & 3 & -1 \\
0 & 1 & 0 & -2 \\
0 & 1 & 1 & 1 \\
0 & 1 & -1 & 1
\end{pmatrix}
\]

### Step 2: Use the second row to eliminate below it
- The pivot is now in row 2, column 2.
- Subtract row 2 from rows 3 and 4 to eliminate the 1's below.

For row 3:

\[
R_3 \rightarrow R_3 - R_2 = (0 - 0), (1 - 1), (1 - 0), (1 - (-2)) = (0, 0, 1, 3)
\]

For row 4:

\[
R_4 \rightarrow R_4 - R_2 = (0 - 0), (1 - 1), (-1 - 0), (1 - (-2)) = (0, 0, -1, 3)
\]

The matrix now becomes:

\[
\begin{pmatrix}
1 & 2 & 3 & -1 \\
0 & 1 & 0 & -2 \\
0 & 0 & 1 & 3 \\
0 & 0 & -1 & 3
\end{pmatrix}
\]

### Step 3: Use the third row to eliminate below it
- The pivot is now in row 3, column 3.
- Add row 3 to row 4 to eliminate the -1 in row 4, column 3.

\[
R_4 \rightarrow R_4 + R_3 = (0 + 0), (0 + 0), (-1 + 1), (3 + 3) = (0, 0, 0, 6)
\]

The matrix becomes:

\[
\begin{pmatrix}
1 & 2 & 3 & -1 \\
0 & 1 & 0 & -2 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & 6
\end{pmatrix}
\]

### Step 4: Normalize the last row
- Divide the last row by 6 to make the pivot 1.

\[
R_4 \rightarrow \frac{1}{6} R_4 = (0, 0, 0, 1)
\]

The matrix now is:

\[
\begin{pmatrix}
1 & 2 & 3 & -1 \\
0 & 1 & 0 & -2 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & 1
\end{pmatrix}
\]

### Step 5: Eliminate above pivots to get reduced row echelon form
Now, we backtrack to clear the values above each pivot:

- Clear the values above the pivot in column 3 by subtracting 3 times row 3 from row 1.

\[
R_1 \rightarrow R_1 - 3R_3 = (1 - 3(0)), (2 - 3(0)), (3 - 3(1)), (-1 - 3(3)) = (1, 2, 0, -10)
\]

- Clear the value above the pivot in column 4 by adding row 4 to row 1 and subtracting -2 times row 4 from row 2.

For row 1:

\[
R_1 \rightarrow R_1 + R_4 = (1 + 0), (2 + 0), (0 + 0), (-10 + 1) = (1, 2, 0, -9)
\]

For row 2:

\[
R_2 \rightarrow R_2 + 2R_4 = (0 + 0), (1 + 0), (0 + 0), (-2 + 2) = (0, 1, 0, 0)
\]

### Final Matrix in Reduced Row Echelon Form:
\[
\begin{pmatrix}
1 & 2 & 0 & -9 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & 1
\end{pmatrix}
\]

This is the matrix in reduced row echelon form (normal form).

Would you like a deeper explanation of any step?

### Related Questions:
1. How do you perform Gaussian elimination on any matrix?
2. What are the differences between Row Echelon Form and Reduced Row Echelon Form?
3. How can matrices be used to solve systems of linear equations?
4. What are the properties of an invertible matrix?
5. What is the significance of pivot elements in matrix reduction?

### Tip:
Always aim to simplify row operations by choosing pivot elements carefully to avoid fractions during calculations when possible.

Reduce the following matrix to its normal form:
[ [1, 2, 3, -1], [-1, -1, -3, -1], [0, 1, 1, 1], [0, 1, -1, 1] ]

Learn how to reduce a given matrix to its normal form through Gaussian elimination. This problem walks through key linear algebra concepts like Row Echelon Form and Reduced Row Echelon Form. Suitable for students in advanced high school or undergraduate courses.

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