1.	Evaluate the Rank of the matrix by reducing it to the Normal form(2,3,1,−1)(−1,−1,−2,−43 )(16 ,33,−20,−7)

To determine the rank of the matrix, we can reduce it to its row echelon form (or normal form). The rank of a matrix is defined as the number of linearly independent rows (or columns), which corresponds to the number of non-zero rows in the row echelon form.

Let's start by writing down the matrix:

\[
\begin{bmatrix}
2 & 3 & 1 & -1 \\
-1 & -1 & -2 & -43 \\
16 & 33 & -20 & -7
\end{bmatrix}
\]

We’ll perform row operations to reduce this matrix to row echelon form.

### Step-by-Step Solution

#### Step 1: Make the first element of the first row 1
Divide the first row by 2:
\[
\begin{bmatrix}
1 & 1.5 & 0.5 & -0.5 \\
-1 & -1 & -2 & -43 \\
16 & 33 & -20 & -7
\end{bmatrix}
\]

#### Step 2: Make the elements below the leading 1 in column 1 equal to 0
Add the first row to the second row:
\[
\begin{bmatrix}
1 & 1.5 & 0.5 & -0.5 \\
0 & 0.5 & -1.5 & -43.5 \\
16 & 33 & -20 & -7
\end{bmatrix}
\]

Subtract 16 times the first row from the third row:
\[
\begin{bmatrix}
1 & 1.5 & 0.5 & -0.5 \\
0 & 0.5 & -1.5 & -43.5 \\
0 & 9 & -28 & 1
\end{bmatrix}
\]

#### Step 3: Make the second pivot element equal to 1
Divide the second row by 0.5:
\[
\begin{bmatrix}
1 & 1.5 & 0.5 & -0.5 \\
0 & 1 & -3 & -87 \\
0 & 9 & -28 & 1
\end{bmatrix}
\]

#### Step 4: Make elements above and below the pivot in column 2 equal to 0
Subtract 1.5 times the second row from the first row:
\[
\begin{bmatrix}
1 & 0 & 5 & 130 \\
0 & 1 & -3 & -87 \\
0 & 9 & -28 & 1
\end{bmatrix}
\]

Subtract 9 times the second row from the third row:
\[
\begin{bmatrix}
1 & 0 & 5 & 130 \\
0 & 1 & -3 & -87 \\
0 & 0 & -1 & 784
\end{bmatrix}
\]

#### Step 5: Make the third pivot element equal to 1
Multiply the third row by -1:
\[
\begin{bmatrix}
1 & 0 & 5 & 130 \\
0 & 1 & -3 & -87 \\
0 & 0 & 1 & -784
\end{bmatrix}
\]

#### Step 6: Make elements above the pivot in column 3 equal to 0
Add 3 times the third row to the second row:
\[
\begin{bmatrix}
1 & 0 & 5 & 130 \\
0 & 1 & 0 & -2339 \\
0 & 0 & 1 & -784
\end{bmatrix}
\]

Subtract 5 times the third row from the first row:
\[
\begin{bmatrix}
1 & 0 & 0 & 4050 \\
0 & 1 & 0 & -2339 \\
0 & 0 & 1 & -784
\end{bmatrix}
\]

### Conclusion
The matrix is now in row echelon form, and each row has a leading 1 in a unique column. Since there are three non-zero rows, the rank of the matrix is **3**.

Let me know if you'd like further details on this solution or have any questions!

---

### Relative Questions
1. How do you find the rank of a matrix using column operations?
2. What is the rank of a zero matrix?
3. Can a matrix have a rank greater than its smallest dimension?
4. How does the determinant relate to the rank of a matrix?
5. What are the practical applications of finding a matrix's rank?

---

**Tip:** The rank of a matrix can also be determined by counting the non-zero rows in its reduced row echelon form (RREF).

Evaluate the Rank of the matrix by reducing it to the Normal form (2,3,1,−1) (−1,−1,−2,−43) (16,33,−20,−7)

Learn how to evaluate the rank of a matrix by reducing it to row echelon form. This step-by-step solution explains row operations, matrix transformations, and the concept of rank in linear algebra. Suitable for high school and college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation