Problem 24 Find the rank of each of the fol... [FREE SOLUTION]

Chapter 3: Problem 24

Find the rank of each of the following matrices. $$ \left(\begin{array}{rrrr} 1 & 0 & 1 & 0 \\ -1 & -2 & -1 & 0 \\ 2 & 2 & 5 & 3 \\ 2 & 4 & 8 & 6 \end{array}\right) $$

Short Answer

Expert verified

The rank of the matrix is 3.

Step by step solution

- Write down the matrix

Consider the given matrix: \[ \left(\begin{array}{rrrr} 1 & 0 & 1 & 0 \ -1 & -2 & -1 & 0 \ 2 & 2 & 5 & 3 \ 2 & 4 & 8 & 6 \end{array}\right) \]

- Perform Row Reduction to Row Echelon Form (REF)

Start by converting the matrix into Row Echelon Form (REF) by performing elementary row operations. 1. Swap $R_1$ and $R_2$: \[ \left(\begin{array}{rrrr} -1 & -2 & -1 & 0 \ 1 & 0 & 1 & 0 \ 2 & 2 & 5 & 3 \ 2 & 4 & 8 & 6 \end{array}\right) \] 2. Multiply $R_1$ by -1: \[ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ -1 & 0 & -1 & 0 \ 2 & 2 & 5 & 3 \ 2 & 4 & 8 & 6 \end{array}\right) \] 3. Add $R_1$ to $R_2$: \[ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 2 & 0 & 0 \ 2 & 2 & 5 & 3 \ 2 & 4 & 8 & 6 \end{array}\right) \]

- Continue Row Reduction

4. Subtract 2*$R_1$ from $R_3$: \[ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 2 & 0 & 0 \ 0 & -2 & 3 & 3 \ 2 & 4 & 8 & 6 \end{array}\right) \] 5. Subtract 2*$R_1$ from $R_4$: \[ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 2 & 0 & 0 \ 0 & -2 & 3 & 3 \ 0 & 0 & 6 & 6 \end{array}\right) \]

- Simplify Rows and Identify Pivots

6. Multiply $R_2$ by $\frac{1}{2}$: \[ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & -2 & 3 & 3 \ 0 & 0 & 6 & 6 \end{array}\right) \] 7. Add 2*$R_2$ to $R_3$: \[ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 3 & 3 \ 0 & 0 & 6 & 6 \end{array}\right) \] 8. Subtract 2*$R_3$ from $R_4$: \[ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 3 & 3 \ 0 & 0 & 0 & 0 \end{array}\right) \]

- Determine the Rank

Identify the number of non-zero rows in the REF. Here, the non-zero rows are: $ \left(1, 2, 1, 0\right) $ $ \left(0, 1, 0, 0\right) $ $ \left(0, 0, 3, 3\right) $. Since there are 3 non-zero rows, the rank of the given matrix is 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Row Echelon Form

Row Echelon Form (REF) is a specific type of matrix that simplifies solving systems of linear equations. In REF, each leading entry (or pivot) of a row is 1, and all entries below each pivot are zeroes. Additionally, each leading 1 is to the right of the leading 1 in the row above it. Here's the given matrix converted to REF:
\
\ \
$ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 3 & 3 \ 0 & 0 & 0 & 0 \end{array}\right) $.

By converting a matrix to REF, you can easily determine its rank and solve related linear system problems.

Elementary Row Operations

To transform a matrix into its Row Echelon Form, we perform Elementary Row Operations (ERO). There are three types of ERO:

Swapping two rows.
Multiplying a row by a non-zero scalar.
Adding or subtracting a scalar multiple of one row to another row.

Let's apply these operations to our given matrix:

1. Swap $ R_1 $ and $ R_2 $:
$ \left(\begin{array}{rrrr} -1 & -2 & -1 & 0 \ 1 & 0 & 1 & 0 \ 2 & 2 & 5 & 3 \ 2 & 4 & 8 & 6 \end{array}\right) $

2. Multiply $ R_1 $ by -1:
$ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 1 & 0 & 1 & 0 \ 2 & 2 & 5 & 3 \ 2 & 4 & 8 & 6 \end{array}\right) $

3. Add $ R_1 $ to $ R_2 $:
$ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 2 & 0 & 0 \ 2 & 2 & 5 & 3 \ 2 & 4 & 8 & 6 \end{array}\right) $

Using these operations, you systematically reduce the matrix to its REF.

Non-zero Rows

After converting a matrix into Row Echelon Form, you can identify its rank by counting the number of non-zero rows. A non-zero row is a row that contains at least one non-zero element. In our example, after transformation, the matrix in REF is:

$ \left(\begin{array}{rrrr} 1 & 2 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 3 & 3 \ 0 & 0 & 0 & 0 \end{array}\right) $

The non-zero rows are:

$ \left(1, 2, 1, 0\right) $
$ \left(0, 1, 0, 0\right) $
$ \left(0, 0, 3, 3\right) $

Therefore, there are three non-zero rows. The rank of the matrix is the number of non-zero rows, which is 3 in this case.

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