Answer :
### Exercise 25
Given matrix [tex]\( A = \begin{pmatrix} 1 & 2 & 1 \\ -1 & 0 & 3 \\ 1 & 5 & 7 \end{pmatrix} \)[/tex]:
1. Find the Row Reduced Echelon Form (RREF):
[tex]\[ \text{RREF} = \begin{pmatrix} 1 & 0 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix} \][/tex]
2. Identify Pivot Columns:
The pivot columns are the first and the second columns.
3. Determine the Basis for the Column Space:
The basis for the column space, [tex]\(R(A)\)[/tex], is formed by taking the corresponding columns of [tex]\(A\)[/tex].
[tex]\[ \text{Basis for } R(A) = \left\{ \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} \right\} \][/tex]
4. Calculate the Rank:
The rank of [tex]\(A\)[/tex] is the number of pivot columns.
[tex]\[ \text{Rank}(A) = 2 \][/tex]
5. Calculate the Nullity:
The nullity is given by the number of columns minus the rank.
[tex]\[ \text{Nullity}(A) = 3 - 2 = 1 \][/tex]
### Summary for Exercise 25:
- Basis for [tex]\( R(A) \)[/tex]: [tex]\(\left\{ \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} \right\}\)[/tex]
- Rank: [tex]\( 2 \)[/tex]
- Nullity: [tex]\( 1 \)[/tex]
---
### Exercise 26
Given matrix [tex]\( A = \begin{pmatrix} 1 & 1 & 2 & 0 \\ 2 & 4 & 2 & 4 \\ 2 & 1 & 5 & -2 \end{pmatrix} \)[/tex]:
1. Find the Row Reduced Echelon Form (RREF):
[tex]\[ \text{RREF} = \begin{pmatrix} 1 & 0 & 3 & -2 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix} \][/tex]
2. Identify Pivot Columns:
The pivot columns are the first and the second columns.
3. Determine the Basis for the Column Space:
The basis for the column space, [tex]\(R(A)\)[/tex], is formed by taking the corresponding columns of [tex]\(A\)[/tex].
[tex]\[ \text{Basis for } R(A) = \left\{ \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 1 \end{pmatrix} \right\} \][/tex]
4. Calculate the Rank:
The rank of [tex]\(A\)[/tex] is the number of pivot columns.
[tex]\[ \text{Rank}(A) = 2 \][/tex]
5. Calculate the Nullity:
The nullity is given by the number of columns minus the rank.
[tex]\[ \text{Nullity}(A) = 4 - 2 = 2 \][/tex]
### Summary for Exercise 26:
- Basis for [tex]\( R(A) \)[/tex]: [tex]\(\left\{ \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 1 \end{pmatrix} \right\}\)[/tex]
- Rank: [tex]\( 2 \)[/tex]
- Nullity: [tex]\( 2 \)[/tex]
Given matrix [tex]\( A = \begin{pmatrix} 1 & 2 & 1 \\ -1 & 0 & 3 \\ 1 & 5 & 7 \end{pmatrix} \)[/tex]:
1. Find the Row Reduced Echelon Form (RREF):
[tex]\[ \text{RREF} = \begin{pmatrix} 1 & 0 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix} \][/tex]
2. Identify Pivot Columns:
The pivot columns are the first and the second columns.
3. Determine the Basis for the Column Space:
The basis for the column space, [tex]\(R(A)\)[/tex], is formed by taking the corresponding columns of [tex]\(A\)[/tex].
[tex]\[ \text{Basis for } R(A) = \left\{ \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} \right\} \][/tex]
4. Calculate the Rank:
The rank of [tex]\(A\)[/tex] is the number of pivot columns.
[tex]\[ \text{Rank}(A) = 2 \][/tex]
5. Calculate the Nullity:
The nullity is given by the number of columns minus the rank.
[tex]\[ \text{Nullity}(A) = 3 - 2 = 1 \][/tex]
### Summary for Exercise 25:
- Basis for [tex]\( R(A) \)[/tex]: [tex]\(\left\{ \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} \right\}\)[/tex]
- Rank: [tex]\( 2 \)[/tex]
- Nullity: [tex]\( 1 \)[/tex]
---
### Exercise 26
Given matrix [tex]\( A = \begin{pmatrix} 1 & 1 & 2 & 0 \\ 2 & 4 & 2 & 4 \\ 2 & 1 & 5 & -2 \end{pmatrix} \)[/tex]:
1. Find the Row Reduced Echelon Form (RREF):
[tex]\[ \text{RREF} = \begin{pmatrix} 1 & 0 & 3 & -2 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix} \][/tex]
2. Identify Pivot Columns:
The pivot columns are the first and the second columns.
3. Determine the Basis for the Column Space:
The basis for the column space, [tex]\(R(A)\)[/tex], is formed by taking the corresponding columns of [tex]\(A\)[/tex].
[tex]\[ \text{Basis for } R(A) = \left\{ \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 1 \end{pmatrix} \right\} \][/tex]
4. Calculate the Rank:
The rank of [tex]\(A\)[/tex] is the number of pivot columns.
[tex]\[ \text{Rank}(A) = 2 \][/tex]
5. Calculate the Nullity:
The nullity is given by the number of columns minus the rank.
[tex]\[ \text{Nullity}(A) = 4 - 2 = 2 \][/tex]
### Summary for Exercise 26:
- Basis for [tex]\( R(A) \)[/tex]: [tex]\(\left\{ \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 1 \end{pmatrix} \right\}\)[/tex]
- Rank: [tex]\( 2 \)[/tex]
- Nullity: [tex]\( 2 \)[/tex]
