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Sagot :
Let's focus on exercise 26:
Given the matrix
[tex]\[ A = \begin{pmatrix} 1 & -2 \\ -3 & 6 \end{pmatrix} \][/tex]
To find the null space (the set of vectors [tex]\(x\)[/tex] such that [tex]\(Ax=0\)[/tex]) and the range (the column space of [tex]\(A\)[/tex]), we can proceed step by step as follows:
### Finding the Null Space
The null space of a matrix [tex]\(A\)[/tex] consists of all vectors [tex]\(x\)[/tex] for which [tex]\(Ax = 0\)[/tex]. Let's denote [tex]\( x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \)[/tex]. Then we solve:
[tex]\[ A \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ -3 & 6 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
This gives us a system of linear equations:
[tex]\[ 1x_1 - 2x_2 = 0 \][/tex]
[tex]\[ -3x_1 + 6x_2 = 0 \][/tex]
Both equations are essentially the same when simplified; the second is just the first multiplied by [tex]\(-3\)[/tex]. So, we can work with the first equation:
[tex]\[ x_1 - 2x_2 = 0 \][/tex]
[tex]\[ x_1 = 2x_2 \][/tex]
This tells us that all vectors in the null space are of the form:
[tex]\[ \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 2x_2 \\ x_2 \end{pmatrix} = x_2 \begin{pmatrix} 2 \\ 1 \end{pmatrix} \][/tex]
So, the null space of [tex]\(A\)[/tex] is spanned by the vector [tex]\( \begin{pmatrix} 2 \\ 1 \end{pmatrix} \)[/tex]. Normalizing this vector, we get:
[tex]\[ \begin{pmatrix} 2 \\ 1 \end{pmatrix} = c \begin{pmatrix} 0.89442719 \\ 0.4472136 \end{pmatrix} \][/tex]
where [tex]\(c\)[/tex] is a scalar.
### Finding the Range
The range (also known as the column space) of a matrix is the span of its column vectors. For matrix [tex]\( A \)[/tex], the columns are:
[tex]\[ \begin{pmatrix} 1 \\ -3 \end{pmatrix} \text{ and } \begin{pmatrix} -2 \\ 6 \end{pmatrix} \][/tex]
Both of these vectors are linearly dependent because [tex]\( \begin{pmatrix} -2 \\ 6 \end{pmatrix} \)[/tex] is a multiple of [tex]\(\begin{pmatrix} 1 \\ -3 \end{pmatrix} \)[/tex].
Thus, the range of [tex]\( A \)[/tex] is spanned by:
[tex]\[ \begin{pmatrix} 1 \\ -3 \end{pmatrix} \][/tex]
### Summary
- The null space of [tex]\( A \)[/tex] is spanned by the vector [tex]\( \begin{pmatrix} 0.89442719 \\ 0.4472136 \end{pmatrix} \)[/tex].
- The range of [tex]\( A \)[/tex] is spanned by the vector [tex]\( \begin{pmatrix} 1 \\ -3 \end{pmatrix} \)[/tex].
Lastly, the rank of matrix [tex]\( A \)[/tex], which is the dimension of the range, is [tex]\( 1 \)[/tex]. This indicates that there is only one independent column vector in [tex]\( A \)[/tex].
Given the matrix
[tex]\[ A = \begin{pmatrix} 1 & -2 \\ -3 & 6 \end{pmatrix} \][/tex]
To find the null space (the set of vectors [tex]\(x\)[/tex] such that [tex]\(Ax=0\)[/tex]) and the range (the column space of [tex]\(A\)[/tex]), we can proceed step by step as follows:
### Finding the Null Space
The null space of a matrix [tex]\(A\)[/tex] consists of all vectors [tex]\(x\)[/tex] for which [tex]\(Ax = 0\)[/tex]. Let's denote [tex]\( x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \)[/tex]. Then we solve:
[tex]\[ A \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ -3 & 6 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
This gives us a system of linear equations:
[tex]\[ 1x_1 - 2x_2 = 0 \][/tex]
[tex]\[ -3x_1 + 6x_2 = 0 \][/tex]
Both equations are essentially the same when simplified; the second is just the first multiplied by [tex]\(-3\)[/tex]. So, we can work with the first equation:
[tex]\[ x_1 - 2x_2 = 0 \][/tex]
[tex]\[ x_1 = 2x_2 \][/tex]
This tells us that all vectors in the null space are of the form:
[tex]\[ \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 2x_2 \\ x_2 \end{pmatrix} = x_2 \begin{pmatrix} 2 \\ 1 \end{pmatrix} \][/tex]
So, the null space of [tex]\(A\)[/tex] is spanned by the vector [tex]\( \begin{pmatrix} 2 \\ 1 \end{pmatrix} \)[/tex]. Normalizing this vector, we get:
[tex]\[ \begin{pmatrix} 2 \\ 1 \end{pmatrix} = c \begin{pmatrix} 0.89442719 \\ 0.4472136 \end{pmatrix} \][/tex]
where [tex]\(c\)[/tex] is a scalar.
### Finding the Range
The range (also known as the column space) of a matrix is the span of its column vectors. For matrix [tex]\( A \)[/tex], the columns are:
[tex]\[ \begin{pmatrix} 1 \\ -3 \end{pmatrix} \text{ and } \begin{pmatrix} -2 \\ 6 \end{pmatrix} \][/tex]
Both of these vectors are linearly dependent because [tex]\( \begin{pmatrix} -2 \\ 6 \end{pmatrix} \)[/tex] is a multiple of [tex]\(\begin{pmatrix} 1 \\ -3 \end{pmatrix} \)[/tex].
Thus, the range of [tex]\( A \)[/tex] is spanned by:
[tex]\[ \begin{pmatrix} 1 \\ -3 \end{pmatrix} \][/tex]
### Summary
- The null space of [tex]\( A \)[/tex] is spanned by the vector [tex]\( \begin{pmatrix} 0.89442719 \\ 0.4472136 \end{pmatrix} \)[/tex].
- The range of [tex]\( A \)[/tex] is spanned by the vector [tex]\( \begin{pmatrix} 1 \\ -3 \end{pmatrix} \)[/tex].
Lastly, the rank of matrix [tex]\( A \)[/tex], which is the dimension of the range, is [tex]\( 1 \)[/tex]. This indicates that there is only one independent column vector in [tex]\( A \)[/tex].
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