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Sagot :
To solve for the matrix [tex]\( X \)[/tex] in the equation [tex]\( X = AX + B \)[/tex], where
[tex]\[ A = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix} \][/tex]
and
[tex]\[ B = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & 3 \end{bmatrix}, \][/tex]
we will go through the following steps:
1. Rewrite the equation:
The given equation is [tex]\( X = AX + B \)[/tex]. We can rewrite this as:
[tex]\[ X - AX = B \][/tex]
2. Factor out [tex]\( X \)[/tex]:
Rewriting [tex]\( X - AX \)[/tex] in matrix form involves factoring out [tex]\( X \)[/tex]:
[tex]\[ X(I - A) = B \][/tex]
Here, [tex]\( I \)[/tex] is the identity matrix of the same dimension as [tex]\( A \)[/tex]:
[tex]\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
3. Calculate [tex]\( I - A \)[/tex]:
Subtract [tex]\( A \)[/tex] from [tex]\( I \)[/tex]:
[tex]\[ I - A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
4. Find the inverse of [tex]\( I - A \)[/tex]:
We need to find the inverse of [tex]\( I - A \)[/tex]:
[tex]\[ (I - A)^{-1} \][/tex]
The inverse of [tex]\( I - A \)[/tex] is calculated as follows. Let's denote [tex]\( I - A \)[/tex] as [tex]\( C \)[/tex]:
[tex]\[ C = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
We can find the inverse of matrix [tex]\( C \)[/tex] using standard formulae, row reduction methods, or any computational aid.
For the matrix [tex]\( C \)[/tex], the inverse [tex]\( C^{-1} \)[/tex] is given by:
[tex]\[ C^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
5. Compute [tex]\( X \)[/tex]:
Now, substituting into the equation [tex]\( X = B (I - A)^{-1} \)[/tex]:
[tex]\[ X = B C^{-1} \][/tex]
Substituting the values of [tex]\( B \)[/tex] and [tex]\( C^{-1} \)[/tex]:
[tex]\[ B = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & 3 \end{bmatrix}, \quad C^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
Performing the matrix multiplication [tex]\( B C^{-1} \)[/tex]:
[tex]\[ X = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + 2 \cdot 0 + 0 \cdot 0 & 1 \cdot (-1) + 2 \cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 2 \cdot (-1) + 0 \cdot 1 \\ 2 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 & 2 \cdot (-1) + 1 \cdot 1 + 0 \cdot 0 & 2 \cdot 1 + 1 \cdot (-1) + 0 \cdot 1 \\ 3 \cdot 1 + 3 \cdot 0 + 0 \cdot 0 & 3 \cdot (-1) + 3 \cdot 1 + 0 \cdot 0 & 3 \cdot 1 + 3 \cdot (-1) + 0 \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 0 & 0 \end{bmatrix} \][/tex]
Thus, the solution matrix [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{bmatrix} 1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 0 & 0 \end{bmatrix} \][/tex]
[tex]\[ A = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix} \][/tex]
and
[tex]\[ B = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & 3 \end{bmatrix}, \][/tex]
we will go through the following steps:
1. Rewrite the equation:
The given equation is [tex]\( X = AX + B \)[/tex]. We can rewrite this as:
[tex]\[ X - AX = B \][/tex]
2. Factor out [tex]\( X \)[/tex]:
Rewriting [tex]\( X - AX \)[/tex] in matrix form involves factoring out [tex]\( X \)[/tex]:
[tex]\[ X(I - A) = B \][/tex]
Here, [tex]\( I \)[/tex] is the identity matrix of the same dimension as [tex]\( A \)[/tex]:
[tex]\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
3. Calculate [tex]\( I - A \)[/tex]:
Subtract [tex]\( A \)[/tex] from [tex]\( I \)[/tex]:
[tex]\[ I - A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
4. Find the inverse of [tex]\( I - A \)[/tex]:
We need to find the inverse of [tex]\( I - A \)[/tex]:
[tex]\[ (I - A)^{-1} \][/tex]
The inverse of [tex]\( I - A \)[/tex] is calculated as follows. Let's denote [tex]\( I - A \)[/tex] as [tex]\( C \)[/tex]:
[tex]\[ C = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
We can find the inverse of matrix [tex]\( C \)[/tex] using standard formulae, row reduction methods, or any computational aid.
For the matrix [tex]\( C \)[/tex], the inverse [tex]\( C^{-1} \)[/tex] is given by:
[tex]\[ C^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
5. Compute [tex]\( X \)[/tex]:
Now, substituting into the equation [tex]\( X = B (I - A)^{-1} \)[/tex]:
[tex]\[ X = B C^{-1} \][/tex]
Substituting the values of [tex]\( B \)[/tex] and [tex]\( C^{-1} \)[/tex]:
[tex]\[ B = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & 3 \end{bmatrix}, \quad C^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
Performing the matrix multiplication [tex]\( B C^{-1} \)[/tex]:
[tex]\[ X = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + 2 \cdot 0 + 0 \cdot 0 & 1 \cdot (-1) + 2 \cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 2 \cdot (-1) + 0 \cdot 1 \\ 2 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 & 2 \cdot (-1) + 1 \cdot 1 + 0 \cdot 0 & 2 \cdot 1 + 1 \cdot (-1) + 0 \cdot 1 \\ 3 \cdot 1 + 3 \cdot 0 + 0 \cdot 0 & 3 \cdot (-1) + 3 \cdot 1 + 0 \cdot 0 & 3 \cdot 1 + 3 \cdot (-1) + 0 \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 0 & 0 \end{bmatrix} \][/tex]
Thus, the solution matrix [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{bmatrix} 1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 0 & 0 \end{bmatrix} \][/tex]
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