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Sagot :
Absolutely, let's walk through the solution step-by-step, but we'll start by clearly outlining how we approach such a problem.
### a. Write the linear system as a matrix equation in the form [tex]\( AX = B \)[/tex].
The given system of equations is:
[tex]\[ \begin{array}{rr} x - y + z &= 0 \\ -2y + z &= -3 \\ -3x - 4y &= -8 \end{array} \][/tex]
We can rewrite this system in matrix form as [tex]\( AX = B \)[/tex], where:
[tex]\[ A = \begin{bmatrix} 1 & -1 & 1 \\ 0 & -2 & 1 \\ -3 & -4 & 0 \end{bmatrix} \][/tex]
is the coefficient matrix,
[tex]\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \][/tex]
is the vector of variables, and
[tex]\[ B = \begin{bmatrix} 0 \\ -3 \\ -8 \end{bmatrix} \][/tex]
is the constants vector.
So the matrix equation [tex]\( AX = B \)[/tex] is:
[tex]\[ \begin{bmatrix} 1 & -1 & 1 \\ 0 & -2 & 1 \\ -3 & -4 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ -3 \\ -8 \end{bmatrix}. \][/tex]
### b. Solve the system using the inverse that is given for the coefficient matrix.
We are given the inverse of the coefficient matrix [tex]\( A \)[/tex], which is:
[tex]\[ A^{-1} = \begin{bmatrix} 4 & -4 & 1 \\ -3 & 3 & -1 \\ -6 & 7 & -2 \end{bmatrix} \][/tex]
To find the solution vector [tex]\( X \)[/tex], we multiply both sides of the equation [tex]\( AX = B \)[/tex] by [tex]\( A^{-1} \)[/tex]:
[tex]\[ A^{-1}AX = A^{-1}B \][/tex]
Since [tex]\( A^{-1}A \)[/tex] is the identity matrix [tex]\( I \)[/tex], this simplifies to:
[tex]\[ X = A^{-1}B \][/tex]
We will now perform the matrix multiplication [tex]\( A^{-1}B \)[/tex]:
[tex]\[ X = \begin{bmatrix} 4 & -4 & 1 \\ -3 & 3 & -1 \\ -6 & 7 & -2 \end{bmatrix} \begin{bmatrix} 0 \\ -3 \\ -8 \end{bmatrix} \][/tex]
Perform the multiplication for each element in [tex]\( X \)[/tex]:
1. For the first element of [tex]\( X \)[/tex]:
[tex]\[ 4 \cdot 0 + (-4) \cdot (-3) + 1 \cdot (-8) = 0 + 12 - 8 = 4 \][/tex]
2. For the second element of [tex]\( X \)[/tex]:
[tex]\[ -3 \cdot 0 + 3 \cdot (-3) + (-1) \cdot (-8) = 0 - 9 + 8 = -1 \][/tex]
3. For the third element of [tex]\( X \)[/tex]:
[tex]\[ -6 \cdot 0 + 7 \cdot (-3) + (-2) \cdot (-8) = 0 - 21 + 16 = -5 \][/tex]
So, the solution vector [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{bmatrix} 4 \\ -1 \\ -5 \end{bmatrix}. \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 4, \quad y = -1, \quad z = -5. \][/tex]
### a. Write the linear system as a matrix equation in the form [tex]\( AX = B \)[/tex].
The given system of equations is:
[tex]\[ \begin{array}{rr} x - y + z &= 0 \\ -2y + z &= -3 \\ -3x - 4y &= -8 \end{array} \][/tex]
We can rewrite this system in matrix form as [tex]\( AX = B \)[/tex], where:
[tex]\[ A = \begin{bmatrix} 1 & -1 & 1 \\ 0 & -2 & 1 \\ -3 & -4 & 0 \end{bmatrix} \][/tex]
is the coefficient matrix,
[tex]\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \][/tex]
is the vector of variables, and
[tex]\[ B = \begin{bmatrix} 0 \\ -3 \\ -8 \end{bmatrix} \][/tex]
is the constants vector.
So the matrix equation [tex]\( AX = B \)[/tex] is:
[tex]\[ \begin{bmatrix} 1 & -1 & 1 \\ 0 & -2 & 1 \\ -3 & -4 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ -3 \\ -8 \end{bmatrix}. \][/tex]
### b. Solve the system using the inverse that is given for the coefficient matrix.
We are given the inverse of the coefficient matrix [tex]\( A \)[/tex], which is:
[tex]\[ A^{-1} = \begin{bmatrix} 4 & -4 & 1 \\ -3 & 3 & -1 \\ -6 & 7 & -2 \end{bmatrix} \][/tex]
To find the solution vector [tex]\( X \)[/tex], we multiply both sides of the equation [tex]\( AX = B \)[/tex] by [tex]\( A^{-1} \)[/tex]:
[tex]\[ A^{-1}AX = A^{-1}B \][/tex]
Since [tex]\( A^{-1}A \)[/tex] is the identity matrix [tex]\( I \)[/tex], this simplifies to:
[tex]\[ X = A^{-1}B \][/tex]
We will now perform the matrix multiplication [tex]\( A^{-1}B \)[/tex]:
[tex]\[ X = \begin{bmatrix} 4 & -4 & 1 \\ -3 & 3 & -1 \\ -6 & 7 & -2 \end{bmatrix} \begin{bmatrix} 0 \\ -3 \\ -8 \end{bmatrix} \][/tex]
Perform the multiplication for each element in [tex]\( X \)[/tex]:
1. For the first element of [tex]\( X \)[/tex]:
[tex]\[ 4 \cdot 0 + (-4) \cdot (-3) + 1 \cdot (-8) = 0 + 12 - 8 = 4 \][/tex]
2. For the second element of [tex]\( X \)[/tex]:
[tex]\[ -3 \cdot 0 + 3 \cdot (-3) + (-1) \cdot (-8) = 0 - 9 + 8 = -1 \][/tex]
3. For the third element of [tex]\( X \)[/tex]:
[tex]\[ -6 \cdot 0 + 7 \cdot (-3) + (-2) \cdot (-8) = 0 - 21 + 16 = -5 \][/tex]
So, the solution vector [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{bmatrix} 4 \\ -1 \\ -5 \end{bmatrix}. \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 4, \quad y = -1, \quad z = -5. \][/tex]
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