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Sagot :
Certainly! Let's solve the given system of linear equations using matrix methods. The system of equations is:
[tex]\[ \left\{\begin{array}{l} -x - 7y - z = -19 \\ 4x + 4y + 4z = 4 \\ 2x + y + 6z = 7 \end{array}\right. \][/tex]
### Step-by-Step Solution:
1. Write the system in matrix form:
The system of equations can be represented in the form [tex]\(AX = B\)[/tex], where [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(X\)[/tex] is the column vector of variables, and [tex]\(B\)[/tex] is the column vector of constants.
[tex]\[ A = \begin{pmatrix} -1 & -7 & -1 \\ 4 & 4 & 4 \\ 2 & 1 & 6 \end{pmatrix} , \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} , \quad B = \begin{pmatrix} -19 \\ 4 \\ 7 \end{pmatrix} \][/tex]
2. Find the inverse of the coefficient matrix [tex]\(A\)[/tex] (if it exists):
For our purposes, it's sufficient to know that the matrix [tex]\(A\)[/tex] is invertible, and we will use the inverse to find [tex]\(X\)[/tex].
3. Multiply both sides of the equation [tex]\(AX = B\)[/tex] by the inverse of [tex]\(A\)[/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]\[ IX = A^{-1}B \quad \text{or simply} \quad X = A^{-1}B \][/tex]
4. Calculate [tex]\(X\)[/tex]:
Using the result obtained from computation:
[tex]\[ X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -4 \\ 3 \\ 2 \end{pmatrix} \][/tex]
### Verification:
To ensure the solution is correct, it's good practice to substitute [tex]\(x = -4\)[/tex], [tex]\(y = 3\)[/tex], and [tex]\(z = 2\)[/tex] back into the original equations:
1. For [tex]\( -x - 7y - z = -19 \)[/tex]:
[tex]\[ -(-4) - 7(3) - 1(2) = 4 - 21 - 2 = -19 \][/tex]
2. For [tex]\( 4x + 4y + 4z = 4 \)[/tex]:
[tex]\[ 4(-4) + 4(3) + 4(2) = -16 + 12 + 8 = 4 \][/tex]
3. For [tex]\( 2x + y + 6z = 7 \)[/tex]:
[tex]\[ 2(-4) + 3 + 6(2) = -8 + 3 + 12 = 7 \][/tex]
The solution satisfies all the original equations, confirming that our solution is correct.
Thus, the solution to the system is:
[tex]\[ x = -4, \quad y = 3, \quad z = 2 \][/tex]
[tex]\[ \left\{\begin{array}{l} -x - 7y - z = -19 \\ 4x + 4y + 4z = 4 \\ 2x + y + 6z = 7 \end{array}\right. \][/tex]
### Step-by-Step Solution:
1. Write the system in matrix form:
The system of equations can be represented in the form [tex]\(AX = B\)[/tex], where [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(X\)[/tex] is the column vector of variables, and [tex]\(B\)[/tex] is the column vector of constants.
[tex]\[ A = \begin{pmatrix} -1 & -7 & -1 \\ 4 & 4 & 4 \\ 2 & 1 & 6 \end{pmatrix} , \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} , \quad B = \begin{pmatrix} -19 \\ 4 \\ 7 \end{pmatrix} \][/tex]
2. Find the inverse of the coefficient matrix [tex]\(A\)[/tex] (if it exists):
For our purposes, it's sufficient to know that the matrix [tex]\(A\)[/tex] is invertible, and we will use the inverse to find [tex]\(X\)[/tex].
3. Multiply both sides of the equation [tex]\(AX = B\)[/tex] by the inverse of [tex]\(A\)[/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]\[ IX = A^{-1}B \quad \text{or simply} \quad X = A^{-1}B \][/tex]
4. Calculate [tex]\(X\)[/tex]:
Using the result obtained from computation:
[tex]\[ X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -4 \\ 3 \\ 2 \end{pmatrix} \][/tex]
### Verification:
To ensure the solution is correct, it's good practice to substitute [tex]\(x = -4\)[/tex], [tex]\(y = 3\)[/tex], and [tex]\(z = 2\)[/tex] back into the original equations:
1. For [tex]\( -x - 7y - z = -19 \)[/tex]:
[tex]\[ -(-4) - 7(3) - 1(2) = 4 - 21 - 2 = -19 \][/tex]
2. For [tex]\( 4x + 4y + 4z = 4 \)[/tex]:
[tex]\[ 4(-4) + 4(3) + 4(2) = -16 + 12 + 8 = 4 \][/tex]
3. For [tex]\( 2x + y + 6z = 7 \)[/tex]:
[tex]\[ 2(-4) + 3 + 6(2) = -8 + 3 + 12 = 7 \][/tex]
The solution satisfies all the original equations, confirming that our solution is correct.
Thus, the solution to the system is:
[tex]\[ x = -4, \quad y = 3, \quad z = 2 \][/tex]
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