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Sagot :
To solve the determinant equation
[tex]\[ \left|\begin{array}{ccc} -1 & 6 & -2 \\ 5 & x & 7 \\ 4 & 1 & -3 \end{array}\right| = 5, \][/tex]
we follow these steps:
1. Find the determinant of the matrix:
[tex]\[ \left|\begin{array}{ccc} -1 & 6 & -2 \\ 5 & x & 7 \\ 4 & 1 & -3 \end{array}\right|. \][/tex]
We will use the cofactor expansion (Laplace expansion) along the first row for simplicity. The determinant of a [tex]\(3 \times 3\)[/tex] matrix [tex]\(\mathbf{A}\)[/tex] is given by:
[tex]\[ \det(\mathbf{A}) = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}, \][/tex]
where [tex]\(\begin{vmatrix} \cdot & \cdot \\ \cdot & \cdot \end{vmatrix}\)[/tex] represents the determinant of a [tex]\(2 \times 2\)[/tex] sub-matrix.
For our matrix,
[tex]\[ \mathbf{A} = \left[\begin{array}{ccc} -1 & 6 & -2 \\ 5 & x & 7 \\ 4 & 1 & -3 \end{array}\right], \][/tex]
the determinant is:
[tex]\[ \det(\mathbf{A}) = (-1) \begin{vmatrix} x & 7 \\ 1 & -3 \end{vmatrix} - 6 \begin{vmatrix} 5 & 7 \\ 4 & -3 \end{vmatrix} + (-2) \begin{vmatrix} 5 & x \\ 4 & 1 \end{vmatrix}. \][/tex]
2. Compute the [tex]\(2 \times 2\)[/tex] sub-determinants:
[tex]\[ \begin{vmatrix} x & 7 \\ 1 & -3 \end{vmatrix} = x(-3) - 7(1) = -3x - 7, \][/tex]
[tex]\[ \begin{vmatrix} 5 & 7 \\ 4 & -3 \end{vmatrix} = 5(-3) - 7(4) = -15 - 28 = -43, \][/tex]
[tex]\[ \begin{vmatrix} 5 & x \\ 4 & 1 \end{vmatrix} = 5(1) - x(4) = 5 - 4x. \][/tex]
3. Substitute these results back into the determinant formula:
[tex]\[ \det(\mathbf{A}) = (-1)(-3x - 7) - 6(-43) + (-2)(5 - 4x). \][/tex]
4. Simplify the expression:
[tex]\[ \det(\mathbf{A}) = 3x + 7 + 258 - 10 + 8x, \][/tex]
[tex]\[ \det(\mathbf{A}) = 11x + 255. \][/tex]
5. Set the determinant equal to 5 and solve for [tex]\(x\)[/tex]:
[tex]\[ 11x + 255 = 5. \][/tex]
Subtract 255 from both sides:
[tex]\[ 11x = 5 - 255, \][/tex]
[tex]\[ 11x = -250. \][/tex]
Divide both sides by 11:
[tex]\[ x = -\frac{250}{11}. \][/tex]
Thus, the solution is:
[tex]\[ \boxed{x = -\frac{250}{11}}. \][/tex]
The determinant simplifies to [tex]\(11x + 255\)[/tex]. Given that we set this equal to 5, solving the equation yields [tex]\(x = -\frac{250}{11}\)[/tex].
[tex]\[ \left|\begin{array}{ccc} -1 & 6 & -2 \\ 5 & x & 7 \\ 4 & 1 & -3 \end{array}\right| = 5, \][/tex]
we follow these steps:
1. Find the determinant of the matrix:
[tex]\[ \left|\begin{array}{ccc} -1 & 6 & -2 \\ 5 & x & 7 \\ 4 & 1 & -3 \end{array}\right|. \][/tex]
We will use the cofactor expansion (Laplace expansion) along the first row for simplicity. The determinant of a [tex]\(3 \times 3\)[/tex] matrix [tex]\(\mathbf{A}\)[/tex] is given by:
[tex]\[ \det(\mathbf{A}) = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}, \][/tex]
where [tex]\(\begin{vmatrix} \cdot & \cdot \\ \cdot & \cdot \end{vmatrix}\)[/tex] represents the determinant of a [tex]\(2 \times 2\)[/tex] sub-matrix.
For our matrix,
[tex]\[ \mathbf{A} = \left[\begin{array}{ccc} -1 & 6 & -2 \\ 5 & x & 7 \\ 4 & 1 & -3 \end{array}\right], \][/tex]
the determinant is:
[tex]\[ \det(\mathbf{A}) = (-1) \begin{vmatrix} x & 7 \\ 1 & -3 \end{vmatrix} - 6 \begin{vmatrix} 5 & 7 \\ 4 & -3 \end{vmatrix} + (-2) \begin{vmatrix} 5 & x \\ 4 & 1 \end{vmatrix}. \][/tex]
2. Compute the [tex]\(2 \times 2\)[/tex] sub-determinants:
[tex]\[ \begin{vmatrix} x & 7 \\ 1 & -3 \end{vmatrix} = x(-3) - 7(1) = -3x - 7, \][/tex]
[tex]\[ \begin{vmatrix} 5 & 7 \\ 4 & -3 \end{vmatrix} = 5(-3) - 7(4) = -15 - 28 = -43, \][/tex]
[tex]\[ \begin{vmatrix} 5 & x \\ 4 & 1 \end{vmatrix} = 5(1) - x(4) = 5 - 4x. \][/tex]
3. Substitute these results back into the determinant formula:
[tex]\[ \det(\mathbf{A}) = (-1)(-3x - 7) - 6(-43) + (-2)(5 - 4x). \][/tex]
4. Simplify the expression:
[tex]\[ \det(\mathbf{A}) = 3x + 7 + 258 - 10 + 8x, \][/tex]
[tex]\[ \det(\mathbf{A}) = 11x + 255. \][/tex]
5. Set the determinant equal to 5 and solve for [tex]\(x\)[/tex]:
[tex]\[ 11x + 255 = 5. \][/tex]
Subtract 255 from both sides:
[tex]\[ 11x = 5 - 255, \][/tex]
[tex]\[ 11x = -250. \][/tex]
Divide both sides by 11:
[tex]\[ x = -\frac{250}{11}. \][/tex]
Thus, the solution is:
[tex]\[ \boxed{x = -\frac{250}{11}}. \][/tex]
The determinant simplifies to [tex]\(11x + 255\)[/tex]. Given that we set this equal to 5, solving the equation yields [tex]\(x = -\frac{250}{11}\)[/tex].
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