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Sagot :
To solve the matrix equation [tex]\(-M + KX = J\)[/tex], we need to find the matrix [tex]\(X\)[/tex].
Given:
[tex]\[ M = \begin{pmatrix} -4 & 0 \\ 4 & -1 \end{pmatrix}, \quad K = \begin{pmatrix} 1 & 3 \\ 4 & 3 \end{pmatrix}, \quad J = \begin{pmatrix} 13 & 12 \\ 23 & 13 \end{pmatrix} \][/tex]
First, we'll rewrite the equation in a more convenient form. We have:
[tex]\[ -M + KX = J \][/tex]
Rearrange to isolate [tex]\(KX\)[/tex]:
[tex]\[ KX = J + M \][/tex]
Now, calculate [tex]\(J + M\)[/tex], the sum of the matrices [tex]\(J\)[/tex] and [tex]\(M\)[/tex]:
[tex]\[ J + M = \begin{pmatrix} 13 & 12 \\ 23 & 13 \end{pmatrix} + \begin{pmatrix} -4 & 0 \\ 4 & -1 \end{pmatrix} = \begin{pmatrix} 13 + (-4) & 12 + 0 \\ 23 + 4 & 13 + (-1) \end{pmatrix} = \begin{pmatrix} 9 & 12 \\ 27 & 12 \end{pmatrix} \][/tex]
Next, we need to solve for [tex]\(X\)[/tex] by isolating it. To do that, we need the inverse of matrix [tex]\(K\)[/tex]. The inverse [tex]\(K^{-1}\)[/tex] satisfies:
[tex]\[ KK^{-1} = I \][/tex]
The inverse of a 2x2 matrix [tex]\(K = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)[/tex] is given by:
[tex]\[ K^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix [tex]\(K\)[/tex]:
[tex]\[ K = \begin{pmatrix} 1 & 3 \\ 4 & 3 \end{pmatrix} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], [tex]\(c = 4\)[/tex], and [tex]\(d = 3\)[/tex]. Now calculate the determinant:
[tex]\[ \det(K) = ad - bc = (1)(3) - (3)(4) = 3 - 12 = -9 \][/tex]
Thus, the inverse [tex]\(K^{-1}\)[/tex] is:
[tex]\[ K^{-1} = \frac{1}{-9} \begin{pmatrix} 3 & -3 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} -0.3333 & 0.3333 \\ 0.4444 & -0.1111 \end{pmatrix} \][/tex]
With [tex]\(K^{-1}\)[/tex] calculated, we can now solve for [tex]\(X\)[/tex]:
[tex]\[ X = K^{-1}(J + M) \][/tex]
Perform the matrix multiplication:
[tex]\[ X = \begin{pmatrix} -0.3333 & 0.3333 \\ 0.4444 & -0.1111 \end{pmatrix} \times \begin{pmatrix} 9 & 12 \\ 27 & 12 \end{pmatrix} = \begin{pmatrix} 6 & 0 \\ 1 & 4 \end{pmatrix} \][/tex]
Thus, the solution for the equation [tex]\(-M + KX = J\)[/tex] is:
[tex]\[ X= \begin{pmatrix} 6 & 0 \\ 1 & 4 \end{pmatrix} \][/tex]
Given:
[tex]\[ M = \begin{pmatrix} -4 & 0 \\ 4 & -1 \end{pmatrix}, \quad K = \begin{pmatrix} 1 & 3 \\ 4 & 3 \end{pmatrix}, \quad J = \begin{pmatrix} 13 & 12 \\ 23 & 13 \end{pmatrix} \][/tex]
First, we'll rewrite the equation in a more convenient form. We have:
[tex]\[ -M + KX = J \][/tex]
Rearrange to isolate [tex]\(KX\)[/tex]:
[tex]\[ KX = J + M \][/tex]
Now, calculate [tex]\(J + M\)[/tex], the sum of the matrices [tex]\(J\)[/tex] and [tex]\(M\)[/tex]:
[tex]\[ J + M = \begin{pmatrix} 13 & 12 \\ 23 & 13 \end{pmatrix} + \begin{pmatrix} -4 & 0 \\ 4 & -1 \end{pmatrix} = \begin{pmatrix} 13 + (-4) & 12 + 0 \\ 23 + 4 & 13 + (-1) \end{pmatrix} = \begin{pmatrix} 9 & 12 \\ 27 & 12 \end{pmatrix} \][/tex]
Next, we need to solve for [tex]\(X\)[/tex] by isolating it. To do that, we need the inverse of matrix [tex]\(K\)[/tex]. The inverse [tex]\(K^{-1}\)[/tex] satisfies:
[tex]\[ KK^{-1} = I \][/tex]
The inverse of a 2x2 matrix [tex]\(K = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)[/tex] is given by:
[tex]\[ K^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix [tex]\(K\)[/tex]:
[tex]\[ K = \begin{pmatrix} 1 & 3 \\ 4 & 3 \end{pmatrix} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], [tex]\(c = 4\)[/tex], and [tex]\(d = 3\)[/tex]. Now calculate the determinant:
[tex]\[ \det(K) = ad - bc = (1)(3) - (3)(4) = 3 - 12 = -9 \][/tex]
Thus, the inverse [tex]\(K^{-1}\)[/tex] is:
[tex]\[ K^{-1} = \frac{1}{-9} \begin{pmatrix} 3 & -3 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} -0.3333 & 0.3333 \\ 0.4444 & -0.1111 \end{pmatrix} \][/tex]
With [tex]\(K^{-1}\)[/tex] calculated, we can now solve for [tex]\(X\)[/tex]:
[tex]\[ X = K^{-1}(J + M) \][/tex]
Perform the matrix multiplication:
[tex]\[ X = \begin{pmatrix} -0.3333 & 0.3333 \\ 0.4444 & -0.1111 \end{pmatrix} \times \begin{pmatrix} 9 & 12 \\ 27 & 12 \end{pmatrix} = \begin{pmatrix} 6 & 0 \\ 1 & 4 \end{pmatrix} \][/tex]
Thus, the solution for the equation [tex]\(-M + KX = J\)[/tex] is:
[tex]\[ X= \begin{pmatrix} 6 & 0 \\ 1 & 4 \end{pmatrix} \][/tex]
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