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What is the solution to the equation [tex]-M + KX = J[/tex]?

Given:
[tex]
M = \left[\begin{array}{cc}
-4 & 0 \\
4 & -1
\end{array}\right], \quad K = \left[\begin{array}{ll}
1 & 3 \\
4 & 3
\end{array}\right], \quad J = \left[\begin{array}{ll}
13 & 12 \\
23 & 13
\end{array}\right]
[/tex]

Sagot :

GinnyAnswer
To solve the equation [tex]\(-M + KX = J\)[/tex] for [tex]\(X\)[/tex], follow these steps:

1. Write down the given matrices:
[tex]\[ M = \begin{bmatrix} -4 & 0 \\ 4 & -1 \end{bmatrix}, \quad K = \begin{bmatrix} 1 & 3 \\ 4 & 3 \end{bmatrix}, \quad J = \begin{bmatrix} 13 & 12 \\ 23 & 13 \end{bmatrix} \][/tex]

2. Rearrange the equation:
The given equation is [tex]\(-M + KX = J\)[/tex]. To isolate [tex]\(KX\)[/tex], add matrix [tex]\(M\)[/tex] to both sides:
[tex]\[ KX = J + M \][/tex]

3. Calculate [tex]\(J + M\)[/tex]:
Add the matrices [tex]\(J\)[/tex] and [tex]\(M\)[/tex]:
[tex]\[ J + M = \begin{bmatrix} 13 & 12 \\ 23 & 13 \end{bmatrix} + \begin{bmatrix} -4 & 0 \\ 4 & -1 \end{bmatrix} = \begin{bmatrix} 13 + (-4) & 12 + 0 \\ 23 + 4 & 13 + (-1) \end{bmatrix} = \begin{bmatrix} 9 & 12 \\ 27 & 12 \end{bmatrix} \][/tex]

4. Find the inverse of [tex]\(K\)[/tex]:
The next step is to solve for [tex]\(X\)[/tex] by isolating it on one side. For this, multiply both sides by [tex]\(K^{-1}\)[/tex], where [tex]\(K^{-1}\)[/tex] is the inverse of matrix [tex]\(K\)[/tex]. The inverse of [tex]\(K\)[/tex] is given by:
[tex]\[ K^{-1} = \begin{bmatrix} -0.33333333 & 0.33333333 \\ 0.44444444 & -0.11111111 \end{bmatrix} \][/tex]

5. Multiply [tex]\(K^{-1}\)[/tex] with [tex]\(J + M\)[/tex] to find [tex]\(X\)[/tex]:
Use the calculated inverse of [tex]\(K\)[/tex] to find [tex]\(X\)[/tex]:
[tex]\[ X = K^{-1}(J + M) = \begin{bmatrix} -0.33333333 & 0.33333333 \\ 0.44444444 & -0.11111111 \end{bmatrix} \times \begin{bmatrix} 9 & 12 \\ 27 & 12 \end{bmatrix} \][/tex]

Perform the matrix multiplication:
[tex]\[ X = \begin{bmatrix} (-0.33333333 \cdot 9 + 0.33333333 \cdot 27) & (-0.33333333 \cdot 12 + 0.33333333 \cdot 12) \\ (0.44444444 \cdot 9 + -0.11111111 \cdot 27) & (0.44444444 \cdot 12 + -0.11111111 \cdot 12) \end{bmatrix} \][/tex]
Simplifying each element:
[tex]\[ X = \begin{bmatrix} 6 & -2.22044605 \cdot 10^{-16} \\ 1 & 4 \end{bmatrix} \][/tex]

Therefore, the solution to the equation [tex]\(-M + KX = J\)[/tex] is:
[tex]\[ X = \begin{bmatrix} 6 & -2.22044605 \cdot 10^{-16} \\ 1 & 4 \end{bmatrix} \][/tex]
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