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Sagot :
Sure! Let's determine the solution for the system of linear equations:
[tex]\[ \begin{aligned} &x_1 + 2x_2 = 11 \\ &3x_1 + 4x_2 = 25 \end{aligned} \][/tex]
First, we write this system in the matrix form [tex]\(\mathbf{A}\mathbf{x} = \mathbf{b}\)[/tex], where:
[tex]\[ \mathbf{A} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix} , \quad \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} , \quad \mathbf{b} = \begin{pmatrix} 11 \\ 25 \\ \end{pmatrix} \][/tex]
Now, we need to solve for [tex]\(\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} \)[/tex].
Given the matrix equation:
[tex]\[ \mathbf{A}\mathbf{x} = \mathbf{b} \][/tex]
we see that:
[tex]\[ \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} = \begin{pmatrix} 11 \\ 25 \\ \end{pmatrix} \][/tex]
To find [tex]\(\mathbf{x}\)[/tex], we use the inverse of [tex]\(\mathbf{A}\)[/tex] (if it exists). The solution can be found by:
[tex]\[ \mathbf{x} = \mathbf{A}^{-1}\mathbf{b} \][/tex]
Given the computations or solving by substitution/elimination, we determine the solutions for [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]. The solution provided is:
[tex]\[ x_1 = 3, \quad x_2 = 4 \][/tex]
So, the detailed solutions to the system of linear equations:
[tex]\[ \begin{aligned} x_1 + 2x_2 &= 11 \\ 3x_1 + 4x_2 &= 25 \end{aligned} \][/tex]
are [tex]\( x_1 \approx 2.9999999999999987 \)[/tex] and [tex]\( x_2 \approx 4.000000000000001 \)[/tex].
[tex]\[ \begin{aligned} &x_1 + 2x_2 = 11 \\ &3x_1 + 4x_2 = 25 \end{aligned} \][/tex]
First, we write this system in the matrix form [tex]\(\mathbf{A}\mathbf{x} = \mathbf{b}\)[/tex], where:
[tex]\[ \mathbf{A} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix} , \quad \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} , \quad \mathbf{b} = \begin{pmatrix} 11 \\ 25 \\ \end{pmatrix} \][/tex]
Now, we need to solve for [tex]\(\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} \)[/tex].
Given the matrix equation:
[tex]\[ \mathbf{A}\mathbf{x} = \mathbf{b} \][/tex]
we see that:
[tex]\[ \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} = \begin{pmatrix} 11 \\ 25 \\ \end{pmatrix} \][/tex]
To find [tex]\(\mathbf{x}\)[/tex], we use the inverse of [tex]\(\mathbf{A}\)[/tex] (if it exists). The solution can be found by:
[tex]\[ \mathbf{x} = \mathbf{A}^{-1}\mathbf{b} \][/tex]
Given the computations or solving by substitution/elimination, we determine the solutions for [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]. The solution provided is:
[tex]\[ x_1 = 3, \quad x_2 = 4 \][/tex]
So, the detailed solutions to the system of linear equations:
[tex]\[ \begin{aligned} x_1 + 2x_2 &= 11 \\ 3x_1 + 4x_2 &= 25 \end{aligned} \][/tex]
are [tex]\( x_1 \approx 2.9999999999999987 \)[/tex] and [tex]\( x_2 \approx 4.000000000000001 \)[/tex].
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