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Sagot :
To solve the system of linear equations given by the matrix equation
[tex]\[ \left[\begin{array}{cc} 2 & 6 \\ 0 & 1 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{c} 2 \\ -3 \end{array}\right], \][/tex]
we need to find the values of [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex].
By inspecting this matrix equation, multiplying out the left side gives us:
[tex]\[ \left[\begin{array}{cc} 2 & 6 \\ 0 & 1 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{c} 2x_1 + 6x_2 \\ x_2 \end{array}\right] . \][/tex]
This must equal the right-hand side:
[tex]\[ \left[\begin{array}{c} 2 \\ -3 \end{array}\right]. \][/tex]
Therefore, we get the following system of equations:
1. [tex]\( 2x_1 + 6x_2 = 2 \)[/tex]
2. [tex]\( x_2 = -3 \)[/tex]
Substitute [tex]\( x_2 = -3 \)[/tex] into the first equation:
[tex]\[ 2x_1 + 6(-3) = 2. \][/tex]
This simplifies to:
[tex]\[ 2x_1 - 18 = 2, \][/tex]
which further simplifies to:
[tex]\[ 2x_1 = 20, \][/tex]
so:
[tex]\[ x_1 = 10. \][/tex]
Thus, the solution [tex]\( \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \)[/tex] is [tex]\( \left[\begin{array}{c} 10 \\ -3 \end{array}\right] \)[/tex].
Now, we need to identify which equation among the given choices correctly matches this solution process. The correct option is:
[tex]\[ \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{cc} 0.5 & -3 \\ 0 & 1 \end{array}\right] \left[\begin{array}{c} 2 \\ -3 \end{array}\right]. \][/tex]
Hence, the correct equation to solve the given problem is:
[tex]\[ \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{cc} 0.5 & -3 \\ 0 & 1 \end{array}\right] \left[\begin{array}{c} 2 \\ -3 \end{array}\right]. \][/tex]
[tex]\[ \left[\begin{array}{cc} 2 & 6 \\ 0 & 1 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{c} 2 \\ -3 \end{array}\right], \][/tex]
we need to find the values of [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex].
By inspecting this matrix equation, multiplying out the left side gives us:
[tex]\[ \left[\begin{array}{cc} 2 & 6 \\ 0 & 1 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{c} 2x_1 + 6x_2 \\ x_2 \end{array}\right] . \][/tex]
This must equal the right-hand side:
[tex]\[ \left[\begin{array}{c} 2 \\ -3 \end{array}\right]. \][/tex]
Therefore, we get the following system of equations:
1. [tex]\( 2x_1 + 6x_2 = 2 \)[/tex]
2. [tex]\( x_2 = -3 \)[/tex]
Substitute [tex]\( x_2 = -3 \)[/tex] into the first equation:
[tex]\[ 2x_1 + 6(-3) = 2. \][/tex]
This simplifies to:
[tex]\[ 2x_1 - 18 = 2, \][/tex]
which further simplifies to:
[tex]\[ 2x_1 = 20, \][/tex]
so:
[tex]\[ x_1 = 10. \][/tex]
Thus, the solution [tex]\( \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \)[/tex] is [tex]\( \left[\begin{array}{c} 10 \\ -3 \end{array}\right] \)[/tex].
Now, we need to identify which equation among the given choices correctly matches this solution process. The correct option is:
[tex]\[ \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{cc} 0.5 & -3 \\ 0 & 1 \end{array}\right] \left[\begin{array}{c} 2 \\ -3 \end{array}\right]. \][/tex]
Hence, the correct equation to solve the given problem is:
[tex]\[ \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{cc} 0.5 & -3 \\ 0 & 1 \end{array}\right] \left[\begin{array}{c} 2 \\ -3 \end{array}\right]. \][/tex]
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