To solve the given system of linear equations:
[tex]\[
\left\{
\begin{array}{l}
3x + 2y + z = 20 \\
x - 4y - z = -10 \\
2x + y + 2z = 15
\end{array}
\right.
\][/tex]
we need to find the values of [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] that satisfy all three equations simultaneously. We shall determine these values by interpreting the equations and checking for consistency. Here's the step-by-step solution:
We denote the system in the matrix form as [tex]\(AX = B\)[/tex], where [tex]\(A\)[/tex] is the matrix of coefficients, [tex]\(X\)[/tex] is the column vector of the variables, and [tex]\(B\)[/tex] is the column vector of constants.
1. Matrix of Coefficients ([tex]\(A\)[/tex]):
[tex]\[
A = \begin{pmatrix}
3 & 2 & 1 \\
1 & -4 & -1 \\
2 & 1 & 2 \\
\end{pmatrix}
\][/tex]
2. Column Vector of Variables ([tex]\(X\)[/tex]):
[tex]\[
X = \begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix}
\][/tex]
3. Column Vector of Constants ([tex]\(B\)[/tex]):
[tex]\[
B = \begin{pmatrix}
20 \\
-10 \\
15 \\
\end{pmatrix}
\][/tex]
Now, the system of linear equations can be represented as:
[tex]\[
\begin{pmatrix}
3 & 2 & 1 \\
1 & -4 & -1 \\
2 & 1 & 2 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix}
=
\begin{pmatrix}
20 \\
-10 \\
15 \\
\end{pmatrix}
\][/tex]
To solve this, we find the inverse of matrix [tex]\(A\)[/tex] and then multiply it by matrix [tex]\(B\)[/tex].
The solution [tex]\(X\)[/tex] such that [tex]\(AX = B\)[/tex] is:
[tex]\[
X = A^{-1} B
\][/tex]
After performing the necessary matrix operations (which might involve finding the determinant of [tex]\(A\)[/tex], its adjugate, and its inverse), and then multiplying the inverse by [tex]\(B\)[/tex], we found that the solution for this system is:
[tex]\[
x = 4.000000000000001 \\
y = 2.999999999999999 \\
z = 2.0000000000000004
\][/tex]
Thus, rounding to the nearest whole number, the solution to the system is:
[tex]\[
x = 4, \quad y = 3, \quad z = 2
\][/tex]
Therefore, the correct option that matches our solution is:
[tex]\((4, 3, 2)\)[/tex]
