Let's consider the given system of linear equations:
[tex]\[
\begin{array}{l}
x + y = 10 \\
2x - z = 9 \\
y - 3z = -5
\end{array}
\][/tex]
We can represent this system in matrix form as [tex]\( A \mathbf{x} = \mathbf{B} \)[/tex], where:
- [tex]\( A \)[/tex] is the coefficient matrix,
- [tex]\( \mathbf{x} \)[/tex] is the vector of variables, and
- [tex]\( \mathbf{B} \)[/tex] is the constant vector.
Here, we have:
[tex]\[
A = \begin{bmatrix}
1 & 1 & 0 \\
2 & 0 & -1 \\
0 & 1 & -3
\end{bmatrix}
\][/tex]
[tex]\[
\mathbf{B} = \begin{bmatrix}
10 \\
9 \\
-5
\end{bmatrix}
\][/tex]
The inverse of the matrix [tex]\( A \)[/tex] is given as:
[tex]\[
A^{-1} = \frac{1}{7} \begin{bmatrix}
1 & 3 & -1 \\
6 & -3 & 1 \\
2 & -1 & -2
\end{bmatrix}
\][/tex]
Now, to find the solution vector [tex]\( \mathbf{x} \)[/tex] which contains [tex]\((x, y, z)\)[/tex], we need to multiply the inverse of [tex]\( A \)[/tex] by the constant vector [tex]\( \mathbf{B} \)[/tex]:
[tex]\[
\mathbf{x} = A^{-1} \mathbf{B}
\][/tex]
Performing the matrix multiplication:
[tex]\[
\mathbf{x} = \frac{1}{7} \begin{bmatrix}
1 & 3 & -1 \\
6 & -3 & 1 \\
2 & -1 & -2
\end{bmatrix}
\begin{bmatrix}
10 \\
9 \\
-5
\end{bmatrix}
\][/tex]
Carrying out the multiplication step by step:
[tex]\[
\mathbf{x}_1 = \frac{1}{7} (1 \cdot 10 + 3 \cdot 9 - 1 \cdot (-5)) = \frac{1}{7} (10 + 27 + 5) = \frac{1}{7} \cdot 42 = 6
\][/tex]
[tex]\[
\mathbf{x}_2 = \frac{1}{7} (6 \cdot 10 - 3 \cdot 9 + 1 \cdot (-5)) = \frac{1}{7} (60 - 27 - 5) = \frac{1}{7} \cdot 28 = 4
\][/tex]
[tex]\[
\mathbf{x}_3 = \frac{1}{7} (2 \cdot 10 - 1 \cdot 9 - 2 \cdot (-5)) = \frac{1}{7} (20 - 9 + 10) = \frac{1}{7} \cdot 21 = 3
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
(x, y, z) = (6, 4, 3)
\][/tex]
So, filling in the drop-down menus:
The solution of this system is [tex]\((x, y, z) = (6, 4, 3)\)[/tex].
