Answer :
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].
[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].