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Solve this system of equations using a matrix inverse.
[tex]\[
\begin{array}{l}
x + y = 10 \\
2x - z = 9 \\
y - 3z = -5
\end{array}
\][/tex]

If [tex]\( A \)[/tex] is the coefficient matrix of the above system of equations and [tex]\( A^{-1} = \frac{1}{7} \left[ \begin{array}{l} R_1 \\ R_2 \\ R_3 \end{array} \right] \)[/tex], then the elements of [tex]\( R_1 \)[/tex] are [tex]\( 1, 3, -1 \)[/tex], the elements of [tex]\( R_2 \)[/tex] are [tex]\( 6, -3, 1 \)[/tex], and the elements of [tex]\( R_3 \)[/tex] are [tex]\( 2, -1, -2 \)[/tex]. The solution of this system is [tex]\( (x, y, z) = \square \)[/tex].



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].