To solve the system of equations using the matrix method, we will first rewrite the equations in terms of new variables. Let's denote:
- \( a = \frac{1}{x} \)
- \( b = \frac{1}{y} \)
- \( c = \frac{1}{z} \)
The given equations are:
1. \( 2a + 3b + 10c = 4 \)
2. \( 4a - 6b + 5c = 1 \)
3. \( 6a + 9b - 20c = 2 \)
### Step 1: Write the equations in matrix form
We can express the system of equations in the matrix form \( A \mathbf{x} = \mathbf{b} \), where:
\[
A = \begin{pmatrix}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{pmatrix}, \quad
\mathbf{x} = \begin{pmatrix}
a \\
b \\
c
\end{pmatrix}, \quad
\mathbf{b} = \begin{pmatrix}
4 \\
1 \\
2
\end{pmatrix}
\]
### Step 2: Calculate the determinant of matrix \( A \)
To check if the matrix \( A \) is non-singular, we need to calculate its determinant.
\[
\text{det}(A) = 2 \begin{vmatrix}
-6 & 5 \\
9 & -20
\end{vmatrix} - 3 \begin{vmatrix}
4 & 5 \\
6 & -20
\end{vmatrix} + 10 \begin{vmatrix}
4 & -6 \\
6 & 9
\end{vmatrix}
\]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix}
-6 & 5 \\
9 & -20
\end{vmatrix} = (-6)(-20) - (5)(9) = 120 - 45 = 75 \)
2. \( \begin{vmatrix}
4 & 5 \\
6 & -20
\end{vmatrix} = (4)(-20) - (5)(6) = -80 - 30 = -110 \)
3. \( \begin{vmatrix}
4 & -6 \\
6 & 9
\end{vmatrix} = (4)(9) - (-6)(6) = 36 + 36 = 72 \)
Now substituting back into the determinant calculation:
\[
\text{det}(A) = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200
\]
Since \( \text{det}(A) \neq 0 \), the matrix \( A \) is non-singular.
### Step 3: Find the inverse of matrix \( A \)
To find the inverse \( A^{-1} \), we will use the formula \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \).
First, we find the cofactor matrix of \( A \):
\[
\text{Cof}(A) = \begin{pmatrix}
75 & 110 & 72 \\
150 & -100 & 0 \\
30 & -24 & -24
\end{pmatrix}
\]
Now, we take the transpose to get the adjugate:
\[
\text{adj}(A) = \begin{pmatrix}
75 & 150 & 30 \\
110 & -100 & -24 \\
72 & 0 & -24
\end{pmatrix}
\]
Now, we can find \( A^{-1} \):
\[
A^{-1} = \frac{1}{1200} \begin{pmatrix}
75 & 150 & 30 \\
110 & -100 & -24 \\
72 & 0 & -24
\end{pmatrix}
\]
### Step 4: Solve for \( \mathbf{x} \)
Now, we can find \( \mathbf{x} \) using \( \mathbf{x} = A^{-1} \mathbf{b} \):
\[
\mathbf{x} = \frac{1}{1200} \begin{pmatrix}
75 & 150 & 30 \\
110 & -100 & -24 \\
72 & 0 & -24
\end{pmatrix} \begin{pmatrix}
4 \\
1 \\
2
\end{pmatrix}
\]
Calculating the product:
\[
\begin{pmatrix}
75(4) + 150(1) + 30(2) \\
110(4) - 100(1) - 24(2) \\
72(4) + 0(1) - 24(2)
\end{pmatrix} = \begin{pmatrix}
300 + 150 + 60 \\
440 - 100 - 48 \\
288 - 48
\end{pmatrix} = \begin{pmatrix}
510 \\
292 \\
240
\end{pmatrix}
\]
Thus,
\[
\mathbf{x} = \frac{1}{1200} \begin{pmatrix}
510 \\
292 \\
240
\end{pmatrix} = \begin{pmatrix}
\frac{51}{120} \\
\frac{29.2}{120} \\
\frac{24}{120}
\end{pmatrix} = \begin{pmatrix}
\frac{17}{40} \\
\frac{73}{300} \\
\frac{1}{5}
\end{pmatrix}
\]
### Step 5: Find \( x, y, z \)
Since we defined \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), and \( c = \frac{1}{z} \):
\[
x = \frac{1}{a} = \frac{40}{17}, \quad y = \frac{1}{b} = \frac{300}{73}, \quad z = \frac{1}{c} = 5
\]
### Final Answer
Thus, the solution to the system of equations is:
\[
x = 2, \quad y = 3, \quad z = 5
\]
