find x, y and z using inverse matrix method with cofactor
x+2y-z=5
3x-y+3z=7
4x-2y+4z=12

To solve the system of equations using the inverse matrix method with cofactors, let's follow these steps:

### Given System of Equations:
1. \( x + 2y - z = 5 \)
2. \( 3x - y + 3z = 7 \)
3. \( 4x - 2y + 4z = 12 \)

### Step 1: Write the System in Matrix Form

We can represent the system of equations in matrix form as:
\[
A \cdot \mathbf{x} = \mathbf{b}
\]
Where:
\[
A = \begin{pmatrix}
1 & 2 & -1 \\
3 & -1 & 3 \\
4 & -2 & 4
\end{pmatrix}, \quad
\mathbf{x} = \begin{pmatrix}
x \\
y \\
z
\end{pmatrix}, \quad
\mathbf{b} = \begin{pmatrix}
5 \\
7 \\
12
\end{pmatrix}
\]

### Step 2: Find the Determinant of Matrix \( A \)

The determinant of \( A \) is calculated as:
\[
\text{det}(A) = 1 \times \begin{vmatrix} -1 & 3 \\ -2 & 4 \end{vmatrix} - 2 \times \begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} - 1 \times \begin{vmatrix} 3 & -1 \\ 4 & -2 \end{vmatrix}
\]

Let's compute each minor determinant:
\[
\begin{vmatrix} -1 & 3 \\ -2 & 4 \end{vmatrix} = (-1)(4) - (3)(-2) = -4 + 6 = 2
\]
\[
\begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} = (3)(4) - (3)(4) = 12 - 12 = 0
\]
\[
\begin{vmatrix} 3 & -1 \\ 4 & -2 \end{vmatrix} = (3)(-2) - (-1)(4) = -6 + 4 = -2
\]

So, the determinant is:
\[
\text{det}(A) = 1(2) - 2(0) - 1(-2) = 2 + 2 = 4
\]

### Step 3: Find the Cofactor Matrix

The cofactor matrix \( C \) is formed by the cofactors of each element in matrix \( A \).

\[
C = \begin{pmatrix}
\text{C}_{11} & \text{C}_{12} & \text{C}_{13} \\
\text{C}_{21} & \text{C}_{22} & \text{C}_{23} \\
\text{C}_{31} & \text{C}_{32} & \text{C}_{33}
\end{pmatrix}
\]

Using the minors calculated earlier:
\[
C = \begin{pmatrix}
2 & 0 & -2 \\
-10 & -4 & -2 \\
0 & 0 & 0
\end{pmatrix}
\]

### Step 4: Find the Adjugate Matrix \( \text{Adj}(A) \)

The adjugate matrix is the transpose of the cofactor matrix \( C \):
\[
\text{Adj}(A) = C^T = \begin{pmatrix}
2 & -10 & 0 \\
0 & -4 & 0 \\
-2 & -2 & 0
\end{pmatrix}
\]

### Step 5: Find the Inverse Matrix \( A^{-1} \)

The inverse matrix is given by:
\[
A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A) = \frac{1}{4} \begin{pmatrix}
2 & -10 & 0 \\
0 & -4 & 0 \\
-2 & -2 & 0
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2} & -\frac{5}{2} & 0 \\
0 & -1 & 0 \\
-\frac{1}{2} & -\frac{1}{2} & 0
\end{pmatrix}
\]

### Step 6: Multiply \( A^{-1} \) by \( \mathbf{b} \) to Find \( \mathbf{x} \)

Now, multiply the inverse matrix by \( \mathbf{b} \):
\[
\mathbf{x} = A^{-1} \cdot \mathbf{b} = \begin{pmatrix}
\frac{1}{2} & -\frac{5}{2} & 0 \\
0 & -1 & 0 \\
-\frac{1}{2} & -\frac{1}{2} & 0
\end{pmatrix} \cdot \begin{pmatrix}
5 \\
7 \\
12
\end{pmatrix}
\]

Carrying out the matrix multiplication:
\[
\mathbf{x} = \begin{pmatrix}
\frac{1}{2}(5) + \left(-\frac{5}{2}\right)(7) + 0(12) \\
0(5) + (-1)(7) + 0(12) \\
-\frac{1}{2}(5) + \left(-\frac{1}{2}\right)(7) + 0(12)
\end{pmatrix} = \begin{pmatrix}
\frac{5}{2} - \frac{35}{2} \\
-7 \\
-\frac{5}{2} - \frac{7}{2}
\end{pmatrix} = \begin{pmatrix}
-15 \\
-7 \\
-6
\end{pmatrix}
\]

### Final Answer:
\[
x = -15, \quad y = -7, \quad z = -6
\]

Would you like more details, or do you have any questions about the process? 

Here are some related questions:
1. What are the steps to calculate the determinant of a 3x3 matrix?
2. How do you calculate the cofactor of a matrix?
3. Why is the inverse of a matrix important in solving systems of equations?
4. Can all matrices be inverted? Why or why not?
5. What does it mean if a matrix has a determinant of zero?

**Tip:** Always check the determinant of a matrix before attempting to find its inverse. If the determinant is zero, the matrix is singular, and an inverse does not exist.

Learn how to solve a system of equations using the inverse matrix method with cofactor expansion. Detailed step-by-step solution for equations x + 2y - z = 5, 3x - y + 3z = 7, and 4x - 2y + 4z = 12. Covers linear algebra concepts, matrix operations, and cofactor calculations. Suitable for advanced high school students.

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