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

To solve the system of linear equations using the matrix inverse method with cofactor, follow these steps:

### Step 1: Write the system of equations in matrix form
The system of equations is:
\[
\begin{aligned}
x + 2y - z &= 5 \quad \text{(Equation 1)} \\
3x - y + 3z &= 7 \quad \text{(Equation 2)} \\
4x - 2y + 4z &= 12 \quad \text{(Equation 3)}
\end{aligned}
\]

This can be written in matrix form as:

\[
A \mathbf{x} = \mathbf{b}
\]

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 inverse of matrix \(A\)
To find the inverse of \(A\), we need to calculate the determinant and the cofactor matrix.

#### 2.1 Calculate the determinant of \(A\)

\[
\text{det}(A) = 1 \cdot \begin{vmatrix}
-1 & 3 \\
-2 & 4
\end{vmatrix}
- 2 \cdot \begin{vmatrix}
3 & 3 \\
4 & 4
\end{vmatrix}
+ (-1) \cdot \begin{vmatrix}
3 & -1 \\
4 & -2
\end{vmatrix}
\]

Calculate each of the 2x2 determinants:

\[
\begin{vmatrix}
-1 & 3 \\
-2 & 4
\end{vmatrix} = (-1)(4) - (3)(-2) = -4 + 6 = 2
\]

\[
\begin{vmatrix}
3 & 3 \\
4 & 4
\end{vmatrix} = 3(4) - 3(4) = 12 - 12 = 0
\]

\[
\begin{vmatrix}
3 & -1 \\
4 & -2
\end{vmatrix} = 3(-2) - (-1)(4) = -6 + 4 = -2
\]

So the determinant is:

\[
\text{det}(A) = 1 \cdot 2 - 2 \cdot 0 - 1 \cdot (-2) = 2 + 2 = 4
\]

#### 2.2 Find the cofactor matrix of \(A\)
The cofactor matrix \(C\) is calculated by finding the minor of each element, followed by applying the sign according to the cofactor matrix pattern.

\[
C = \begin{pmatrix}
\begin{vmatrix}
-1 & 3 \\
-2 & 4
\end{vmatrix} & -\begin{vmatrix}
3 & 3 \\
4 & 4
\end{vmatrix} & \begin{vmatrix}
3 & -1 \\
4 & -2
\end{vmatrix} \\
-\begin{vmatrix}
2 & -1 \\
-2 & 4
\end{vmatrix} & \begin{vmatrix}
1 & -1 \\
4 & 4
\end{vmatrix} & -\begin{vmatrix}
1 & 2 \\
4 & -2
\end{vmatrix} \\
\begin{vmatrix}
2 & -1 \\
-1 & 3
\end{vmatrix} & -\begin{vmatrix}
1 & 2 \\
3 & 3
\end{vmatrix} & \begin{vmatrix}
1 & 2 \\
3 & -1
\end{vmatrix}
\end{pmatrix}
\]

Calculate each of these 2x2 minors:

\[
\text{Minor}_{11} = \begin{vmatrix} -1 & 3 \\ -2 & 4 \end{vmatrix} = 2, \quad \text{Minor}_{12} = \begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} = 0, \quad \text{Minor}_{13} = \begin{vmatrix} 3 & -1 \\ 4 & -2 \end{vmatrix} = -2
\]

\[
\text{Minor}_{21} = \begin{vmatrix} 2 & -1 \\ -2 & 4 \end{vmatrix} = 6, \quad \text{Minor}_{22} = \begin{vmatrix} 1 & -1 \\ 4 & 4 \end{vmatrix} = 8, \quad \text{Minor}_{23} = \begin{vmatrix} 1 & 2 \\ 4 & -2 \end{vmatrix} = -10
\]

\[
\text{Minor}_{31} = \begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} = 5, \quad \text{Minor}_{32} = \begin{vmatrix} 1 & 2 \\ 3 & 3 \end{vmatrix} = -3, \quad \text{Minor}_{33} = \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = -7
\]

Therefore, the cofactor matrix is:

\[
C = \begin{pmatrix}
2 & 0 & -2 \\
-6 & 8 & 10 \\
5 & 3 & -7
\end{pmatrix}
\]

#### 2.3 Find the adjugate of \(A\)
The adjugate of \(A\) is the transpose of the cofactor matrix \(C\):

\[
\text{Adj}(A) = C^T = \begin{pmatrix}
2 & -6 & 5 \\
0 & 8 & 3 \\
-2 & 10 & -7
\end{pmatrix}
\]

#### 2.4 Calculate the inverse of \(A\)
The inverse of \(A\) is given by:

\[
A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A) = \frac{1}{4} \begin{pmatrix}
2 & -6 & 5 \\
0 & 8 & 3 \\
-2 & 10 & -7
\end{pmatrix}
\]

So:

\[
A^{-1} = \begin{pmatrix}
\frac{1}{2} & -\frac{3}{2} & \frac{5}{4} \\
0 & 2 & \frac{3}{4} \\
-\frac{1}{2} & \frac{5}{2} & -\frac{7}{4}
\end{pmatrix}
\]

### Step 3: Multiply the inverse of \(A\) by \(\mathbf{b}\) to find \(\mathbf{x}\)

\[
\mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix}
\frac{1}{2} & -\frac{3}{2} & \frac{5}{4} \\
0 & 2 & \frac{3}{4} \\
-\frac{1}{2} & \frac{5}{2} & -\frac{7}{4}
\end{pmatrix}
\begin{pmatrix}
5 \\
7 \\
12
\end{pmatrix}
\]

Perform the matrix multiplication:

\[
\mathbf{x} = \begin{pmatrix}
\frac{1}{2} \cdot 5 + -\frac{3}{2} \cdot 7 + \frac{5}{4} \cdot 12 \\
0 \cdot 5 + 2 \cdot 7 + \frac{3}{4} \cdot 12 \\
-\frac{1}{2} \cdot 5 + \frac{5}{2} \cdot 7 + -\frac{7}{4} \cdot 12
\end{pmatrix}
\]

\[
\mathbf{x} = \begin{pmatrix}
\frac{5}{2} - \frac{21}{2} + 15 \\
14 + 9 \\
-\frac{5}{2} + \frac{35}{2} - 21
\end{pmatrix} = \begin{pmatrix}
-8 + 15 \\
23 \\
\frac{30}{2} - 21
\end{pmatrix} = \begin{pmatrix}
7 \\
23 \\
-6
\end{pmatrix}
\]

### Final Answer
\[
x = 7, \quad y = 23, \quad z = -6
\]

Do you want details or have any questions? Here are some related questions:

1. How do you compute the determinant of a matrix?
2. What are cofactors, and how are they calculated

Learn how to solve a system of linear equations using the matrix inverse method with cofactor. This problem involves matrix operations, determinants, and cofactor calculations. Suitable for college-level students.

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