2x-3y+5z=16,
3x+ 2y-4z= -4,
x + y - 2z =- 3.

Answer

To solve the system of linear equations given by:

1. \( 2x - 3y + 5z = 16 \)
2. \( 3x + 2y - 4z = -4 \)
3. \( x + y - 2z = -3 \)

we will use the matrix method. Here are the steps to find the values of \(x\), \(y\), and \(z\):

### Step 1: Write the equations in matrix form

We can express the system of equations in the form \( AX = B \), where:

\[ A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 16 \\ -4 \\ -3 \end{pmatrix} \]

### Step 2: Find the determinant of matrix \(A\)

To find the inverse of matrix \(A\), we first need to calculate its determinant (\( |A| \)):

\[ |A| = 2 \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} - (-3) \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} + 5 \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (-4)(1) = -4 + 4 = 0 \)
2. \( \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} = (3)(-2) - (-4)(1) = -6 + 4 = -2 \)
3. \( \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = (3)(1) - (2)(1) = 3 - 2 = 1 \)

Now substituting back into the determinant formula:

\[ |A| = 2(0) + 3(-2) + 5(1) = 0 - 6 + 5 = -1 \]

### Step 3: Find the adjoint of matrix \(A\)

Next, we need to find the adjoint of matrix \(A\). The adjoint is obtained by finding the cofactor matrix and then taking its transpose.

Calculating the cofactors:

1. \( C_{11} = \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} = 0 \)
2. \( C_{12} = -\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} = 2 \)
3. \( C_{13} = \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = 1 \)
4. \( C_{21} = -\begin{vmatrix} -3 & 5 \\ 1 & -2 \end{vmatrix} = -(-6 - 5) = 11 \)
5. \( C_{22} = \begin{vmatrix} 2 & 5 \\ 1 & -2 \end{vmatrix} = -4 - 5 = -9 \)
6. \( C_{23} = -\begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = -(-2 + 3) = -1 \)
7. \( C_{31} = \begin{vmatrix} -3 & 5 \\ 2 & -4 \end{vmatrix} = 12 - 10 = 2 \)
8. \( C_{32} = -\begin{vmatrix} 2 & 5 \\ 3 & -4 \end{vmatrix} = -(-8 - 15) = 23 \)
9. \( C_{33} = \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} = 4 + 9 = 13 \)

Thus, the cofactor matrix is:

\[ \text{Cofactor}(A) = \begin{pmatrix} 0 & 2 & 1 \\ 11 & -9 & -1 \\ 2 & 23 & 13 \end{pmatrix} \]

Taking the transpose gives us the adjoint:

\[ \text{Adj}(A) = \begin{pmatrix} 0 & 11 & 2 \\ 2 & -9 & 23 \\ 1 & -1 & 13 \end{pmatrix} \]

### Step 4: Find the inverse of matrix \(A\)

The inverse of matrix \(A\) is given by:

\[ A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A) = -1 \cdot \begin{pmatrix} 0 & 11 & 2 \\ 2 & -9 & 23 \\ 1 & -1 & 13 \end{pmatrix} = \begin{pmatrix} 0 & -11 & -2 \\ -2 & 9 & -23 \\ -1 & 1 & -13 \end{pmatrix} \]

### Step 5: Multiply \(A^{-1}\) by \(B\)

Now we can find \(X\) by multiplying \(A^{-1}\) with \(B\):

\[ X = A^{-1}B = \begin{pmatrix} 0 & -11 & -2 \\ -2 & 9 & -23 \\ -1 & 1 & -13 \end{pmatrix} \begin{pmatrix} 16 \\ -4 \\ -3 \end{pmatrix} \]

Calculating the multiplication:

1. For \(x\):
\[ 0 \cdot 16 + (-11) \cdot (-4) + (-2) \cdot (-3) = 0 + 44 + 6 = 50 \]

2. For \(y\):
\[ -2 \cdot 16 + 9 \cdot (-4) + (-23) \cdot (-3) = -32 - 36 + 69 = 1 \]

3. For \(z\):
\[ -1 \cdot 16 + 1 \cdot (-4) + (-13) \cdot (-3) = -16 - 4 + 39 = 19 \]

Thus, we have:

\[ X = \begin{pmatrix} 50 \\ 1 \\ 19 \end{pmatrix} \]

### Final Result

The values of \(x\), \(y\), and \(z\) are:

\[ x = 50, \quad y = 1, \quad z = 19 \]