Step 1: Solve for \( x \) and \( y \) in terms of \( z \).
From Equation 1:
\[
x = 5 - z.
\]
From Equation 2:
\[
y = 7 - z.
\]
Step 2: Substitute \( x \) and \( y \) into Equation 3.
Substitute \( x = 5 - z \) and \( y = 7 - z \) into Equation 3:
\[
(5 - z) + (7 - z) + z = 9.
\]
Simplify the equation:
\[
5 + 7 - z - z + z = 9 $\Rightarrow$ 12 - z = 9 $\Rightarrow$ z = 3.
\]
Step 3: Find \( x \) and \( y \).
Now that we know \( z = 3 \), substitute this value into the expressions for \( x \) and \( y \):
\[
x = 5 - 3 = 2,
\]
\[
y = 7 - 3 = 4.
\]
Conclusion:
The values of \( x \), \( y \), and \( z \) are:
\[
x = 2, y = 4, z = 3.
\]
If \( x, y, z \) are all different and
\[ \Delta = \begin{vmatrix} x^2 & x^3 + 1 \\ y^2 & y^3 + 1 \\ z^2 & z^3 + 1 \end{vmatrix} = 0, \text{ show that } xyz = -1. \]
If \[ A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}, \] \(\text{then prove that}\) \[ A \cdot \text{adj} \, A = |A| \cdot I. \]
If \[ A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}, \] \(\text{then prove that}\) \[ A \cdot \text{adj}(A) = |A| \cdot I. \text{ Also, find } A^{-1}. \]