If \( A = \begin{bmatrix} -1 & a & 2 \\ 1 & 2 & x \\ 3 & 1 & 1 \end{bmatrix} \) and \( A^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ b & y & 3 \end{bmatrix} \), find the value of \( (a + x) - (b + y) \).
Step 1: Utilize the property of matrix inverses.
The product of a matrix \( A \) and its inverse \( A^{-1} \) results in the identity matrix:
\[ A \cdot A^{-1} = I_3, \] where \( I_3 \) denotes the \( 3 \times 3 \) identity matrix.
Step 2: Calculate the product of \( A \) and \( A^{-1} \).
Compute the matrix product \( A \cdot A^{-1} \):
\[ \begin{bmatrix} -1 & a & 2 \\ 1 & 2 & x \\ 3 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ b & y & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. \]
Step 3: Examine each element of the resulting matrix.
From the first row of the product:
\[ [-1(1) + a(-8) + 2(b)] = 1, \quad [-1(-1) + a(7) + 2(y)] = 0, \quad [-1(1) + a(-5) + 2(3)] = 0. \]
Simplify these equations:
From the third row of the product:
\[ [3(1) + 1(-8) + 1(b)] = 0, \quad [3(-1) + 1(7) + 1(y)] = 0, \quad [3(1) + 1(-5) + 1(3)] = 1. \]
Simplify these equations:
Step 4: Calculate \((a + x) - (b + y)\).
Substitute the determined values: \( a = 1 \), \( x = 3 \), \( b = 5 \), and \( y = -4 \):
\[ (a + x) - (b + y) = (1 + 3) - (5 + (-4)). \]
Perform the simplification:
\[ (a + x) - (b + y) = 4 - (5 - 4) = 4 - 1 = 3. \]
Final Result:
\[ (a + x) - (b + y) = 3. \]