Step 1: Use the property of matrix inverses. The product of a matrix \(A\) and its inverse \(A^{-1}\) is the identity matrix: \[ A \cdot A^{-1} = I_3, \] where \(I_3\) is the \(3 \times 3\) identity matrix.
Step 2: Multiply \(A\) and \(A^{-1}\). Compute the 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: Analyze each element of the product. 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 each equation: \[ -8a + 2b = 2, \quad 7a + 2y = -1, \quad -5a = -5. \] \[ a = 1, \quad 4a - b = -1, \quad 7(1) + 2y = -1 \Rightarrow y = -4. \] From the second row of the product: \[ [1(1) + 2(-8) + x(b)] = 0, \quad [1(-1) + 2(7) + x(y)] = 1, \quad [1(1) + 2(-5) + x(3)] = 0. \] Simplify each equation: \[ x(5) = 15 \Rightarrow x = 3, \quad x(-4) = -12. \] 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 each equation: \[ b = 5, \quad -3 + 7 + y = 0 \Rightarrow y = -4. \]
Step 4: Compute \((a + x) - (b + y)\). Substitute the values \(a = 1\), \(x = 3\), \(b = 5\), and \(y = -4\): \[ (a + x) - (b + y) = (1 + 3) - (5 + (-4)). \] \[ = 4 - (5 - 4) = 4 - 1 = 3. \]
Final Answer: \[ (a + x) - (b + y) = 3. \]
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.