Question

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 \((a + x) - (b + y)\).

Hint:

To solve problems involving matrix inverses, use the property \(A \cdot A^{-1} = I\), and analyze the resulting equations row by row. Simplify systematically to find unknown elements.

Answer 1

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: \[ (a + x) - (b + y) = 3. \]