To find the inverse of the given $2 \times 2$ matrix $A = \begin{bmatrix} 8 & 5 \\ -8 & 5 \end{bmatrix}$, we can use the formula for the inverse of a $2 \times 2$ matrix:

$A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

where for matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant $\text{det}(A) = ad - bc$.

Step 1: Compute the Determinant

For $A = \begin{bmatrix} 8 & 5 \\ -8 & 5 \end{bmatrix}$:

• $a = 8$
• $b = 5$
• $c = -8$
• $d = 5$

Then, the determinant is: $\text{det}(A) = (8)(5) - (5)(-8) = 40 + 40 = 80$

Step 2: Set up the Inverse Matrix

Now, apply the inverse formula: $A^{-1} = \frac{1}{80} \begin{bmatrix} 5 & -5 \\ 8 & 8 \end{bmatrix}$

Step 3: Simplify Each Element

Multiply each element by $\frac{1}{80}$: $A^{-1} = \begin{bmatrix} \frac{5}{80} & \frac{-5}{80} \\ \frac{8}{80} & \frac{8}{80} \end{bmatrix} = \begin{bmatrix} \frac{1}{16} & -\frac{1}{16} \\ \frac{1}{10} & \frac{1}{10} \end{bmatrix}$

Final Answer

Thus, the elements are:

• $b_{11} = \frac{1}{16}$
• $b_{12} = -\frac{1}{16}$
• $b_{21} = \frac{1}{10}$
• $b_{22} = \frac{1}{10}$

Would you like a further breakdown of these steps?

Here are five related questions:

1. How is the determinant of a $2 \times 2$ matrix calculated?
2. What does it mean if the determinant of a matrix is zero?
3. How does the inverse of a matrix relate to solving systems of linear equations?
4. Can you calculate the inverse of any matrix, or are there exceptions?
5. What are some applications of matrix inverses in real-world problems?

Tip: Remember that only matrices with a non-zero determinant have inverses.