Find the sum of the entries in the inverse of \(\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}\)

To solve this problem, it helps to understand two main ideas: how to find the inverse of a \(2 \times 2\) matrix, and how to sum up its entries.

Given a matrix

\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]

the formula for its inverse (as long as the determinant is not zero) is:

\[ A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

In this question, the matrix is:

\[ M = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} \]

Step 1: Compute the determinant

\[ \text{det}(M) = (1)(5) - (2)(3) = 5 - 6 = -1 \]

Step 2: Write out the inverse using the formula

\[ M^{-1} = \frac{1}{-1} \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix} \]

Step 3: Add up the entries of the inverse

Add all four numbers together:

• \(-5 + 2 + 3 + (-1)\)

Break this down:

• \(-5 + 2 = -3\)
• \(-3 + 3 = 0\)
• \(0 + (-1) = -1\)

So, the sum of the entries in the inverse matrix is \(-1\).

Key Concept: The answer is based on correctly applying the formula for the inverse of a \(2 \times 2\) matrix and careful addition.