JAMB - Mathematics (2025 - No. 13)
Find the sum of the entries in the inverse of \(\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}\)
1
-1
2
0.5
Explanation
Let the matrix be \(A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}\)
The determinant of \(A\) is calculated as follows:
\(\text{Determinant} = 1 \cdot 5 - 2 \cdot 3 = 5 - 6 = -1\)
To find the inverse of \(A\), we first determine the adjoint matrix:
\(\text{Adjoint} = \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix}\)
The inverse is given by
\(A^{-1} = \frac{1}{\text{Determinant}} \cdot \text{Adjoint} = \frac{1}{-1} \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix}\)
Next, we calculate the sum of the elements in the inverse matrix:
\(\text{Sum} = -5 + 2 + 3 + (-1) = -1\)
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