Problem 58 - Algebra and Trigonometry

Question

Explain how to find the multiplicative inverse for a \(2 \times 2\) invertible matrix.

Step-by-Step Solution

Answer

The multiplicative inverse of a \(2 \times 2\) matrix \(A = \begin{bmatrix}a & b\ c & d\end{bmatrix}\) is \(A^{-1} = \frac{1}{ad-bc} \times \begin{bmatrix}d & -b\ -c & a\end{bmatrix}\).

Step 1: Calculate the Determinant

Assume the matrix as \(A = \begin{bmatrix}a & b\ c & d\end{bmatrix}\). The determinant (det \(A\)) is calculated as: \[det(A) = ad - bc\]. Note that the determinant should not be equal to 0. If the determinant is 0, the matrix does not have a multiplicative inverse.

Step 2: Finding the Adjugate

To find the adjugate, swap the a and d elements while changing the sign of b and c. This results in the adjugate matrix \(adj(A)\), \[\begin{bmatrix}d & -b\ -c & a\end{bmatrix}\].

Step 3: Calculate the Multiplicative Inverse

The multiplicative inverse of a matrix \( A^{-1}\) is obtained by dividing the adjugate matrix by the determinant as follows: \(A^{-1} = \frac{1}{det(A)} \times adj(A) = \frac{1}{ad-bc} \times \begin{bmatrix}d & -b\ -c & a\end{bmatrix}\). Thus, you get the multiplicative inverse of matrix A.\n