Problem 59 - 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 \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is found by the formula \(A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\), only if the determinant \(ad-bc\) is not zero.

Step 1: Identify the original \(2 \times 2\) matrix

Let's say the \(2 \times 2\) matrix is denoted by \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). The elements \(a, b, c, d\) represent any real numbers.

Step 2: Find the determinant of the matrix

The determinant of this \(2 \times 2\) matrix is calculated as the product of the elements on the main diagonal minus the product of the elements off the main diagonal. It is denoted as \(\det(A)\), and it is calculated as \(\det(A) = ad - bc\). The determinant must not be equal to zero for the matrix to be invertible.

Step 3: Create the adjugate matrix

The adjugate of \(A\) is obtained by interchanging the elements on the main diagonal and multiplying the off-diagonal elements by \(-1\). This results in the matrix \(adj(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\).

Step 4: Multiply the adjugate by \(1/\)determinant

The multiplicative inverse of the matrix \(A\), denoted \(A^{-1}\), is found by multiplying each element of the adjugate matrix by \(\frac{1}{\det(A)}\). This results in \(A^{-1} = \frac{1}{\det(A)} \times adj(A) = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\).