What is $A^{-1}$ ?
$A=\left[\begin{array}{cc}
1 & 9 \\
-2 & -24
\end{array}\right]$
To find the inverse of the matrix \(A = \begin{bmatrix} 1 & 9 \ -2 & -24 \end{bmatrix}\), we denote it as \(A^{-1} = \begin{bmatrix} a & b \ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). First, we identify the elements: \(a=1\), \(b=9\), \(c=-2\), and \(d=-24\).
The determinant of a \(2 \times 2\) matrix is given by \(\det(A) = ad - bc\). Substituting our values, we get \(\det(A) = (1)(-24) - (9)(-2) = -24 - (-18) = -24 + 18 = -6\). Since \(\det(A) eq 0\), the inverse matrix exists.
The adjugate matrix is formed by swapping the diagonal elements and changing the signs of the off-diagonal elements: \(\text{adj}(A) = \begin{bmatrix} -24 & -9 \ 2 & 1 \end{bmatrix}\). To find the inverse, we multiply the adjugate matrix by the reciprocal of the determinant: \(A^{-1} = \frac{1}{-6} \begin{bmatrix} -24 & -9 \ 2 & 1 \end{bmatrix}\). Multiplying each entry by \(-\frac{1}{6}\), we get \(A^{-1} = \begin{bmatrix} \frac{-24}{-6} & \frac{-9}{-6} \ \frac{2}{-6} & \frac{1}{-6} \end{bmatrix} = \begin{bmatrix} 4 & \frac{3}{2} \ -\frac{1}{3} & -\frac{1}{6} \end{bmatrix}\).
In data analysis and computer graphics, matrices are used to transform coordinates. If a transformation is represented by matrix \(A\), its inverse \(A^{-1}\) allows us to reverse the operation (e.g., reverting an object to its original position). Understanding how to calculate the inverse of a matrix is essential for solving systems of linear equations and performing complex geometric transformations accurately.