91Ó°ÊÓ

The multiplicative inverse of a matrix \(A\) is another matrix \(A^{-1}\) such that \(A \times A^{-1} = I\), where \(I\) is the identity matrix. This inverse exists if and only if the determinant of \(A\) is not zero.

To find the inverse, first calculate the determinant of the matrix:\[A = \left[\begin{array}{rrr}1 & 0 & 6 \ -2 & 1 & 7 \ 3 & 0 & 2\end{array}\right]\]The determinant \(\det(A)\) is computed by expanding along the first row:\[\det(A) = 1 \cdot (1 \cdot 2 - 7 \cdot 0) - 0 \cdot (-2 \cdot 2 - 7 \cdot 3) + 6 \cdot (-2 \cdot 0 - 1 \cdot 3)\]\[= 1 \cdot 2 + 0 + 6 \cdot (-3) = 2 - 18 = -16\]Since the determinant is \(-16\), which is not zero, the matrix is invertible.

The adjunct matrix (or adjugate matrix) is found by calculating the cofactor of each element in \(A\) and transposing the result. This includes the following cofactors:\[C_{11} = (1 \cdot 2 - 7 \cdot 0) = 2, \quad C_{12} = -(0 \cdot 2 - 7 \cdot 3) = 21, \quad C_{13} = (0 \cdot 0 - 1 \cdot 3) = -3\]\[C_{21} = -(-2 \cdot 2 - 7 \cdot 3) = 25, \quad C_{22} = (1 \cdot 2 - 6 \cdot 3) = -16, \quad C_{23} = -(-6 \cdot 0 - 1 \cdot 3) = 3\]\[C_{31} = (0 \cdot 7 + 1 \cdot 3) = 3, \quad C_{32} = -(0 \cdot 7 + 6 \cdot 3) = -18, \quad C_{33} = (1 \cdot 1 - (-2) \cdot 0) = 1\]Then, the cofactor matrix is:\[\left[\begin{array}{ccc}2 & -21 & -3 \ -25 & -16 & -3 \ 3 & 18 & 1\end{array}\right]\]Transpose it to get the adjunct matrix:\[\left[\begin{array}{ccc}2 & -25 & 3 \ -21 & -16 & 18 \ -3 & -3 & 1\end{array}\right]\]

Multiply each element of the adjunct matrix by \(1/\det(A)\), which is \(1/(-16)\) to get the inverse:\[A^{-1} = \frac{1}{-16} \times \left[\begin{array}{ccc}2 & -25 & 3 \ -21 & -16 & 18 \ -3 & -3 & 1\end{array}\right]\]\[= \left[\begin{array}{ccc}-\frac{1}{8} & \frac{25}{16} & -\frac{3}{16} \ \frac{21}{16} & \frac{1}{1} & -\frac{9}{8} \ \frac{3}{16} & \frac{3}{16} & -\frac{1}{16}\end{array}\right]\]

Multiply \(A\) and \(A^{-1}\) to ensure the product is an identity matrix \(I\). If \(A \times A^{-1} = I\), then the inverse is correct.