I have the following matrix:
$A = \begin{bmatrix}2 & 1 & 1\\ 1 & 2 & 1\\1 & 1 & 2 \end{bmatrix}$
If I use the Excel MINVERSE function it returns:
$A^{-1} = \begin {bmatrix}0.75 & -0.25 & -0.25\\ -0.25 & 0.75 & -0.25 \\ -0.25 & -0.25 & 0.75 \end{bmatrix}$
However if I calculate the minors and cofactors then $|A| = 10$. Using the Adjoint method yields:
$\frac{1}{10}\begin{bmatrix}5 & -3 & 3\\ -3 & 5 & -3\\ 3 & -3 & 5 \end{bmatrix}$ = $\begin{bmatrix}0.5 & -0.3 & 0.3\\ -0.3 & 0.5 & -0.3\\ 0.3 & -0.3 & 0.5 \end{bmatrix}$
Am I making a mistake in the calculation to derive the Adjoint or is this related to the fact that the matrix A is symmetrical?
Best Answer
Determinant:
$det(A) = (2\cdot2\cdot 2 + 1\cdot1\cdot1 + 1\cdot1\cdot 1) - (1\cdot2\cdot1 + 1\cdot1\cdot2 + 2\cdot1\cdot1) = (8+1+1)-(2+2+2) = 10-6 = 4$
Cofactor Matrix:
$C_{11} = C_{22} = C_{33} = (-1)^{1+1}\,(2\cdot2 - 1\cdot1) = 3$
$C_{12} = C_{21} = C_{32} = C_{23} = (-1)^{1+2}\,(2\cdot1 - 1\cdot1) = -1$
$C_{13} = C_{31} = (-1)^{1+3}\,(1\cdot1 - 1\cdot2) = -1$
$C = \begin{bmatrix}3&-1&-1\\-1&3&-1\\-1&-1&3\end{bmatrix}$
$adj(A) = C^T = C$
Inverse Matrix:
$A^{-1} = \frac{1}{det(A)}\,adj(A) = \frac{1}{4}\begin{bmatrix}3&-1&-1\\-1&3&-1\\-1&-1&3\end{bmatrix} = \begin{bmatrix}0.75&-0.25&-0.25\\-0.25&0.75&-0.25\\-0.25&-0.25&0.75\end{bmatrix}$