Problem:
Find the inverse of the following matrix by finding its adjoint:
$$
\begin{bmatrix}
-1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
\end{bmatrix} $$
Answer:
The first step is to find the determinant of the matrix. \begin{align*} \begin{vmatrix} -1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix} &= -1 \begin{vmatrix} 5 & 6 \\ 8 & 9 \\ \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \\ \end{vmatrix} \\ \begin{vmatrix} 5 & 6 \\ 8 & 9 \\ \end{vmatrix} &= 45 - 48 = -3 \\ \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix} &= 36 - 42 = -6 \\ % \begin{vmatrix} -1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix} &= 3 - 2(-6) + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \\ \end{vmatrix} \\ % \begin{vmatrix} -1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix} &= 3 + 12 + 3( 32 - 35) \\ \begin{vmatrix} -1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix} &= 15 - 3(3) = 6 \end{align*} We are now going to find the cofactors. \begin{align*} C_{11} &=\begin{vmatrix} 5 & 6 \\ 8 & 9 \\ \end{vmatrix} = 45 - 48 \\ C_{11} &= -3 \\ C_{12} &= - \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix} = -(28 - 42) = -28 + 42 \\ C_{12} &= 14 \\ C_{13} &=\begin{vmatrix} 4 & 5 \\ 7 & 8 \\ \end{vmatrix} = 32 - 35 \\ C_{13} &= -3 \\ C_{21} &= - \begin{vmatrix} 2 & 3 \\ 8 & 9 \\ \end{vmatrix} = -( 18 - 24) \\ C_{21} &= 6 \\ C_{22} &=\begin{vmatrix} -1 & 3 \\ 7 & 9 \\ \end{vmatrix} = -9 - 21 \\ C_{22} &= -30 \\ C_{23} &= - \begin{vmatrix} -1 & 3 \\ 7 & 8 \\ \end{vmatrix} = -( -8 - 21 ) \\ C_{23} &= 29 \\ C_{31} &= \begin{vmatrix} 2 & 3 \\ 5 & 6 \\ \end{vmatrix} = 12 - 15 \\ C_{31} &= -3 \\ C_{32} &= - \begin{vmatrix} -1 & 3 \\ 4 & 6 \\ \end{vmatrix} = -( -6 - 12 ) \\ C_{32} &= 18 \\ C_{33} &= \begin{vmatrix} -1 & 2 \\ 4 & 5 \\ \end{vmatrix} = -5 - 8 \\ C_{33} &= 13 \\ \end{align*} Now we need to find the adjoint of the matrix. $$ C = \begin{bmatrix} -1 & 14 & 3 \\ 6 & -30 & 29 \\ -3 & 18 & 13 \\ \end{bmatrix} $$ Now, here is the adjoint of the original matrix: $$ \begin{bmatrix} -1 & 6 & -3 \\ 14 & -30 & 18 \\ 3 & 29 & 13 \\ \end{bmatrix} $$ Now to find the inverse of the original matrix we divide the adjoint by the determinate. This gives us the following matrix: $$ \begin{bmatrix} -\frac{1}{6} & \frac{6}{6} & -\frac{3}{6} \\ \frac{14}{6} & - \frac{30}{6} & \frac{18}{6} \\ \frac{3}{6} & \frac{29}{6} & \frac{13}{6} \\ \end{bmatrix} $$ Simplyfing the matrix we get: $$ \begin{bmatrix} -\frac{1}{6} & 1 & -\frac{1}{2} \\ \frac{7}{3} & -5 & 3 \\ \frac{1}{2} & \frac{29}{6} & \frac{13}{6} \\ \end{bmatrix} $$ However, SciLab gets the following matrix for the invese. Where did I go wrong? $$ \begin{bmatrix} -0.5 &1.& -0.5 \\ \frac{7}{3} & -5 & 3 \\ \frac{1}{2} & \frac{29}{6} & \frac{13}{6} \\ \end{bmatrix} $$
Based upon comments from the group, I have updated my answer. I know believe it is correct. I am hoping that somebody can confirm that or tell me why I am wrong. Here is my updated answer.
The first step is to find the determinant of the matrix. \begin{align*} \begin{vmatrix} -1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix} &= -1 \begin{vmatrix} 5 & 6 \\ 8 & 9 \\ \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \\ \end{vmatrix} \\ \begin{vmatrix} 5 & 6 \\ 8 & 9 \\ \end{vmatrix} &= 45 - 48 = -3 \\ \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix} &= 36 - 42 = -6 \\ % \begin{vmatrix} -1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix} &= 3 - 2(-6) + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \\ \end{vmatrix} \\ % \begin{vmatrix} -1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix} &= 3 + 12 + 3( 32 - 35) \\ \begin{vmatrix} -1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix} &= 15 - 3(3) = 6 \end{align*} We are now going to find the cofactors. \begin{align*} C_{11} &=\begin{vmatrix} 5 & 6 \\ 8 & 9 \\ \end{vmatrix} = 45 - 48 \\ C_{11} &= -3 \\ C_{12} &= - \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix} = -(36 - 42) = -36 + 42 \\ C_{12} &= 6 \\ % C_{13} &=\begin{vmatrix} 4 & 5 \\ 7 & 8 \\ \end{vmatrix} = 32 - 35 \\ C_{13} &= -3 \\ C_{21} &= - \begin{vmatrix} 2 & 3 \\ 8 & 9 \\ \end{vmatrix} = -( 18 - 24) \\ C_{21} &= 6 \\ C_{22} &=\begin{vmatrix} -1 & 3 \\ 7 & 9 \\ \end{vmatrix} = -9 - 21 \\ C_{22} &= -30 \\ % C_{23} &= - \begin{vmatrix} -1 & 2 \\ 7 & 8 \\ \end{vmatrix} = -( -8 - 14 ) = 8 + 14 \\ C_{23} &= 22 \\ C_{31} &= \begin{vmatrix} 2 & 3 \\ 5 & 6 \\ \end{vmatrix} = 12 - 15 \\ C_{31} &= -3 \\ C_{32} &= - \begin{vmatrix} -1 & 3 \\ 4 & 6 \\ \end{vmatrix} = -( -6 - 12 ) \\ C_{32} &= 18 \\ C_{33} &= \begin{vmatrix} -1 & 2 \\ 4 & 5 \\ \end{vmatrix} = -5 - 8 \\ C_{33} &= -13 \\ \end{align*} Now we need to find the adjoint of the matrix. $$ C = \begin{bmatrix} 3 & 6 & -3 \\ 6 & -30 & 22 \\ -3 & 18 & -13 \\ \end{bmatrix} $$ Now, here is the adjoint of the original matrix: $$ \begin{bmatrix} 3 & 6 & -3 \\ 6 & -30 & 18 \\ -3 & 22 & -13 \\ \end{bmatrix} $$ Now to find the inverse of the original matrix we divide the adjoint by the determinate. This gives us the following matrix: $$ \begin{bmatrix} \frac{-3}{6} & \frac{6}{6} & -\frac{3}{6} \\ \frac{6}{6} & - \frac{30}{6} & \frac{18}{6} \\ \frac{-3}{6} & \frac{22}{6} & -\frac{13}{6} \\ \end{bmatrix} $$ Simplyfing the matrix we get: $$ \begin{bmatrix} -\frac{1}{2} & 1 & -\frac{1}{2} \\ 1 & -5 & 3 \\ -\frac{1}{2} & \frac{11}{3} & -\frac{13}{6} \\ \end{bmatrix} $$
It should be
$$C_{12} = - \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix} = -(\color{red}{36} - 42) = 6 $$