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:
$$ \begin{bmatrix}
-\frac{1}{2} & 1 & -\frac{1}{2} \\
1 & -5 & 3 \\
-\frac{1}{2} & \frac{11}{3} & -\frac{13}{6} \\
\end{bmatrix} $$
Best Answer
It should be
$$C_{12} = - \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix} = -(\color{red}{36} - 42) = 6 $$