Solution.
A = ( 1 2 3 2 0 1 2 3 4 ) A=\begin{pmatrix}
1& 2&3\\
2& 0&1\\
2&3&4
\end{pmatrix} A = ⎝ ⎛ 1 2 2 2 0 3 3 1 4 ⎠ ⎞ The elements of the minor of the matrix A will be given by
M 1 , 1 = ∣ 0 1 3 4 ∣ = − 3 M_{1,1}=\begin{vmatrix}
0 & 1\\
3& 4
\end{vmatrix}=-3 M 1 , 1 = ∣ ∣ 0 3 1 4 ∣ ∣ = − 3
M 1 , 2 = ∣ 2 1 2 4 ∣ = 6 M_{1,2}=\begin{vmatrix}
2& 1\\
2& 4
\end{vmatrix}=6 M 1 , 2 = ∣ ∣ 2 2 1 4 ∣ ∣ = 6
M 1 , 3 = ∣ 2 0 2 3 ∣ = 6 M_{1,3}=\begin{vmatrix}
2 & 0\\
2& 3
\end{vmatrix}=6 M 1 , 3 = ∣ ∣ 2 2 0 3 ∣ ∣ = 6
M 2 , 1 = ∣ 2 3 3 4 ∣ = − 1 M_{2,1}=\begin{vmatrix}
2& 3\\
3& 4
\end{vmatrix}=-1 M 2 , 1 = ∣ ∣ 2 3 3 4 ∣ ∣ = − 1
M 2 , 2 = ∣ 1 3 2 4 ∣ = − 2 M_{2,2}=\begin{vmatrix}
1 & 3\\
2& 4
\end{vmatrix}=-2 M 2 , 2 = ∣ ∣ 1 2 3 4 ∣ ∣ = − 2
M 2 , 3 = ∣ 1 2 2 3 ∣ = − 1 M_{2,3}=\begin{vmatrix}
1 & 2\\
2& 3
\end{vmatrix}=-1 M 2 , 3 = ∣ ∣ 1 2 2 3 ∣ ∣ = − 1
M 3 , 1 = ∣ 2 3 0 1 ∣ = 2 M_{3,1}=\begin{vmatrix}
2 & 3\\
0& 1
\end{vmatrix}=2 M 3 , 1 = ∣ ∣ 2 0 3 1 ∣ ∣ = 2
M 3 , 2 = ∣ 1 3 2 1 ∣ = − 5 M_{3,2}=\begin{vmatrix}
1& 3\\
2& 1
\end{vmatrix}=-5 M 3 , 2 = ∣ ∣ 1 2 3 1 ∣ ∣ = − 5
M 3 , 3 = ∣ 1 2 2 0 ∣ = − 4 M_{3,3}=\begin{vmatrix}
1& 2\\
2& 0
\end{vmatrix}=-4 M 3 , 3 = ∣ ∣ 1 2 2 0 ∣ ∣ = − 4
Hence, M = ( − 3 6 6 − 1 − 2 − 1 2 − 5 − 4 ) M=\begin{pmatrix}
-3& 6&6 \\
-1& -2&-1\\
2&-5&-4
\end{pmatrix} M = ⎝ ⎛ − 3 − 1 2 6 − 2 − 5 6 − 1 − 4 ⎠ ⎞
The cofactor matrix will be given by C i , j = ( − 1 ) i + j M i , j . C_{i,j}=(-1)^{i+j}M_{i,j}. C i , j = ( − 1 ) i + j M i , j .
C = ( − 3 − 6 6 1 − 2 1 2 5 − 4 ) C=\begin{pmatrix}
-3& -6&6 \\
1& -2&1\\
2&5&-4
\end{pmatrix} C = ⎝ ⎛ − 3 1 2 − 6 − 2 5 6 1 − 4 ⎠ ⎞
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