A ( 8 ) B ( 14 ) C ( 13 ) P 1 2 3 Q 2 3 2 R 1 2 2 \def\arraystretch{1.5}
\begin{array}{c|c:c:c}
& A(8) & B(14) & C(13)\\ \hline
P & 1& 2& 3\\
\hdashline
Q& 2 & 3& 2\\
\hdashline
R & 1& 2 & 2
\end{array} P Q R A ( 8 ) 1 2 1 B ( 14 ) 2 3 2 C ( 13 ) 3 2 2
A = ( 1 2 1 2 3 2 3 2 2 ) , A=\begin{pmatrix}
1& 2 &1 \\
2& 3 & 2\\
3& 2 & 2
\end{pmatrix}, A = ⎝ ⎛ 1 2 3 2 3 2 1 2 2 ⎠ ⎞ ,
B = ( 8 14 13 ) , B=\begin{pmatrix}
8 \\
14\\
13
\end{pmatrix}, B = ⎝ ⎛ 8 14 13 ⎠ ⎞ ,
X = ( P Q R ) . X=\begin{pmatrix}
P \\
Q\\
R
\end{pmatrix}. X = ⎝ ⎛ P Q R ⎠ ⎞ .
∣ A ∣ = ∣ 1 2 1 2 3 2 3 2 2 ∣ = 1 ⋅ ( 6 − 4 ) − 2 ⋅ ( 4 − 6 ) + 1 ⋅ ( 4 − 9 ) = 1 ≠ 0. |A|=\begin{vmatrix}
1& 2 &1 \\
2& 3 & 2\\
3& 2 & 2
\end{vmatrix}=1\cdot(6-4)-2\cdot(4-6)+1\cdot(4-9)=1\neq 0. ∣ A ∣ = ∣ ∣ 1 2 3 2 3 2 1 2 2 ∣ ∣ = 1 ⋅ ( 6 − 4 ) − 2 ⋅ ( 4 − 6 ) + 1 ⋅ ( 4 − 9 ) = 1 = 0.
a 11 = ∣ 3 2 2 2 ∣ = 2 a_{11}=\begin{vmatrix}
3&2 \\
2 & 2
\end{vmatrix}=2 a 11 = ∣ ∣ 3 2 2 2 ∣ ∣ = 2 , a 12 = − ∣ 2 2 3 2 ∣ = 2 a_{12}=-\begin{vmatrix}
2&2 \\
3 & 2
\end{vmatrix}=2 a 12 = − ∣ ∣ 2 3 2 2 ∣ ∣ = 2 , a 13 = ∣ 2 3 3 2 ∣ = − 5 , a_{13}=\begin{vmatrix}
2&3 \\
3 & 2
\end{vmatrix}=-5, a 13 = ∣ ∣ 2 3 3 2 ∣ ∣ = − 5 ,
a 21 = − ∣ 2 1 2 2 ∣ = − 2 , a_{21}=-\begin{vmatrix}
2&1 \\
2 & 2
\end{vmatrix}=-2, a 21 = − ∣ ∣ 2 2 1 2 ∣ ∣ = − 2 , a 22 = ∣ 1 1 3 2 ∣ = − 1 , a_{22}=\begin{vmatrix}
1&1 \\
3 & 2
\end{vmatrix}=-1, a 22 = ∣ ∣ 1 3 1 2 ∣ ∣ = − 1 , a 23 = − ∣ 1 2 3 2 ∣ = 4 , a_{23}=-\begin{vmatrix}
1&2 \\
3 & 2
\end{vmatrix}=4, a 23 = − ∣ ∣ 1 3 2 2 ∣ ∣ = 4 ,
a 31 = ∣ 2 1 3 2 ∣ = 1 , a_{31}=\begin{vmatrix}
2&1 \\
3 & 2
\end{vmatrix}=1, a 31 = ∣ ∣ 2 3 1 2 ∣ ∣ = 1 , a 32 = − ∣ 1 1 2 2 ∣ = 0 , a_{32}=-\begin{vmatrix}
1&1 \\
2 & 2
\end{vmatrix}=0, a 32 = − ∣ ∣ 1 2 1 2 ∣ ∣ = 0 , a 33 = ∣ 1 2 2 3 ∣ = − 1. a_{33}=\begin{vmatrix}
1&2 \\
2 & 3
\end{vmatrix}=-1. a 33 = ∣ ∣ 1 2 2 3 ∣ ∣ = − 1.
a d j A = ( 2 2 − 5 − 2 − 1 4 1 0 − 1 ) T = ( 2 − 2 1 2 − 1 0 − 5 4 − 1 ) . adjA={\begin{pmatrix}
2 & 2 &-5 \\
-2 & -1 &4 \\
1& 0& -1
\end{pmatrix}}^{T}={\begin{pmatrix}
2& -2 &1 \\
2& -1& 0\\
-5&4 & -1
\end{pmatrix}}. a d j A = ⎝ ⎛ 2 − 2 1 2 − 1 0 − 5 4 − 1 ⎠ ⎞ T = ⎝ ⎛ 2 2 − 5 − 2 − 1 4 1 0 − 1 ⎠ ⎞ .
A − 1 = a d j A ∣ A ∣ = ( 2 − 2 1 2 − 1 0 − 5 4 − 1 ) . A^{-1}=\frac{adjA}{|A|}=\begin{pmatrix}
2& -2 &1 \\
2& -1& 0\\
-5&4 & -1
\end{pmatrix}. A − 1 = ∣ A ∣ a d j A = ⎝ ⎛ 2 2 − 5 − 2 − 1 4 1 0 − 1 ⎠ ⎞ .
X = A − 1 B . X=A^{-1}B. X = A − 1 B .
( P Q R ) = ( 2 − 2 1 2 − 1 0 − 5 4 − 1 ) ⋅ ( 8 14 13 ) = ( 16 − 28 + 13 16 − 14 − 40 + 56 − 13 ) = ( 1 2 3 ) . \begin{pmatrix}
P \\
Q\\
R
\end{pmatrix}=\begin{pmatrix}
2& -2 &1 \\
2& -1& 0\\
-5&4 & -1
\end{pmatrix}\cdot \begin{pmatrix}
8 \\
14\\
13
\end{pmatrix}=\begin{pmatrix}
16-28+13 \\
16-14\\
-40+56-13
\end{pmatrix}=\begin{pmatrix}
1 \\
2\\
3
\end{pmatrix}. ⎝ ⎛ P Q R ⎠ ⎞ = ⎝ ⎛ 2 2 − 5 − 2 − 1 4 1 0 − 1 ⎠ ⎞ ⋅ ⎝ ⎛ 8 14 13 ⎠ ⎞ = ⎝ ⎛ 16 − 28 + 13 16 − 14 − 40 + 56 − 13 ⎠ ⎞ = ⎝ ⎛ 1 2 3 ⎠ ⎞ .
P = 1 , Q = 2 , R = 3. P=1, Q=2, R=3. P = 1 , Q = 2 , R = 3.
Comments