B = ( 2 2 0 1 0 1 0 1 1 ) B=\begin{pmatrix}
2 & 2 & 0 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{pmatrix} B = ⎝ ⎛ 2 1 0 2 0 1 0 1 1 ⎠ ⎞
det B = ∣ 2 2 0 1 0 1 0 1 1 ∣ \text{det}B=\begin{vmatrix}
2 & 2 & 0 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{vmatrix} det B = ∣ ∣ 2 1 0 2 0 1 0 1 1 ∣ ∣
= ( − 1 ) 1 + 1 ⋅ 2 ⋅ ∣ 0 1 1 1 ∣ + ( − 1 ) 2 + 1 ⋅ 1 ⋅ ∣ 2 0 1 1 ∣ =(-1)^{1+1}\cdot 2\cdot\begin{vmatrix}
0 & 1 \\
1 & 1
\end{vmatrix}+(-1)^{2+1}\cdot 1\cdot\begin{vmatrix}
2 & 0 \\
1 & 1
\end{vmatrix} = ( − 1 ) 1 + 1 ⋅ 2 ⋅ ∣ ∣ 0 1 1 1 ∣ ∣ + ( − 1 ) 2 + 1 ⋅ 1 ⋅ ∣ ∣ 2 1 0 1 ∣ ∣
+ ( − 1 ) 3 + 1 ⋅ 0 ⋅ ∣ 0 1 1 1 ∣ = +(-1)^{3+1}\cdot 0\cdot\begin{vmatrix}
0 & 1 \\
1 & 1
\end{vmatrix}= + ( − 1 ) 3 + 1 ⋅ 0 ⋅ ∣ ∣ 0 1 1 1 ∣ ∣ =
= 2 ( 0 − 1 ) − ( 2 − 0 ) = − 4 ≠ 0 =2(0-1)-(2-0)=-4\not=0 = 2 ( 0 − 1 ) − ( 2 − 0 ) = − 4 = 0 Therefore the matrix B B B is invertible and B − 1 B^{-1} B − 1 exists.
Augment the matrix with the identity matrix:
[ 2 2 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 1 ] \begin{bmatrix}
2 & 2 & 0 & & 1 & 0 & 0 \\
1 & 0 & 1 & & 0 & 1 & 0 \\
0 & 1 & 1 & & 0 & 0 & 1 \\
\end{bmatrix} ⎣ ⎡ 2 1 0 2 0 1 0 1 1 1 0 0 0 1 0 0 0 1 ⎦ ⎤
R 1 = R 1 2 R_1=\dfrac{R_1}{2} R 1 = 2 R 1
[ 1 1 0 1 / 2 0 0 1 0 1 0 1 0 0 1 1 0 0 1 ] \begin{bmatrix}
1 & 1 & 0 & & 1/2 & 0 & 0 \\
1 & 0 & 1 & & 0 & 1 & 0 \\
0 & 1 & 1 & & 0 & 0 & 1 \\
\end{bmatrix} ⎣ ⎡ 1 1 0 1 0 1 0 1 1 1/2 0 0 0 1 0 0 0 1 ⎦ ⎤
R 2 = R 2 − R 1 R_2=R_2-R_1 R 2 = R 2 − R 1
[ 1 1 0 1 / 2 0 0 0 − 1 1 − 1 / 2 1 0 0 1 1 0 0 1 ] \begin{bmatrix}
1 & 1 & 0 & & 1/2 & 0 & 0 \\
0 & -1 & 1 & & -1/2 & 1 & 0 \\
0 & 1 & 1 & & 0 & 0 & 1 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 1 − 1 1 0 1 1 1/2 − 1/2 0 0 1 0 0 0 1 ⎦ ⎤
R 2 = − R 2 R_2=-R_2 R 2 = − R 2
[ 1 1 0 1 / 2 0 0 0 1 − 1 1 / 2 − 1 0 0 1 1 0 0 1 ] \begin{bmatrix}
1 & 1 & 0 & & 1/2 & 0 & 0 \\
0 & 1 & -1 & & 1/2 & -1 & 0 \\
0 & 1 & 1 & & 0 & 0 & 1 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 1 1 1 0 − 1 1 1/2 1/2 0 0 − 1 0 0 0 1 ⎦ ⎤
R 1 = R 1 − R 2 R_1=R_1-R_2 R 1 = R 1 − R 2
[ 1 0 1 0 1 0 0 1 − 1 1 / 2 − 1 0 0 1 1 0 0 1 ] \begin{bmatrix}
1 & 0 & 1 & & 0 & 1 & 0 \\
0 & 1 & -1 & & 1/2 & -1 & 0 \\
0 & 1 & 1 & & 0 & 0 & 1 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 1 1 − 1 1 0 1/2 0 1 − 1 0 0 0 1 ⎦ ⎤
R 3 = R 3 − R 2 R_3=R_3-R_2 R 3 = R 3 − R 2
[ 1 0 1 0 1 0 0 1 − 1 1 / 2 − 1 0 0 0 2 − 1 / 2 1 1 ] \begin{bmatrix}
1 & 0 & 1 & & 0 & 1 & 0 \\
0 & 1 & -1 & & 1/2 & -1 & 0 \\
0 & 0 & 2 & & -1/2 & 1 & 1 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 1 − 1 2 0 1/2 − 1/2 1 − 1 1 0 0 1 ⎦ ⎤
R 3 = R 3 / 2 R_3=R_3/2 R 3 = R 3 /2
[ 1 0 1 0 1 0 0 1 − 1 1 / 2 − 1 0 0 0 1 − 1 / 4 1 / 2 1 / 2 ] \begin{bmatrix}
1 & 0 & 1 & & 0 & 1 & 0 \\
0 & 1 & -1 & & 1/2 & -1 & 0 \\
0 & 0 & 1 & & -1/4 & 1/2 & 1/2 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 1 − 1 1 0 1/2 − 1/4 1 − 1 1/2 0 0 1/2 ⎦ ⎤
R 1 = R 1 − R 3 R_1=R_1-R_3 R 1 = R 1 − R 3
[ 1 0 0 1 / 4 1 / 2 − 1 / 2 0 1 − 1 1 / 2 − 1 0 0 0 1 − 1 / 4 1 / 2 1 / 2 ] \begin{bmatrix}
1 & 0 & 0 & & 1/4 & 1/2 & -1/2 \\
0 & 1 & -1 & & 1/2 & -1 & 0 \\
0 & 0 & 1 & & -1/4 & 1/2 & 1/2 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 0 − 1 1 1/4 1/2 − 1/4 1/2 − 1 1/2 − 1/2 0 1/2 ⎦ ⎤
R 2 = R 2 + R 3 R_2=R_2+R_3 R 2 = R 2 + R 3
[ 1 0 0 1 / 4 1 / 2 − 1 / 2 0 1 0 1 / 4 − 1 / 2 1 / 2 0 0 1 − 1 / 4 1 / 2 1 / 2 ] \begin{bmatrix}
1 & 0 & 0 & & 1/4 & 1/2 & -1/2 \\
0 & 1 & 0 & & 1/4 & -1/2 & 1/2 \\
0 & 0 & 1 & & -1/4 & 1/2 & 1/2 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 0 0 1 1/4 1/4 − 1/4 1/2 − 1/2 1/2 − 1/2 1/2 1/2 ⎦ ⎤ On the left is the identity matrix. On the right is the inverse matrix.
B − 1 = ( 1 / 4 1 / 2 − 1 / 2 1 / 4 − 1 / 2 1 / 2 − 1 / 4 1 / 2 1 / 2 ) B^{-1}=\begin{pmatrix}
1/4 & 1/2 & -1/2 \\
1/4 & -1/2 & 1/2 \\
-1/4 & 1/2 & 1/2
\end{pmatrix} B − 1 = ⎝ ⎛ 1/4 1/4 − 1/4 1/2 − 1/2 1/2 − 1/2 1/2 1/2 ⎠ ⎞
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