わかりやすい３D数学理論入門-2章解答

(1) $A \times B = 3 \times 4 + (-2) \times 1 = 12 – 2 = \boxed{10}$

(2) $A \times B = (-1) \times 2 + 5 \times (-3) = -2 – 15 = \boxed{-17}$

(3) $A \times B = 0 \times (-2) + 6 \times 1 = 0 + 6 = \boxed{6}$

(4) $A \times B = 7 \times 1 + (-4) \times 5 = 7 – 20 = \boxed{-13}$

(5) $A \times B = (-3) \times (-2) + (-3) \times (-2) = 6 + 6 = \boxed{12}$

① $ A \times B =
\begin{bmatrix}
1\times5 + 2\times7 & 1\times6 + 2\times8 \\
3\times5 + 4\times7 & 3\times6 + 4\times8
\end{bmatrix}
=
\begin{bmatrix}
19 & 22 \\
43 & 50
\end{bmatrix} $

② $ A \times B =
\begin{bmatrix}
(-1)\times4 + 0\times1 & (-1)\times(-2) + 0\times5 \\
2\times4 + 3\times1 & 2\times(-2) + 3\times5
\end{bmatrix}
=
\begin{bmatrix}
-4 & 2 \\
11 & 11
\end{bmatrix} $

③ $ A \times B =
\begin{bmatrix}
2\times(-3) + (-1)\times5 & 2\times4 + (-1)\times1 \\
0\times(-3) + 3\times5 & 0\times4 + 3\times1
\end{bmatrix}
=
\begin{bmatrix}
-11 & 7 \\
15 & 3
\end{bmatrix} $

④ $ A \times B =
\begin{bmatrix}
6\times0 + 2\times5 & 6\times3 + 2\times(-2) \\
(-1)\times0 + 4\times5 & (-1)\times3 + 4\times(-2)
\end{bmatrix}
=
\begin{bmatrix}
10 & 14 \\
20 & -11
\end{bmatrix} $

⑤ $ A \times B =
\begin{bmatrix}
1\times(-2) + (-3)\times3 & 1\times4 + (-3)\times1 \\
2\times(-2) + 0\times3 & 2\times4 + 0\times1
\end{bmatrix}
=
\begin{bmatrix}
-11 & 1 \\
-4 & 8
\end{bmatrix} $

① $A\vec{v} = (1×2 + 2×1,\ 3×2 + 4×1) = (4,\ 10)$

② $A\vec{v} = (0×3 + (-1)×4,\ 1×3 + 0×4) = (-4,\ 3)$

③ $A\vec{v} = (2×5 + 0×(-3),\ 0×5 + 2×(-3)) = (10,\ -6)$

④ $A\vec{v} = (-1×(-2) + 0×7,\ 0×(-2) + (-1)×7) = (2,\ -7)$

⑤ $A\vec{v} = (3×1 + 1×2,\ -2×1 + 4×2) = (5,\ 6)$