線形代数-1-基礎問題-2

どうも、いもけんぴぃです。

今回は設問数と計算量が多いので１題のみです！

目次

• 問1.3
• 次回予告

問1.3

<解説>
今回は計算を頑張る回です.
面倒かもしれませんが耐えましょう.

$(1)$${}^tAA$$=$$\begin{eqnarray} \left( \begin{array}{cc} 1 & 3 \\ 2 & 1 \\ -1 & 0 \\ \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} 1 & 2 & -1 \\ 3 & 1 & 0 \\ \end{array} \right) \end{eqnarray}$

$=\begin{eqnarray} \left( \begin{array}{cc} 1\times1 + 3\times3 & 1\times2 + 3\times1 & 1\times(-1) + 3\times0 \\ 2\times1 + 1\times3 & 2\times2 + 1\times1 & 2\times(-1) + 1\times0 \\ -1\times1 & -1\times2 & -1\times(-1) \\ \end{array} \right) \end{eqnarray}$

$=\begin{eqnarray} \left( \begin{array}{cc} 10 & 5 & -1 \\ 5 & 5 & -2 \\ -1 & -2 & 1 \\ \end{array} \right) \end{eqnarray}$

$(2)$$A{}^tA$$=$$ \begin{eqnarray} \left( \begin{array}{cc} 1 & 2 & -1 \\ 3 & 1 & 0 \\ \end{array} \right) \end{eqnarray}$$\begin{eqnarray} \left( \begin{array}{cc} 1 & 3 \\ 2 & 1 \\ -1 & 0 \\ \end{array} \right) \end{eqnarray} $

$=$$\begin{eqnarray} \left( \begin{array}{cc} 1\times1 + 2\times2 + (-1)\times(-1) & 1\times3 + 2\times1 + (-1)\times0 \\ 3\times1 + 1\times2 + 0\times(-1) & 3\times3 + 1\times1 +0\times0 \\ \end{array} \right) \end{eqnarray} $

$=$$\begin{eqnarray} \left( \begin{array}{cc} 6 & 5 \\ 5 & 10 \\ \end{array} \right) \end{eqnarray} $

$(3)$$B^2$$=$$\begin{eqnarray} \left( \begin{array}{cc} 2 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 3 \\ \end{array} \right) \end{eqnarray}$$\begin{eqnarray} \left( \begin{array}{cc} 2 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 3 \\ \end{array} \right) \end{eqnarray}$

$=\begin{eqnarray} \left( \begin{array}{cc} 2\times2 + 0\times(-1) + 1\times0 & 2\times0 + 0\times1 + 1\times1 & 2\times1 + 0\times0 + 1\times3 \\ -1\times2 + 1\times(-1) + 0\times1 & -1\times0 + 1\times1 + 0\times1 & -1\times1 + 1\times0 + 0\times3 \\ 0\times2 + 1\times(-1) + 3\times0 & 0\times0 + 1\times1 + 3\times1 & 0\times1 + 1\times0 + 3\times3 \\ \end{array} \right) \end{eqnarray}$

$=$$\begin{eqnarray} \left( \begin{array}{cc} 4 & 1 & 5 \\ -3 & 1 & -1 \\ -1 & 4 & 9 \\ \end{array} \right) \end{eqnarray}$

$(4)$$AB$$=$$\begin{eqnarray} \left( \begin{array}{cc} 1 & 2 & -1 \\ 3 & 1 & 0 \\ \end{array} \right) \end{eqnarray}$$\begin{eqnarray} \left( \begin{array}{cc} 2 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 3 \\ \end{array} \right) \end{eqnarray}$

$=$$\begin{eqnarray} \left( \begin{array}{cc} 1\times2 + 2\times(-1) + -1\times0 & 1\times0 + 2\times1 + -1\times1 & 1\times1 + 2\times0 + -1\times3 \\ 3\times2 + 1\times(-1) + 0\times0 & 3\times0 + 1\times1 + 0\times1 & 3\times1 + 1\times0 + 0\times3 \\ \end{array} \right) \end{eqnarray}$

$=$$\begin{eqnarray} \left( \begin{array}{cc} 0 & 1 & -2 \\ 5 & 1 & 3 \\ \end{array} \right) \end{eqnarray}$

$(5)$${}^taBa$$=$$\begin{eqnarray} \left( \begin{array}{cc} 1 & -2 & 1 \\ \end{array} \right) \end{eqnarray}$$\begin{eqnarray} \left( \begin{array}{cc} 2 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 3 \\ \end{array} \right) \end{eqnarray}$$\begin{eqnarray} \left( \begin{array}{cc} 1 \\ -2 \\ 1 \\ \end{array} \right) \end{eqnarray}$

$=$$\begin{eqnarray} \left( \begin{array}{cc} 1\times2 + (-2)\times(-1) + 1\times0 & 1\times0 + (-2)\times1 + 1\times1 & 1\times1 + (-2)\times0 + 1\times3 \\ \end{array} \right) \end{eqnarray}$$\begin{eqnarray} \left( \begin{array}{cc} 1 \\ -2 \\ 1 \\ \end{array} \right) \end{eqnarray}$

$=$$\begin{eqnarray} \left( \begin{array}{cc} 4 & 3 & 4 \\ \end{array} \right) \end{eqnarray}$$\begin{eqnarray} \left( \begin{array}{cc} 1 \\ -2 \\ 1 \\ \end{array} \right) \end{eqnarray}$

$=$$4\times1 + 3\times(-2) + 4\times1$

$=$$10$

$(6)$$ab$$=$$\begin{eqnarray} \left( \begin{array}{cc} 1 \\ -2 \\ 1 \\ \end{array} \right) \end{eqnarray}$$\begin{eqnarray} \left( \begin{array}{cc} 3 & 2 \\ \end{array} \right) \end{eqnarray}$

$=$$\begin{eqnarray} \left( \begin{array}{cc} 1\times3 & 1\times2 \\ -2\times3 & -2\times2 \\ 1\times3 & 1\times2 \\ \end{array} \right) \end{eqnarray}$

$=$$\begin{eqnarray} \left( \begin{array}{cc} 3 & 2 \\ -6 & -4 \\ 3 & 2 \\ \end{array} \right) \end{eqnarray}$

次回予告

次回は問1.4の1本のみでお送りします。
次回もまた見てくださいね！
じゃ〜んけ〜ん
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ぱー！