B=(a,b,c)C=(c,d))B∩C=(c)A=(1,2)A×(c)=(1,2)×(c)B=(a,b,c)\\ C=(c,d))\\ B\cap C=(c)\\ A=(1,2)\\ A\times (c)=(1,2)\times(c)B=(a,b,c)C=(c,d))B∩C=(c)A=(1,2)A×(c)=(1,2)×(c)
a) c=(1001)c=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}c=(1001)
A×c=(12)⋅(1001)=(12)A\times c=\begin{pmatrix} 1 &2 \end{pmatrix}\cdot\begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 \end{pmatrix}A×c=(12)⋅(1001)=(12)
b)
c=(1111)A×c=(12)⋅(1111)=(33)c=\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\\ A\times c=\begin{pmatrix} 1 &2 \end{pmatrix}\cdot\begin{pmatrix} 1 &1 \\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 3 & 3 \end{pmatrix}c=(1111)A×c=(12)⋅(1111)=(33)
c)
c=(−111−1)A×c=(12)⋅(−111−1)=(1−1)c=\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\\ A\times c=\begin{pmatrix} 1 &2 \end{pmatrix}\cdot\begin{pmatrix} -1 &1 \\ 1 & -1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \end{pmatrix}c=(−111−1)A×c=(12)⋅(−111−1)=(1−1)
d)
c=(0110)A×c=(12)⋅(0110)=(21)c=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\\ A\times c=\begin{pmatrix} 1 &2 \end{pmatrix}\cdot\begin{pmatrix} 0 &1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} 2 & 1 \end{pmatrix}c=(0110)A×c=(12)⋅(0110)=(21)
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