Answer to Question #116063 in Linear Algebra for Joshua

Question

Let A = \(\\mathrm{\\{}\)1,2\(\\mathrm{\\}}\), B = \(\\mathrm{\\{}\)a,b,c\(\\mathrm{\\}}\), C = \(\\mathrm{\\{}\)c,d\(\\mathrm{\\}}\). Find \(A\\ \\times\\ (B\\cap\\ C). \)
a.\\(\\left[\\begin{array}{cc} 1 & 0\\\\ 0 & 1 \\end{array}\\right]\\)
b.\\(\\left[\\begin{array}{cc} 1 & 1\\\\ 1 & 1 \\end{array}\\right]\\)
c.\\(\\left[\\begin{array}{cc} -1 & 1\\\\ 1 & -1 \\end{array}\\right]\\)
d.\\(\\left[\\begin{array}{cc} 0 & 1\\\\ 1 & 0 \\end{array}\\right]\\)

Answer 1

$B=(a,b,c)\\ C=(c,d))\\ B\cap C=(c)\\ A=(1,2)\\ A\times (c)=(1,2)\times(c)$

a) $c=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$A\times c=\begin{pmatrix} 1 &2 \end{pmatrix}\cdot\begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 \end{pmatrix}$

b)

$c=\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\\ A\times c=\begin{pmatrix} 1 &2 \end{pmatrix}\cdot\begin{pmatrix} 1 &1 \\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 3 & 3 \end{pmatrix}$

c)

$c=\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\\ A\times c=\begin{pmatrix} 1 &2 \end{pmatrix}\cdot\begin{pmatrix} -1 &1 \\ 1 & -1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \end{pmatrix}$

d)

$c=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\\ A\times c=\begin{pmatrix} 1 &2 \end{pmatrix}\cdot\begin{pmatrix} 0 &1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} 2 & 1 \end{pmatrix}$