B = {1} ∅ {1} {2} {3} {4} {1, 2} {2, 3} {3, 4} {1, 3} {1, 4} {2, 4} {1, 2, 3} {1, 2, 4} {2, 3, 4} {1, 3, 4} {1, 2, 3, 4}
Added by Victor J.
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Step 1: Identify the given set B = {1}. Show more…
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\begin{aligned} &A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} -3 & 5 \\ 2 & -1 \end{array}\right]\\ &\mathbf{C}=\left[\begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array}\right], \quad \mathbf{D}=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\\ &\mathbf{E}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right], \quad \mathbf{F}=\left[\begin{array}{rr} 3 & 3 \\ -1 & -1 \end{array}\right]\\ &0=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right], \quad \quad \mathbf{I}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned} $$A B$$
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\begin{aligned} &A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} -3 & 5 \\ 2 & -1 \end{array}\right]\\ &\mathbf{C}=\left[\begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array}\right], \quad \mathbf{D}=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\\ &\mathbf{E}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right], \quad \mathbf{F}=\left[\begin{array}{rr} 3 & 3 \\ -1 & -1 \end{array}\right]\\ &0=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right], \quad \quad \mathbf{I}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned} $$2 \mathrm{A}$$
\begin{aligned} &A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} -3 & 5 \\ 2 & -1 \end{array}\right]\\ &\mathbf{C}=\left[\begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array}\right], \quad \mathbf{D}=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\\ &\mathbf{E}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right], \quad \mathbf{F}=\left[\begin{array}{rr} 3 & 3 \\ -1 & -1 \end{array}\right]\\ &0=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right], \quad \quad \mathbf{I}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned} $$(-1) \mathbf{D}$$
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