2) Let 4 be a group \( a, b \in G \). Show that the equatim \( x_{a}=b \) has a unique solutur
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We need to show that the equation \( xa = b \) has a unique solution for \( x \) in a group \( G \), where \( a, b \in G \). Show more…
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If . $A=\left[\begin{array}{lll}a & 0 & 1 \\ 1 & c & b \\ 1 & d & b\end{array}\right], B=\left[\begin{array}{lll}a & 1 & 1 \\ 0 & d & c \\ f & g & h\end{array}\right], U=\left[\begin{array}{l}f \\ g \\ h\end{array}\right], V=\left[\begin{array}{c}a^{2} \\ 0 \\ 0\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $A X=U$ has infinitely many solutions, prove that $B X=V$ has no unique solution. Also show that if $\neq$ fd 0 , then $B X=V$ has no solution.
Theorem 11.3.9 Linear Equations in a Group. If G is a group and a, b ∈ G, the equation a * x = b has a unique solution, x = a⁻¹ * b. In addition, the equation x * a = b has a unique solution, x = b * a⁻¹. Proof. We prove the theorem only for a * x = b, since the second statement is proven identically. a * x = b = e * b = (a * a⁻¹) * b = a * (a⁻¹ * b) By the cancellation law, we can conclude that x = a⁻¹ * b. If c and d are two solutions of the equation a * x = b, then a * c = b = a * d and, by the cancellation law, c = d. This verifies that a⁻¹ * b is the only solution of a * x = b. Note 11.3.10 Our proof of Theorem 11.3.9 was analogous to solving the concrete equation 4x = 9 in the following way: 4x = 9 = (4 · 1/4)9 = 4(1/4 9) Therefore, by cancelling 4, x = 1/4 · 9 = 9/4
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Solve each equation for $x$. See Example $d(c)$ $$ a x+b^{2}=b x-a^{2} $$
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