Solved ex2 from algebra hw3

2 files changed, 106 insertions, 1 deletions

diff --git a/Year 1/Algebra/Homework3/Exercise2.pdf b/Year 1/Algebra/Homework3/Exercise2.pdf
index b165c39..734b6da 100644
--- a/Year 1/Algebra/Homework3/Exercise2.pdf
+++ b/Year 1/Algebra/Homework3/Exercise2.pdf
diff --git a/Year 1/Algebra/Homework3/Exercise2.tex b/Year 1/Algebra/Homework3/Exercise2.tex
index 00cb24a..e368736 100644
--- a/Year 1/Algebra/Homework3/Exercise2.tex
+++ b/Year 1/Algebra/Homework3/Exercise2.tex
@@ -56,6 +56,111 @@
 
 \paragraph{Решение:}
 
-[TODO]
+Нека $e' = e T', \quad e = e' T, \quad f = f'S$ и $A'$ е матрицата на $\varphi$ спрямо базиса $e'$ и $f'$.
+\begin{align*}
+	& \text{От условие, извеждаме че } T' = \left( \begin{array}{rr}
+		1 & 2 \\
+		3 & 5
+	\end{array} \right).
+	\quad e = e'T = eT'T \quad E_2 = T'T \quad \left( \begin{array}{rr}
+		1 & 2 \\
+		3 & 5
+	\end{array} \right) T = \left( \begin{array}{rr}
+		1 & 0 \\
+		0 & 1
+	\end{array} \right) \\
+	& \left( \begin{array}{rr|rr}
+		1 & 2 & 1 & 0 \\
+		3 & 5 & 0 & 1
+	\end{array} \right)
+	\quad \begin{aligned}
+		& \text{Умножаваме първи ред по -3 и събираме с втори. Умножаваме втори по 2 и} \\
+		& \text{събираме с първи. Умножаваме втори по -1.}
+	\end{aligned} \\
+	& \sim \left( \begin{array}{rr|rr}
+		1 & 0 & -5 & 2 \\
+		0 & 1 & 3 & -1
+	\end{array} \right)
+	\quad \text{Следователно } T = \left( \begin{array}{rr}
+		-5 & 2 \\
+		3 & -1
+	\end{array} \right)
+\end{align*}
+\begin{align*}
+	& \text{От условие, извеждаме че } S = \left( \begin{array}{rrr}
+		1 & 0 & -1 \\
+		3 & -1 & -4 \\
+		6 & -2 & -7
+	\end{array} \right).
+	\quad SS^{-1} = E_3 \quad \left( \begin{array}{rrr}
+		1 & 0 & -1 \\
+		3 & -1 & -4 \\
+		6 & -2 & -7
+	\end{array} \right) S^{-1} = \left( \begin{array}{rrr}
+		1 & 0 & 0 \\
+		0 & 1 & 0 \\
+		0 & 0 & 1
+	\end{array} \right) \\
+	& \left( \begin{array}{rrr|rrr}
+		1 & 0 & -1 & 1 & 0 & 0 \\
+		3 & -1 & -4 & 0 & 1 & 0 \\
+		6 & -2 & -7 & 0 & 0 & 1
+	\end{array} \right)
+	\quad \begin{aligned}
+		& \text{Умножаваме втори ред по -2 и събираме с трети.} \\
+		& \text{Умножаваме първи по -3 и събираме с втори.}
+	\end{aligned} \\
+	& \left( \begin{array}{rrr|rrr}
+		1 & 0 & -1 & 1 & 0 & 0 \\
+		0 & -1 & -1 & -3 & 1 & 0 \\
+		0 & 0 & 1 & 0 & -2 & 1
+	\end{array} \right)
+	\quad \begin{aligned}
+		& \text{Събираме трети ред с първи и втори. Умножаваме втори по -1.}
+	\end{aligned} \\
+	& \left( \begin{array}{rrr|rrr}
+		1 & 0 & 0 & 1 & -2 & 1 \\
+		0 & 1 & 0 & 3 & 1 & -1 \\
+		0 & 0 & 1 & 0 & -2 & 1
+	\end{array} \right)
+	\quad \text{Следователно } S^{-1} = \left( \begin{array}{rrr}
+		1 & -2 & 1 \\
+		3 & 1 & -1 \\
+		0 & -2 & 1
+	\end{array} \right)
+\end{align*}
+\begin{align*}
+	& \text{От условие, извеждаме че } A' = \left( \begin{array}{rr}
+		2 & 4 \\
+		6 & 12 \\
+		14 & 27
+	\end{array} \right).
+	\quad \text{По Теорема, } A = S^{-1}AT = \\
+	& = \left( \begin{array}{rrr}
+		1 & -2 & 1 \\
+		3 & 1 & -1 \\
+		0 & -2 & 1
+	\end{array} \right) \left( \begin{array}{rr}
+		2 & 4 \\
+		6 & 12 \\
+		14 & 27
+	\end{array} \right) \left( \begin{array}{rr}
+		-5 & 2 \\
+		3 & -1
+	\end{array} \right) = \left( \begin{array}{rr}
+		4 & 7 \\
+		-2 & -3 \\
+		2 & 3
+	\end{array} \right) \left( \begin{array}{rr}
+		-5 & 2 \\
+		3 & -1
+	\end{array} \right) = \answerbox{
+		\left( \begin{array}{rr}
+			1 & 1 \\
+			1 & -1 \\
+			-1 & 1
+		\end{array} \right)
+	}
+\end{align*}
 
 \end{document}
\ No newline at end of file