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I am reading Bruce Cooperstein's book: Advanced Linear Algebra ... ...
I am focused on Section 1.2 The Space \(\displaystyle \mathbb{F}^n\) ...
I need help with Exercise 12 ... since I do not get the same answer as the author ...
Exercise 12 reads as follows:
View attachment 5103My attempt at a solution to this apparently simple problem is as follows:
Let \(\displaystyle \underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}\) ... ... We need to find \(\displaystyle v_1\) and \(\displaystyle v_2\) such that:
\(\displaystyle \begin{pmatrix} 2v_1 \\ 2v_2 \end{pmatrix} + \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\) (mod 5) ... ... ... (1)Thus, given (1), we must have, for \(\displaystyle v_1\) :
\(\displaystyle 2 v_1 +3 = 1\) (mod 5)
so we must have
\(\displaystyle 2 v_1 = 1 - 3 = -2\) (mod 5)
and so
\(\displaystyle v_1 = -1 \equiv 4\) (mod 5)
----------------------------------------------
For \(\displaystyle v_2\), given (1)we must have
\(\displaystyle 2 v_2 + 4 = 3\) (mod 5)
So then we must have
\(\displaystyle 2 v_2 = 3 - 4 = -1 \equiv 4\) (mod 5)
and so
\(\displaystyle v_2 = 2\) Thus ...
\(\displaystyle \underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}\)BUT ... in the "Hints to Selected Problems, Cooperstein gives the answer as\(\displaystyle \underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\)
So, my problem is ... why is there a discrepancy ... where is my error ...?
Hope someone can help ...
Peter
I am focused on Section 1.2 The Space \(\displaystyle \mathbb{F}^n\) ...
I need help with Exercise 12 ... since I do not get the same answer as the author ...
Exercise 12 reads as follows:
View attachment 5103My attempt at a solution to this apparently simple problem is as follows:
Let \(\displaystyle \underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}\) ... ... We need to find \(\displaystyle v_1\) and \(\displaystyle v_2\) such that:
\(\displaystyle \begin{pmatrix} 2v_1 \\ 2v_2 \end{pmatrix} + \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\) (mod 5) ... ... ... (1)Thus, given (1), we must have, for \(\displaystyle v_1\) :
\(\displaystyle 2 v_1 +3 = 1\) (mod 5)
so we must have
\(\displaystyle 2 v_1 = 1 - 3 = -2\) (mod 5)
and so
\(\displaystyle v_1 = -1 \equiv 4\) (mod 5)
----------------------------------------------
For \(\displaystyle v_2\), given (1)we must have
\(\displaystyle 2 v_2 + 4 = 3\) (mod 5)
So then we must have
\(\displaystyle 2 v_2 = 3 - 4 = -1 \equiv 4\) (mod 5)
and so
\(\displaystyle v_2 = 2\) Thus ...
\(\displaystyle \underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}\)BUT ... in the "Hints to Selected Problems, Cooperstein gives the answer as\(\displaystyle \underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\)
So, my problem is ... why is there a discrepancy ... where is my error ...?
Hope someone can help ...
Peter