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I am reading Bruce Cooperstein's book: Advanced Linear Algebra ... ...
I am focused on Section 1.2 The Space $$\mathbb{F}^n$$ ...
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 $$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$$ ... ... We need to find $$v_1$$ and $$v_2$$ such that:
$$\begin{pmatrix} 2v_1 \\ 2v_2 \end{pmatrix} + \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$$ (mod 5) ... ... ... (1)Thus, given (1), we must have, for $$v_1$$ :
$$2 v_1 +3 = 1$$ (mod 5)
so we must have
$$2 v_1 = 1 - 3 = -2$$ (mod 5)
and so
$$v_1 = -1 \equiv 4$$ (mod 5)
----------------------------------------------
For $$v_2$$, given (1)we must have
$$2 v_2 + 4 = 3$$ (mod 5)
So then we must have
$$2 v_2 = 3 - 4 = -1 \equiv 4$$ (mod 5)
and so
$$v_2 = 2$$ Thus ...
$$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$$BUT ... in the "Hints to Selected Problems, Cooperstein gives the answer as$$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$
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 $$\mathbb{F}^n$$ ...
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 $$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$$ ... ... We need to find $$v_1$$ and $$v_2$$ such that:
$$\begin{pmatrix} 2v_1 \\ 2v_2 \end{pmatrix} + \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$$ (mod 5) ... ... ... (1)Thus, given (1), we must have, for $$v_1$$ :
$$2 v_1 +3 = 1$$ (mod 5)
so we must have
$$2 v_1 = 1 - 3 = -2$$ (mod 5)
and so
$$v_1 = -1 \equiv 4$$ (mod 5)
----------------------------------------------
For $$v_2$$, given (1)we must have
$$2 v_2 + 4 = 3$$ (mod 5)
So then we must have
$$2 v_2 = 3 - 4 = -1 \equiv 4$$ (mod 5)
and so
$$v_2 = 2$$ Thus ...
$$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$$BUT ... in the "Hints to Selected Problems, Cooperstein gives the answer as$$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$
So, my problem is ... why is there a discrepancy ... where is my error ...?
Hope someone can help ...
Peter