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joseestrada27
23.10.2020 •
Mathematics
(a) Let
\[\bold{A} = \begin{pmatrix} 3 & -2 & 3 \\ 1 & 2 & 1 \\ 1 & 3 & 0 \end{pmatrix},\]and let
\[\bold{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \bold{v}_2 = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}, \quad \bold{v}_3 = \begin{pmatrix} 11 \\ 1 \\ -14 \end{pmatrix}.\]Show that $\mathbf{A}$ sends each of $\mathbf{v}_1, \mathbf{v}_2,$ and $\mathbf{v}_3$ to scalar multiples of themselves, and find the value of these scalars.
(b) Let $n$ be a positive integer. Use part (a) to find the vectors \[\mathbf{A}^n \mathbf{v}_1, \mathbf{A}^n \mathbf{v}_2, \mathbf{A}^n\mathbf{v}_3.\](c) Write the vector $\begin{pmatrix} 10 \\ 4 \\ -11 \end{pmatrix}$ as a linear combination of $\mathbf{v}_1, \mathbf{v}_2,$ and $\mathbf{v}_3$.
(d) Using parts (a), (b), and (c), calculate
\[\bold{A}^{10} \begin{pmatrix} 10 \\ 4 \\ -11 \end{pmatrix}.\]
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Ответ:
Option A) Inconsistent is correct
Step-by-step explanation:
Given equations are![y=5x+2\hfill (1)](/tpl/images/0458/8748/ad43f.png)
and![10x-2y=12\hfill (2)](/tpl/images/0458/8748/8ec26.png)
Rewritting the equation y=5x+2 as below
5x-y=-2
Now multiplying the above equation into 2 we get
Now subtracting the equations (2) and (3)
10x-2y=12
10x-2y=-4
________
Therefore the given system of equations has no solution
Therefore it is inconsistent
Therefore given two lines are parallel. Two parallel lines never intersect, so the given system has no solution and is inconsistent.
Option A) Inconsistent is correct