(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
\[\bold{A}^{10} \begin{pmatrix} 10 \\ 4 \\ -11 \end{pmatrix}.\]

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

justin bieber ml.

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step-by-step explanation:

4.

step-by-step explanation:

it have the same area.brainliest