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klmjnmkj:

math problem

klmjnmkj:

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klmjnmkj:

@florisalreadytaken

klmjnmkj:

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Florisalreadytaken:

just do some multiplication and you will be good to go

Florisalreadytaken:

\[ \begin{aligned} A &=\left[\begin{array}{ccc} -2 & -2 & 0 \\ -2 & 3 & -2 \\ -4 & 3 & -4 \end{array}\right] \end{aligned} \] \[ \begin{aligned} (-9) \times A &= (-9) \times \left[\begin{array}{ccc} -2 & -2 & 0 \\ -2 & 3 & -2 \\ -4 & 3 & -4 \end{array}\right] \end{aligned} \] \[ \begin{aligned} (-9) \times A &= \left[\begin{array}{ccc} (-9) \times -2 & (-9) \times -2 & (-9) \times 0 \\ (-9) \times -2 & (-9) \times 3 & (-9) \times -2 \\ (-9) \times -4 & (-9) \times 3 & (-9) \times -4 \end{array}\right] \end{aligned} \] yeah pretty straightforward as for the 2nd one, just change the operators of the numbers to their opposite -- when youre done, post it here so i check it.

Florisalreadytaken:

\[ \begin{aligned} (-9) \times A &= \left[\begin{array}{ccc} 18 & 18 & 0 \\ 18 & -27 & 18 \\ 36 & -27 & 36 \end{array}\right] \end{aligned} \] \[ \begin{aligned} (-1) \times B &= \left[\begin{array}{ccc} -2 & 3 & 3 \\ 0 & 4& 0 \\ -5 & 2 & 5 \end{array}\right] \end{aligned} \Rightarrow \begin{aligned} (-1) \times B &= \left[\begin{array}{ccc} 2 & -3 & -3 \\ 0 & -4& 0 \\ 5 & -2 & -5 \end{array}\right] \end{aligned} \] now that we did the two, we have got this condition \( (-9)A+(-1)B \) thus, we add the 2 together: \[ (-9)A+(-1)B = \begin{aligned} \left[\begin{array}{ccc} 18 & 18 & 0 \\ 18 & -27 & 18 \\ 36 & -27 & 36 \end{array}\right] \end{aligned} \begin{aligned} + \left[\begin{array}{ccc} 2 & -3 & -3 \\ 0 & -4& 0 \\ 5 & -2 & -5 \end{array}\right] \end{aligned} \] \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Updownarrow \] \[ (-9)A+(-1)B = \left[\begin{array}{ccc} 20 & 15 & -3 \\ 18 & -31 & 18 \\ 41 & -29 & 31 \end{array}\right] \] that took a while :\

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