Question

math problem

Answer 1

@florisalreadytaken

Answer 2

just do some multiplication and you will be good to go

Answer 3

\[ \begin{aligned}
A &=\left[\begin{array}{ccc}
-2 & -2 & 0 \\
-2 & 3 & -2 \\
-4 & 3 & -4
\end{array}\right]
\end{aligned} \]

\[ \begin{aligned}
(-9) \times A &= (-9) \times \left[\begin{array}{ccc}
-2 & -2 & 0 \\
-2 & 3 & -2 \\
-4 & 3 & -4
\end{array}\right]
\end{aligned} \]

\[ \begin{aligned}
(-9) \times A &= \left[\begin{array}{ccc}
(-9) \times -2 & (-9) \times -2 & (-9) \times 0 \\
(-9) \times -2 & (-9) \times 3 & (-9) \times -2 \\
(-9) \times -4 & (-9) \times 3 & (-9) \times -4
\end{array}\right]
\end{aligned} \]

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.

Answer 4

\[ \begin{aligned}
(-9) \times A &= \left[\begin{array}{ccc}
18 & 18 & 0 \\
18 & -27 & 18 \\
36 & -27 & 36
\end{array}\right]
\end{aligned} \]

\[ \begin{aligned}
(-1) \times B &= \left[\begin{array}{ccc}
-2 & 3 & 3 \\
0 & 4& 0 \\
-5 & 2 & 5
\end{array}\right]
\end{aligned} \Rightarrow
\begin{aligned}
(-1) \times B &= \left[\begin{array}{ccc}
2 & -3 & -3 \\
0 & -4& 0 \\
5 & -2 & -5
\end{array}\right]
\end{aligned} \]

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 :\