Answer: a) AB = ![\left[\begin{array}{ccc}-14&26\\-11&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-14%2626%5C%5C-11%2622%5Cend%7Barray%7D%5Cright%5D)
; b) ![\left[\begin{array}{ccc}5&-5\\13&9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-5%5C%5C13%269%5Cend%7Barray%7D%5Cright%5D)
; c) No, they are different; d) No, they are never the same;
Step-by-step explanation:
a) A = ![\left[\begin{array}{ccc}-2&4\\1&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%264%5C%5C1%263%5Cend%7Barray%7D%5Cright%5D)
, B = ![\left[\begin{array}{ccc}-2&1\\-3&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%261%5C%5C-3%267%5Cend%7Barray%7D%5Cright%5D)
. This multiplication is possible, because matrix A has 2 columns and matrix B has 2 rows. When those value are equal, the multiplication is possible.
AB = . =
b) BA = . =
c) As, we can see, parts a and b are differents.
d) With matrices, multiplication is not commutative, which means AB≠BA
Y is being multiplied by 0.6 so you have to divide by 0.6 on both sides to leave y alone.
y=10.3/0.6
Answer:
6
Step-by-step explanation:
Using Euclid's algorithm, we divide the larger by the smaller. If the remainder is zero, the divisor is the GCF. Otherwise, we replace the larger with the remainder and repeat.
18 ÷ 12 = 1 r 6
12 ÷ 6 = 2 r 0 . . . . the GCF is 6
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You can also factor the numbers and see what the common factors are.
18 = 2·3·3
12 = 2·2·3
The common factors are 2·3 = 6.
In the factorizations, we see 2 to powers of 1 and 2, and we see 3 to powers of 1 and 2. The GCF is the product of the common factors to their lowest powers: (2^1)(3^1) = (2)(3) = 6
-infinity, positive infinity
