Answer:
AB = ![\left[\begin{array}{ccc}-3&4&6\\-6&3&5\\5&0&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%264%266%5C%5C-6%263%265%5C%5C5%260%26-4%5Cend%7Barray%7D%5Cright%5D)
Each column of AB is written as a linear combination of columns of Matrix A in the explanation below.
Step-by-step explanation:
A = ![\left[\begin{array}{ccc}-2&2&1\\-3&1&1\\2&0&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%262%261%5C%5C-3%261%261%5C%5C2%260%26-1%5Cend%7Barray%7D%5Cright%5D)
B= ![\left[\begin{array}{ccc}2&-1&0\\1&2&1\\-1&-2&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-1%260%5C%5C1%262%261%5C%5C-1%26-2%264%5Cend%7Barray%7D%5Cright%5D)
We need to write each column of AB as a linear combination of the columns of A so we will multiply each column of A with each column element of B to get the column of AB. So,
AB Column 1 = 2 * ![\left[\begin{array}{ccc}-2\\-3\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%5C%5C-3%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
+ 1 ![\left[\begin{array}{ccc}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
+ (-1) ![\left[\begin{array}{ccc}1\\1\\-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5C%5C-1%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-3\\-6\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%5C%5C-6%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
AB Column 2 = (-1) + 2 + (-2) = ![\left[\begin{array}{ccc}4\\3\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C3%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
AB Column 3 = (0) + (1) + 4 = ![\left[\begin{array}{ccc}6\\5\\-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%5C%5C5%5C%5C-4%5Cend%7Barray%7D%5Cright%5D)
Finally, we can combine all three columns of AB to form the 3x3 matrix AB.
AB =
Answer:
Step 1: 8 + 16x Step 2: 4(4) + 4(12x) Step 3: 4(4 + 12x) Step 4: Dimensions of the rectangle are 4 and 4 + 12x In which step did the student first make an error and what is the correct step? Step 3; 4 + (4 + 12x) Step 3; 4 + (4 ⋅ 12x) Step 2; 4(2) + 4(4x) Step 2; 4(2) + 4x(2)
Answer:
Error:
not 
Solution:x=0 and 3
Step-by-step explanation:
We have to find the error and correct answer
Given:![2ln x=ln(3x)-[ln9-2ln(3)]](https://tex.z-dn.net/?f=2ln%20x%3Dln%283x%29-%5Bln9-2ln%283%29%5D)
![lnx^2=ln(3x)-[ln9-ln3^2]](https://tex.z-dn.net/?f=lnx%5E2%3Dln%283x%29-%5Bln9-ln3%5E2%5D)
Using the formula
![lnx^2=ln(3x)-[ln9-ln9]](https://tex.z-dn.net/?f=lnx%5E2%3Dln%283x%29-%5Bln9-ln9%5D)
Therefore, x=0 and x=3
But last step in the given solution
It is wrong this property is used when
then
Hence, the student wrote 
instead of 
and solution is given by
x=0 and x=3
I’m not sure but since no one else commented I would think it’s D, I think you take the first number like 6 for a and subtract the first number for b like 3
I know this ...... it is false
