Answer:
Remember, a basis for the row space of a matrix A is the set of rows different of zero of the echelon form of A.
We need to find the echelon form of the matrix augmented matrix of the system A2x=b2
![B=\left[\begin{array}{cccc}1&2&3&1\\4&5&6&1\\7&8&9&1\\3&2&4&1\\6&5&4&1\\9&8&7&1\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C4%265%266%261%5C%5C7%268%269%261%5C%5C3%262%264%261%5C%5C6%265%264%261%5C%5C9%268%267%261%5Cend%7Barray%7D%5Cright%5D)
We apply row operations:
1.
- To row 2 we subtract row 1, 4 times.
- To row 3 we subtract row 1, 7 times.
- To row 4 we subtract row 1, 3 times.
- To row 5 we subtract row 1, 6 times.
- To row 6 we subtract row 1, 9 times.
We obtain the matrix
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&-6&-12&-6\\0&-4&-5&-2\\0&-7&-14&-5\\0&-10&-20&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%26-6%26-12%26-6%5C%5C0%26-4%26-5%26-2%5C%5C0%26-7%26-14%26-5%5C%5C0%26-10%26-20%26-8%5Cend%7Barray%7D%5Cright%5D)
2.
- We subtract row two twice to row three of the previous matrix.
- we subtract 4/3 from row two to row 4.
- we subtract 7/3 from row two to row 5.
- we subtract 10/3 from row two to row 6.
We obtain the matrix
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&0&0\\0&0&3&2\\0&0&0&2\\0&0&0&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%260%260%260%5C%5C0%260%263%262%5C%5C0%260%260%262%5C%5C0%260%260%262%5Cend%7Barray%7D%5Cright%5D)
3.
we exchange rows three and four of the previous matrix and obtain the echelon form of the augmented matrix.
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&3&2\\0&0&0&0\\0&0&0&2\\0&0&0&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%260%263%262%5C%5C0%260%260%260%5C%5C0%260%260%262%5C%5C0%260%260%262%5Cend%7Barray%7D%5Cright%5D)
Since the only nonzero rows of the augmented matrix of the coefficient matrix are the first three, then the set
![\{\left[\begin{array}{c}1\\2\\3\end{array}\right],\left[\begin{array}{c}0\\-3\\-6\end{array}\right],\left[\begin{array}{c}0\\0\\3\end{array}\right] \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5C%5C3%5Cend%7Barray%7D%5Cright%5D%2C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C-3%5C%5C-6%5Cend%7Barray%7D%5Cright%5D%2C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5Cend%7Barray%7D%5Cright%5D%20%5C%7D)
is a basis for Row (A2)
Now, observe that the last two rows of the echelon form of the augmented matrix have the last coordinate different of zero. Then, the system is inconsistent. This means that the system has no solutions.
The expression would be j / 9. J is a variable.
Answer:
Mark up = 33.33%
Step-by-step explanation:
Given that:
buying price = $75
selling price = $100
First we will find the difference between two prices:
Difference = 100 - 75 = $25
We will find now that 25 is what percent of 75 so:
=25/75* 100
= 0.333333 * 100
= 33.33%
So the store marked the price by 33.33% instead of 30%
i hope it will help you!
1,938 is the answer but you have to make an estimate
Yes, it is a function as for every x value there is only one y value, you could also see that it is not a function by doing the vertical line test, this equation passes it.