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.
Answer:
the 2nd option is the right one on edgenuiity
Step-by-step explanation:
Answer: It is the distance that -5 is form 0 on the number line
Step-by-step explanation:
We can define the absolute value of any number:
IxI as:
"The distance between the point x and the origin in the coordinate axis"
If we are in one dimension, x is just a value, and the coordinate axis will be a number line.
Then we could rewrite this as:
"is the distance between the value x and the 0 in the number line"
So:
IxI = x if x > 0
IxI = -x if x < 0.
Then:
I -5 I
Is the distance between the number -5, and the 0 in the number line, and this is:
I - 5 I = 5.
The correct option is:
"It is the distance that -5 is form 0 on the number line"
Answer: 3i
Sq rt. of -9
Can break into sq rt of 9 and -1
-1 = i
Sq rt. of 9 is 3
3i