The row echelon form of the matrix is presented as follows;
![\begin{bmatrix}1 &-2 &-5 \\ 0& 1 & -7\\ 0&0 &1 \\\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%20%26-2%20%20%26-5%20%20%5C%5C%200%26%201%20%26%20%20-7%5C%5C%200%260%20%20%261%20%20%5C%5C%5Cend%7Bbmatrix%7D)
<h3>What is the row echelon form of a matrix?</h3>
The row echelon form of a matrix has the rows consisting entirely of zeros at the bottom, and the first entry of a row that is not entirely zero is a one.
The given matrix is presented as follows;
![\begin{bmatrix}-3 &6 &15 \\ 2& -6 & 4\\ 1&0 &-1 \\\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D-3%20%266%20%20%2615%20%20%5C%5C%202%26%20-6%20%26%20%204%5C%5C%201%260%20%20%26-1%20%20%5C%5C%5Cend%7Bbmatrix%7D)
The conditions of a matrix in the row echelon form are as follows;
- There are row having nonzero entries above the zero rows.
- The first nonzero entry in a nonzero row is a one.
- The location of the leading one in a nonzero row is to the left of the leading one in the next lower rows.
Dividing Row 1 by -3 gives:
![\begin{bmatrix}1 &-2 &-5 \\ 2& -6 & 4\\ 1&0 &-1 \\\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%20%26-2%20%20%26-5%20%20%5C%5C%202%26%20-6%20%26%20%204%5C%5C%201%260%20%20%26-1%20%20%5C%5C%5Cend%7Bbmatrix%7D)
Multiplying Row 1 by 2 and subtracting the result from Row 2 gives;
![\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 1&0 &-1 \\\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%20%26-2%20%20%26-5%20%20%5C%5C%200%26%20-2%20%26%20%2014%5C%5C%201%260%20%20%26-1%20%20%5C%5C%5Cend%7Bbmatrix%7D)
Subtracting Row 1 from Row 3 gives;
![\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 0&2 &4 \\\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%20%26-2%20%20%26-5%20%20%5C%5C%200%26%20-2%20%26%20%2014%5C%5C%200%262%20%20%264%20%20%5C%5C%5Cend%7Bbmatrix%7D)
Adding Row 2 to Row 3 gives;
![\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 0&0 &18 \\\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%20%26-2%20%20%26-5%20%20%5C%5C%200%26%20-2%20%26%20%2014%5C%5C%200%260%20%20%2618%20%20%5C%5C%5Cend%7Bbmatrix%7D)
Dividing Row 2 by -2, and Row 3 by 18 gives;
![\begin{bmatrix}1 &-2 &-5 \\ 0& 1 & -7\\ 0&0 &1 \\\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%20%26-2%20%20%26-5%20%20%5C%5C%200%26%201%20%26%20%20-7%5C%5C%200%260%20%20%261%20%20%5C%5C%5Cend%7Bbmatrix%7D)
The above matrix is in the row echelon form
Learn more about the row echelon form here:
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Interest equation,,, I=prt,,4485=6500*0.115 t...can you solve for t??
37 . i think :D .! if not sorry:(
Answer:
A
Step-by-step explanation:
There is an x in all for terms that means that x = 0 is one of the factors and H is incorrect.
One of the remaining factors is x^2 - 9 which factors into (x - 3)(x + 3)
You are told that x = 1 is a zero which means x - 1 is a factor.
The answer is g(x) = x (x - 3)(x + 3) (x - 1) which is A