The solution of the system is x = -2, y = -9 and, z = 1.
How to solve the system when the row-echelon form is given?
We can convert the row-echelon form into a normal system of equations using matrix multiplication.
We can solve the system as shown below:
The row-echelon form can also be written as:
[tex]\left[\begin{array}{ccc}1&0&-1\\0&1&5\\0&0&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}-3\\-4\\1\end{array}\right][/tex]
We can use matrix multiplication to get:
x - z = -3
y + 5z = -4
z = 1
Now substitute the value of z in the first equation:
x - 1 = -3
⇒ x = -2
Now substitute the value of z in the second equation:
y + 5 = -4
⇒ y = -9
We have found the values of x, y, and z to be -2, -9, and 1 respectively.
We have solved the system of equations to get x = -2, y = -9 and, z = 1.
Therefore, we have found the solution of the system to be x = -2, y = -9 and, z = 1. The correct answer is option C.
Learn more about the row-echelon form here: https://brainly.com/question/11736291
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