Question

Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.

Answer 1

C

Explanation:

EDGE BOY (or girl)

Answer 2

c

Explanation:

First, we can transform this into a matrix. The x coefficients will be the first ones for each row, the y coefficients the second column, etc.

`\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right]`

Next, we can define a reduced row echelon form matrix as follows:

With the leading entry being the first non zero number in the first row, the leading entry in each row must be 1. Next, there must only be 0s above and below the leading entry. After that, the leading entry of a row must be to the left of the leading entry of the next row. Finally, rows with all zeros should be at the bottom of the matrix.

Because there are 3 rows and we want to solve for 3 variables, making the desired matrix of form

`\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]` for the first three rows and columns. This would make the equation translate to

x= something

y= something

z = something, making it easy to solve for x, y, and z.

Going back to our matrix,

`\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right]` ,

we can start by removing the nonzero values from the first column for rows 2 and 3 to reach the first column of the desired matrix. We can do this by multiplying the first row by -6 and adding it to the second row, as well as multiplying the first row by -1 and adding it to the third row. This results in

`\left[\begin{array}{cccc}1&-2&3&-2\\0&14&-16&-36\\0&6&0&-36\end{array}\right]`

as our matrix. * Next, we can reach the second column of our desired matrix by first multiplying the second row by (2/14) and adding it to the first row as well as multiplying the second row by (-6/14) and adding it to the third row. This eliminates the nonzero values from all rows in the second column except for the second row. This results in

`\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&14&-16&-36\\0&0&96/14&-288/14\end{array}\right]`

After that, to reach the desired second column, we can divide the second row by 14, resulting in

`\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&1&-16/14&-36/14\\0&0&96/14&-288/14\end{array}\right]`

Finally, to remove the zeros from all rows in the third column outside of the third row, we can multiply the third row by (16/96) and adding it to the second row as well as multiplying the third row by (-10/96) and adding it to the first row. This results in

`\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&96/14&-288/14\end{array}\right]`

We can then divide the third row by -96/14 to reach the desired third column, making the reduced row echelon form of the matrix

`\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&1&-3\end{array}\right]`

Therefore,

x=-5

y=-6

z=-3

* we could also switch the second and third rows here to make the process a little simpler