Question

Solve the system of equations by transforming a matrix representing the system of equation into reduced row echelon form.

2x + y − 3z= −20
x + 2y + z= −3
x − y + 5z= 19

What is the solution to the system of equations?

Answer 1

1 0 0 -3

0 1 0 -2

0 0 1 4

AND

(-3, -2, 4)

Explanation:

I don't have a step-by-step explanation because I still don't understand this sh*t myself. The reason I know it's the right answer is because I just got a 30% on my quiz and it shows you the right answers after. Good luck in this class! You're gonna need it.

Answer 2

Take the augmented matrix,

`\left[\begin{array}c2&1&-3&-20\\1&2&1&-3\\1&-1&5&19\end{array}\right]`

Swap the row 1 and row 2:

`\left[\begin{array}ccc1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]`

Add -2(row 1) to row 2, and -1(row 1) to row 3:

`\left[\begin{array}ccc1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]`

Add -1(row 2) to row 3:

`\left[\begin{array}ccc1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]`

Multiply through row 3 by 1/9:

`\left[\begin{array}c1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]`

Add 5(row 3) to row 2:

`\left[\begin{array}c1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]`

Multiply through row 2 by -1/3:

`\left[\begin{array}ccc1&2&1&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]`

Add -2(row 2) and -1(row 3) to row 1:

`\left[\begin{array}ccc1&0&0&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]`

So we have
`\boxed{x=-3,y=-2,z=4}`.