Question

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

Answer 1

Write the system in augmented-matrix form:

$\left[\begin{array}{ccc|c}2&2&4&16\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]$

Multiply through row 1 by 1/2:

$\left[\begin{array}{ccc|c}1&1&2&8\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]$

Add -1(row 1) to row 3, and add -5(row 1) to row 2:

$\left[\begin{array}{ccc|c}1&1&2&8\\0&-7&-7&-41\\0&1&-5&-17\end{array}\right]$

Swap rows 2 and 3:

$\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&-7&-7&-41\end{array}\right]$

Add -7(row 2) to row 3:

$\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&-42&-160\end{array}\right]$

Multiply through row 3 by -1/42:

$\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&1&\frac{80}{21}\end{array}\right]$

Add 5(row 3) to row 2:

$\left[\begin{array}{ccc|c}1&1&2&8\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]$

Add -1(row 2) and -2(row3) to row 1:

$\left[\begin{array}{ccc|c}1&0&0&-\frac53\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]$

So the solution to the system is

$x=-\dfrac53,y=\dfrac{43}{21},z=\dfrac{80}{21}$

Answer 2

Answer:

what the other person said

Step-by-step explanation:

i need more points/brainlys and this is an old question soo