Question

What is the solution to this system of equations? x+3y−z=6 4x−2y+2z=−10 6x+z=−12 (−4, 0, 12) (0, −2, −12) (2, 1, −3) (−3, 5, 6)

Answer 1

Answer:

Solution :  (− 3, 5, 6)

Step-by-step explanation:

We have the following system of equations that we have to solve for,

$\begin{bmatrix}x+3y-z=6\\ 4x-2y+2z=-10\\ 6x+z=-12\end{bmatrix}$

To solve this problem we can start by writing the matrix with their respective coefficients --- (1)

$\begin{bmatrix}1&3&-1&|&6\\ 4&-2&2&|&-10\\ 6&0&1&|&-12\end{bmatrix}$

Now we can reduce this to row echelon form, receiving our solution --- (2)

$\begin{pmatrix}1&3&-1&6\\ 4&-2&2&-10\\ 6&0&1&-12\end{pmatrix}$ Swap row 1 and 3,

$\begin{pmatrix}6&0&1&-12\\ 4&-2&2&-10\\ 1&3&-1&6\end{pmatrix}$ Cancel leading coefficient in row 3,

$\begin{pmatrix}6&0&1&-12\\ 0&-2&\frac{4}{3}&-2\\ 0&3&-\frac{7}{6}&8\end{pmatrix}$ Swap row 2 and 3

$\begin{pmatrix}6&0&1&-12\\ 0&3&-\frac{7}{6}&8\\ 0&-2&\frac{4}{3}&-2\end{pmatrix}$ Cancel leading coefficient in row 3

$\begin{pmatrix}6&0&1&-12\\ 0&3&-\frac{7}{6}&8\\ 0&0&\frac{5}{9}&\frac{10}{3}\end{pmatrix}$

At this point you can see that we have to cancel the leading coefficient in each row, to row echelon form. Continuing this pattern we have the following matrix,

$\begin{bmatrix}1&0&0&|&-3\\ 0&1&0&|&5\\ 0&0&1&|&6\end{bmatrix}$

As you can see, x = - 3, y = 5, and z = 6, giving us a solution of (− 3, 5, 6). This is the fourth option.