Not yet answered Marked out of 1 Remove flag Write the augmented matrix for the system of linear equations. \[ \left\{\begin{array}{r} 6 x-2 z=20 \\ 9 y+7 z=56 \\ 2 x+8 y+9 z=53 \end{array}\right. \] Select one: a. $\quad\left[\begin{array}{rrr|r}6 & 0 & -2 & 20 \\ 0 & 9 & 7 & 56 \\ 2 & 8 & 9 & 53\end{array}\right]$ b. $\quad\left[\begin{array}{rrr|r}6 & -2 & 0 & 20 \\ 9 & 7 & 0 & 56 \\ 2 & 8 & 9 & 53\end{array}\right]$ C. $\quad\left[\begin{array}{rrr|r}6 & 0 & 2 & 20 \\ 0 & 9 & 8 & 56 \\ -2 & 7 & 9 & 53\end{array}\right]$ d. $\quad\left[\begin{array}{rrr|r}0 & -2 & 6 & 20 \\ 9 & 0 & 7 & 56 \\ 2 & 8 & 9 & 53\end{array}\right]$

Step 1 ：The question is asking for the augmented matrix of the given system of linear equations. An augmented matrix is a matrix obtained by appending the columns of two given matrices, usually for the purpose of performing the same elementary row operations on each of the given matrices. In this case, the augmented matrix will be a 3x4 matrix, where the first three columns represent the coefficients of the variables x, y, and z in each equation, and the fourth column represents the constants on the right side of each equation.

Step 2 ：To form the augmented matrix, we simply need to write down the coefficients of x, y, and z and the constants from each equation in the order they appear.

Step 3 ：From the first equation, we have 6 as the coefficient of x, 0 as the coefficient of y (since y does not appear in the equation), -2 as the coefficient of z, and 20 as the constant.

Step 4 ：From the second equation, we have 0 as the coefficient of x (since x does not appear in the equation), 9 as the coefficient of y, 7 as the coefficient of z, and 56 as the constant.

Step 5 ：From the third equation, we have 2 as the coefficient of x, 8 as the coefficient of y, 9 as the coefficient of z, and 53 as the constant.

Step 6 ：So, the augmented matrix should be: \[\left[\begin{array}{rrr|r}6 & 0 & -2 & 20 \ 0 & 9 & 7 & 56 \ 2 & 8 & 9 & 53\end{array}\right]\]

Step 7 ：Comparing this with the options given in the question, the correct answer is option a. \(\boxed{\left[\begin{array}{rrr|r}6 & 0 & -2 & 20 \ 0 & 9 & 7 & 56 \ 2 & 8 & 9 & 53\end{array}\right]}\)