Question

fill in the blank by performing the indicated elementary row operation(s). \\(\left\

$$\begin{array}{cc|c} 3 & 1 & 4 \\\\ 1 & 8 & -4 \\end{array}$$

\
ight \stackrel{-3r_2 + r_1}{\longrightarrow} \underline{\quad?\quad}\\)\
answer \\(\left\

$$\begin{array}{cc|c} \\boxed{} & \\boxed{} & \\boxed{} \\\\ \\boxed{} & \\boxed{} & \\boxed{} \\end{array}$$

\
ight\\)

Explanation:

Step1: Identify row operation rule

The operation $-3R_2+R_1$ means replace $R_1$ with $-3R_2 + R_1$; $R_2$ stays unchanged.

Step2: Calculate new $R_1$ first element

$\text{New } R_1(1) = -3(1) + 3 = -3 + 3 = 0$

Step3: Calculate new $R_1$ second element

$\text{New } R_1(2) = -3(8) + 1 = -24 + 1 = -23$

Step4: Calculate new $R_1$ third element

$\text{New } R_1(3) = -3(-4) + 4 = 12 + 4 = 16$

Step5: Keep $R_2$ unchanged

$R_2 = [1 \quad 8 \mid -4]$

Answer:

$$\begin{bmatrix} 0 & -23 & 16 \\ 1 & 8 & -4 \end{bmatrix}$$