Question

question 5 of 10, step 1 of 1
fill in the blank by performing the indicated elementary row operation(s).
\\(\

$$\begin{bmatrix}-3 & 1 & -8\\\\-6 & 2 & 2\\\\15 & -5 & -9\\\\15 & -5 & 7\\end{bmatrix}$$

\\)\\(\xrightarrow{3r_2+r_1, -15r_2+r_3,?}\\)\\(\

$$\begin{bmatrix}\\square & \\square & \\square\\\\\\square & \\square & \\square\\\\\\square & \\square & \\square\\\\\\square & \\square & \\square\\end{bmatrix}$$

\\)
answer

Explanation:

Step1: Define row operations

The operations are $3R_2+R_1
ightarrow R_1$ and $-15R_2+R_3
ightarrow R_3$; $R_2$ stays unchanged.

Step2: Calculate new $R_1$

For each entry:
First column: $3(1)+(-3)=0$
Second column: $3(2)+(-6)=0$
Third column: $3(-5)+15=0$
Fourth column: $3(-5)+15=0$
New $R_1$:

$$\begin{bmatrix}0&0&0&0\end{bmatrix}$$

Step3: Keep $R_2$ unchanged

$R_2$:

$$\begin{bmatrix}1&2&-5&-5\end{bmatrix}$$

Step4: Calculate new $R_3$

For each entry:
First column: $-15(1)+(-8)=-23$
Second column: $-15(2)+2=-28$
Third column: $-15(-5)+(-9)=66$
Fourth column: $-15(-5)+7=82$
New $R_3$:

$$\begin{bmatrix}-23&-28&66&82\end{bmatrix}$$

Answer:

$$\begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & 2 & -5 & -5 \\ -23 & -28 & 66 & 82 \end{bmatrix}$$