[tex]\begin{gathered} \begin{bmatrix}{1} & {0.8} & {-0.1} \\ {-3} & {2} & {1.1} \\ {4} & {-0.2} & {9}\end{bmatrix}\rightarrow0.3r_1+r_2 \\ \\ This\text{ gives} \\ \begin{bmatrix}{1} & {0.8} & {-0.1} \\ {(0.3\times1)-0.3} & {(0.8\times0.3)+2} & {(0.3\times-1)+1.1} \\ {4} & {-0.2} & {9}\end{bmatrix} \\ \begin{bmatrix}{1} & {0.8} & {-0.1} \\ 0{} & {2.24} & {1.07} \\ {4} & {-0.2} & {9}\end{bmatrix} \\ \end{gathered}[/tex]
The above shows when 0.3r1+r2 is applied to the second row.
Step 2
we apply -4r1+r3 to the third row
[tex]\begin{gathered} \begin{bmatrix}{1} & {0.8} & {-0.1} \\ 0{} & {2.24} & {1.07} \\ {4} & {-0.2} & {9}\end{bmatrix}-4r_1+r_{3\text{ }}to\text{ the third row} \\ \\ \begin{bmatrix}{1} & {0.8} & {-0.1} \\ 0{} & {2.24} & {1.07} \\ {(-4\times1)+4} & {(-4\times0.8)-0.2} & {(-4\times-0.1)+9}\end{bmatrix} \\ \begin{bmatrix}{1} & {0.8} & {-0.1} \\ 0{} & {2.24} & {1.07} \\ {0} & {-3.4} & {9.4}\end{bmatrix} \\ \end{gathered}[/tex]
step 3
[tex]\begin{gathered} \text{Therefore the final answer is the first option.} \\ \begin{bmatrix}{1} & {0.8} & {-0.1} \\ 0{} & {2.24} & {1.07} \\ {0} & {-3.4} & {9.4}\end{bmatrix} \end{gathered}[/tex]
