1 Answer

Ask a Question

NeoNexus DeMortis · Answer 1 · 2023-01-17T10:58:54+0000

ANSWER:

$-2R_2+3R_1=\begin{pmatrix}-12 & 20 & 8 \\ -8 & 1 & -3\end{pmatrix}$

Explanation:

We have the following matrix:

$A=\begin{pmatrix}-3 & 5 & 2 \\ 8 & -1 & 3\end{pmatrix}$

We apply the operation where R1 is the first row and R2 is the second row, therefore:

$\begin{gathered} -2R_2=\begin{pmatrix}-3 & \:5 & \:2 \\ \:\:-2\cdot8 & -2\cdot-1 & -2\cdot3\end{pmatrix}=\begin{pmatrix}-3 & \:5 & \:2 \\ \:\:-16 & 2 & -6\end{pmatrix} \\ \\ 3R_1=\begin{pmatrix}3\cdot-3 & 3\cdot5 & 3\cdot2 \\ \:8 & -1 & 3\end{pmatrix}=\begin{pmatrix}-9 & 15 & 6 \\ \:8 & -1 & 3\end{pmatrix} \\ \\ -2R_2+3R_1=\begin{pmatrix}-3 & \:5 & \:2 \\ \:\:-16 & 2 & -6\end{pmatrix}+\begin{pmatrix}-9 & 15 & 6 \\ \:8 & -1 & 3\end{pmatrix}=\begin{pmatrix}-3+-9 & 5+15 & 2+6 \\ -16+\:8 & 2+-1 & -6+3\end{pmatrix} \\ \\ -2R_2+3R_1=\begin{pmatrix}-12 & 20 & 8 \\ -8 & 1 & -3\end{pmatrix} \end{gathered}$

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Other Questions