+701 i have tried over 6 times and ive been stuck on it for the past hour, can someone please explain how this can be solved?
\(\begin{bmatrix} 1 & -1&-2 |-6\\ 3 & 2&0 |25\\ -4&1&-1 |12 \end{bmatrix}\)
sorry if this is not very legible, here is the equation for it, i just tried to convert it into an augmented matrix :')
\(x-y-2z=-6 \)
\(3x+2y=-25\)
\(-4x+y-z=12\)
Thank you so much.
edit: i have been trying for the past 3 hours and i still cannot solve this lol
+6251 well for one you've got 25 in the matrix where it should be -25.
\(\begin{pmatrix} 1&-1&-2&|&-6\\ 3&2&0&|&-25\\ -4&1&-1&|&12 \end{pmatrix}\\ \begin{pmatrix} 1&-1&-2&|&-6\\ 0&5&6&|&-7\\ -4&1&-1&|&12 \end{pmatrix}\\ \begin{pmatrix} 1&-1&-2&|&-6\\ 0&5&6&|&-7\\ 0&-3&-9&|&-12 \end{pmatrix}\\\)
Now exchange rows 2 and 3 and normalize row 2
\(\begin{pmatrix} 1 & -1 & -2 & -6 \\ 0 & 1 & 3 & 4 \\ 0 & 5 & 6 & -7 \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 5 & 6 & -7 \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & -9 & -27 \end{pmatrix}\\ \)
Normalize row 3 and continue
\(\begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \)
\(\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-5\\-5\\3\end{pmatrix}\)
