Solve for X and Y using matrices

[Ti-Pro.doc.]

  [[-3  4] * [[2 -3]
  [y   1]]   [x -2]] = [[-34 1]
                       [-3 -17]]

 

[soroban]

Hello, Ti-Pro.doc.!

Is there a typo?
As written, there is no solution.

  [[-3  4] * [[2 -3]
  [y   1]]   [x -2]] = [[-34 1]
                       [-3 -17]]

Perhaps you meant: \(\displaystyle \L\:\begin{bmatrix}-3 & 4 \\ y & 1\end{bmatrix}\,\cdot\,
\begin{bmatrix}2 & -3 \\ x & -2\end{bmatrix} \:=\:
\begin{bmatrix}-34 & 1 \\ 3 & -17\end{bmatrix}\)

Multiply: \(\displaystyle \L\:\begin{bmatrix}-6\,+\,4x & 9\,-\,8 \\ 2y\,+\,x & -3y\,-\,2\end{bmatrix}\:=\:\begin{bmatrix}-34 & 1 \\ 3 & -17\end{bmatrix}\)

Equate components: \(\displaystyle \L\:\begin{array}{ccc}-6\,+\,4x\:=\:-34 & \qquad & 9\,-\,8\:=\:1 \\ 2y\,+\,x\:=\:3 & \qquad & -3y\,-\,2\:=\:-17\end{array}\)


The first equation is: \(\displaystyle \:-6\,+\,4x\:=\:-34\;\;\Rightarrow\;\;4x\:=\:-38\;\;\Rightarrow\;\;\L\fbox{x\,=\,-7}\)

The fourth is: \(\displaystyle \:-3y\,-\,2\:=\:-17\;\;\Rightarrow\;\;-3y\,=\,-15\;\;\Rightarrow\;\;\L\fbox{y\,=\,5}\)

And they satisfy the third equation: \(\displaystyle \:2(5)\,+\,(-7)\:=\:3\)

 

[Ti-Pro.doc.]

yeah i didn't know how to do that fancy layout.
i new to the whole system here.
thank 4 your help.