Question

3. Consider the system below.

Solve the system by using a matrix equation. Show your work.

Picture of the equation here: https://gyazo.com/ec01a413615226cc46968e79f970ac74

Answer 1

Given the system


`4x-2y=-12\ .\ .\ .\ (1) \\ 3x-y=-3\ .\ .\ .\ (2)`

A matrix equation is of the form AX = B, where A is the matrix of the coefficients of the variables, X is the matrix of the variables, and B is the matrix of the constants.

Thus, rewriting the system into matrix equation we have:


`\left[\begin{array}{cc}4&-2\\3&-1\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}-12\\-3\end{array}\right] \\ \\ \Rightarrow\left[\begin{array}{c}x\\y\end{array}\right]= \left[\begin{array}{cc}4&-2\\3&-1\end{array}\right]^(-1)\left[\begin{array}{c}-12\\-3\end{array}\right] \\ \\ = (1)/(-4-(-6)) \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]\left[\begin{array}{c}-12\\-3\end{array}\right]= (1)/(-4+6) \left[\begin{array}{c}12-6\\36-12\end{array}\right]`

`(1)/(2) \left[\begin{array}{c}6\\24\end{array}\right]=\left[\begin{array}{c}3\\12\end{array}\right]`

Therefore, x = 3 and y = 12.