Question

Consider the system {6q+3r=-1.5
q-5r=21.75
a) write the system as a matrix equation
b) solve the system

Answer 1

\[\begin{pmatrix}
6&3\\1&-5
\end{pmatrix}\begin{pmatrix}q\\r\end{pmatrix}=\begin{pmatrix}-1.5\\21.75\end{pmatrix}\]

Answer 2

I got that part but what about b?

Answer 3

You have an equation of the form
\[{\bf A}{\bf x}={\bf b}\]
where \(\bf A\) is the coefficient matrix (leftmost above), \(\bf x\) is the variable matrix, and \(\bf b\) is the constant matrix. What you can do is multiply both sides by the inverse of \(\bf A\):
\[{\bf A}^{-1}{\bf A}{\bf x}={\bf A}^{-1}{\bf b}\]
and you have \({\bf A}^{-1}{\bf A}={\bf I}\), where \(\bf I\) is the identity matrix, and you're left with
\[{\bf x}={\bf A}^{-1}{\bf b}\]
In the context of your problem, you have
\[\begin{pmatrix}
q\\r
\end{pmatrix}=\begin{pmatrix}
6&3\\1&-5
\end{pmatrix}^{-1}\begin{pmatrix}
-1.5\\21.75
\end{pmatrix}\]
Do you know how to find the inverse of a 2x2 matrix?

Answer 4

yes

Answer 5

@SithsAndGiggles

Answer 6

I believe you said you know how to find the inverse...

Answer 7

okay for the first system?

Answer 8

What did you get for the solution? q & r ??

Answer 9

i dont know

Answer 10

I assume they want you to solve it using the matrix method.

Answer 11

solve for r in a similar manner by substituting the constants for the coefficients of r