Question

Consider the following two systems. (a) $\begin{cases} 6x - 5y = 1 \ -2x + 5y = 3 \end{cases}$ (b) $\begin{cases} 6x - 5y = -3 \ -2x + 5y = -4 \end{cases}$ (i) Find the inverse of the (common) coefficient matrix of the two systems. $A^{-1} = \begin{bmatrix} \\ \\ \end{bmatrix}$ (ii) Find the solutions to the two systems by using the inverse, i.e. by evaluating $A^{-1}B$ where $B$ represents the right hand side (i.e. $B = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ for system (a) and $B = \begin{bmatrix} -3 \\ -4 \end{bmatrix}$ for system (b)). Solution to system (a): $x = \Box, y = \Box$ Solution to system (b): $x = \Box, y = \Box$

          Consider the following two systems.
(a)
$\begin{cases} 6x - 5y = 1 \ -2x + 5y = 3 \end{cases}$
(b)
$\begin{cases} 6x - 5y = -3 \ -2x + 5y = -4 \end{cases}$
(i) Find the inverse of the (common) coefficient matrix of the two systems.
$A^{-1} = \begin{bmatrix} \\ \\ \end{bmatrix}$
(ii) Find the solutions to the two systems by using the inverse, i.e. by evaluating $A^{-1}B$ where $B$ represents the right hand side (i.e. $B = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ for system (a) and $B = \begin{bmatrix} -3 \\ -4 \end{bmatrix}$ for system (b)).
Solution to system (a): $x = \Box, y = \Box$
Solution to system (b): $x = \Box, y = \Box$

Consider the following two systems.
(a)
6x - 5y = 1 -2x + 5y = 3
(b)
6x - 5y = -3 -2x + 5y = -4
(i) Find the inverse of the (common) coefficient matrix of the two systems.
A^-1 =
< b m a t r i x >
(ii) Find the solutions to the two systems by using the inverse, i.e. by evaluating A^-1B where B represents the right hand side (i.e. B =
< b m a t r i x > for system (a) and B =
< b m a t r i x > for system (b)).
Solution to system (a): x = , y =
Solution to system (b): x = , y =

Added by Julia R.

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