Question

(1 point) Consider the following two systems. (a) $\begin{cases} 6x + 4y = -1 \\ 2x + 7y = -1 \end{cases}$ (b) $\begin{cases} 6x + 4y = 3 \\ 2x + 7y = 3 \end{cases}$ (i) Find the inverse of the (common) coefficient matrix of the two systems. $A^{-1} = \[ \]$ (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. $B = egin{bmatrix} -1 \\ -1 \end{bmatrix}$ for system (a) and $B = egin{bmatrix} 3 \\ 3 \end{bmatrix}$ for system (b)). $B = egin{bmatrix} 3 \\ 3 \end{bmatrix}$ for system (b)). Solution to system (a): $x = \[ \]$, $y = \[ \]$ Solution to system (b): $x = \[ \]$, $y = \[ \]$

          (1 point) Consider the following two systems.
(a)
$\begin{cases} 6x + 4y = -1 \\ 2x + 7y = -1 \end{cases}$
(b)
$\begin{cases} 6x + 4y = 3 \\ 2x + 7y = 3 \end{cases}$
(i) Find the inverse of the (common) coefficient matrix of the two systems.
$A^{-1} = \[

\]$
(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. $B = egin{bmatrix} -1 \\ -1 \end{bmatrix}$ for system (a) and $B = egin{bmatrix} 3 \\ 3 \end{bmatrix}$ for system (b)).
$B = egin{bmatrix} 3 \\ 3 \end{bmatrix}$ for system (b)).
Solution to system (a): $x = \[
\]$, $y = \[
\]$
Solution to system (b): $x = \[
\]$, $y = \[
\]$

(1 point) Consider the following two systems.
(a)
6x + 4y = -1
2x + 7y = -1
(b)
6x + 4y = 3
2x + 7y = 3
(i) Find the inverse of the (common) coefficient matrix of the two systems.
A^-1 =
(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 = eginbmatrix -1
-1 for system (a) and B = eginbmatrix 3
3 for system (b)).
B = eginbmatrix 3
3 for system (b)).
Solution to system (a): x =, y =
Solution to system (b): x =, y =

Added by Jerry D.

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