MHB -5.5 Solve the matrix equation AX=B to find x and y

[karush]

2000
5.1 Suppose that we know that
$A^{-1}=\begin{bmatrix}1&3\\2&5 \end{bmatrix}$
Solve the matrix equation $AX=B$ to find $x$ and $y$ where
$X=\begin{bmatrix}x\\y \end{bmatrix}\& \quad B=\begin{bmatrix}1\\3 \end{bmatrix}$
ok well first find A
$A=\begin{bmatrix}1&3\\2&5 \end{bmatrix}^{-1}
=\left[ \begin{array}{rr|rr}1&3&1&0 \\ 2&5&0&1\end{array}\right]
=\left[ \begin{array}{rr|rr}1&0&-5&3 \\ 0&1&2&-1\end{array}\right]
=\left[ \begin{array}{rr} -5 & 3 \\ 2 & -1 \end{array} \right]$
then we have
$\left[ \begin{array}{rr} -5 & 3 \\ 2 & -1 \end{array} \right]
\begin{bmatrix}x\\y \end{bmatrix}
=\begin{bmatrix}1\\3 \end{bmatrix}$
ok just seeing If I am going the right direction on this .. if so the rest would be a simultaneous equation

 

[skeeter]

$AX=B$

$A^{-1}AX = A^{-1}B$

$X = A^{-1}B$

$\begin{bmatrix}
x \\ y
\end {bmatrix} = \begin{bmatrix}
1 & 3\\
2 & 5
\end{bmatrix} \cdot \begin{bmatrix}
1\\3
\end{bmatrix}$

 

[karush]

ok so they both give the same answer as

$$x=10,\:y=17$$

so assume your equation is easier since you derive x and y directly?

 

[skeeter]

ok so they both give the same answer as

$$x=10,\:y=17$$

so assume your equation is easier since you derive x and y directly?

no need to determine matrix A

 

[karush]

total awesome