- #1
karush
Gold Member
MHB
- 3,269
- 5
Suppose we are given
$A^{-1} =
\begin{bmatrix}
1 & 4 & 0 \\
2 & 3 & 0 \\
4 & 2 & 2
\end{bmatrix}
=
\left[\begin{array}{rrr|rrr}
1&4&0&1&0&0\\
2&3&0&0&1&0\\
4&2&2&0&0&1
\end{array}\right]$
then
$A=
\begin{bmatrix}
-\frac{3}{5}&\frac{4}{5}&0\\ \frac{2}{5}&-\frac{1}{5}&0\\ \frac{4}{5}&-\frac{7}{5}&\frac{1}{2}
\end{bmatrix}$
Solve the matrix equation $AX=B$ to find $x$, $y$ and $z$ where $X =
\begin{bmatrix}
x \\y \\z
\end{bmatrix},
\textit{ and }
B = \begin{bmatrix}
-1 \\2 \\3
\end{bmatrix}$
$\therefore AX=B$ is
$\begin{bmatrix}
-\frac{3}{5}&\frac{4}{5}&0\\ \frac{2}{5}&-\frac{1}{5}&0\\ \frac{4}{5}&-\frac{7}{5}&\frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
x \\y \\z
\end{bmatrix}
=\begin{bmatrix}
-1 \\2 \\3
\end{bmatrix}$
$W\vert A$ returned
$x=7,y=4,z=6$ok my question is that A has 2 zero's in $C_3$
I was able to get x=7 and y=4 by a simple double simultanious equation
but not sure what the proper methed is take advantage of the 2 zeros
$A^{-1} =
\begin{bmatrix}
1 & 4 & 0 \\
2 & 3 & 0 \\
4 & 2 & 2
\end{bmatrix}
=
\left[\begin{array}{rrr|rrr}
1&4&0&1&0&0\\
2&3&0&0&1&0\\
4&2&2&0&0&1
\end{array}\right]$
then
$A=
\begin{bmatrix}
-\frac{3}{5}&\frac{4}{5}&0\\ \frac{2}{5}&-\frac{1}{5}&0\\ \frac{4}{5}&-\frac{7}{5}&\frac{1}{2}
\end{bmatrix}$
Solve the matrix equation $AX=B$ to find $x$, $y$ and $z$ where $X =
\begin{bmatrix}
x \\y \\z
\end{bmatrix},
\textit{ and }
B = \begin{bmatrix}
-1 \\2 \\3
\end{bmatrix}$
$\therefore AX=B$ is
$\begin{bmatrix}
-\frac{3}{5}&\frac{4}{5}&0\\ \frac{2}{5}&-\frac{1}{5}&0\\ \frac{4}{5}&-\frac{7}{5}&\frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
x \\y \\z
\end{bmatrix}
=\begin{bmatrix}
-1 \\2 \\3
\end{bmatrix}$
$W\vert A$ returned
$x=7,y=4,z=6$ok my question is that A has 2 zero's in $C_3$
I was able to get x=7 and y=4 by a simple double simultanious equation
but not sure what the proper methed is take advantage of the 2 zeros