MHB 5.2 another Ax=B with an inverse

[karush]

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

 

[topsquark]

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

If you are asking if it makes sense to write out the two equations in x and y that don't involve z and solving the simpler system for x and y then I think you are doing just fine.

-Dan

 

[HOI]

You are given that A^{-1} exists and is equal to \begin{bmatrix}1& 4 & 0 \\ 2 & 3 & 0 \\ 4 & 2 & 2 \end{bmatrix}. So the solution to AX= B is just X= A^{-1}B.

With B given as \begin{bmatrix}-1 \\ 2 \\ 3\end{bmatrix}, as here, we have immediately that X= \begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}1& 4 & 0 \\ 2 & 3 & 0 \\ 4 & 2 & 2 \end{bmatrix}\begin{bmatrix}-1 \\ 2 \\ 3\end{bmatrix}= \begin{bmatrix}7 \\ 4 \\ 6\end{bmatrix}. That is, x= 7, y= 4, and z= 6. But we can write the initial "AX= B" as the system of equstions -\frac{3}{5}x+ \frac{4}{5}y= -1, \frac{2}{5}x- \frac{1}{5}y= 2, and \frac{4}{5}x- \frac{7}{5}y+ \frac{1}{2}z= 3. As topsquark said, the fact that the first two entries in the last column are "0" mean that the first two equations have no "z" and can be solved for x and y indpendently of z. I would first multiply each equation by 5 to get -3x+ 4y= -5 and 2x- y= 10. Multiply the second equation by 4 to geta 8x- 4y= 40 and add that to the first equation to eliminate y and get 5x= 35 so that x= 7. Setting x= 7 in -3x+ 4y= -5 gives -21+ 4y= -5 so that 4y= 16 and y= 4. Finally, put x= 7, y= 4 into the last equation to get \frac{28}{5}- \frac{28}{5}+ \frac{1}{2}z= \frac{1}{2}z= 3 so that z= 6. We get x= 7, y= 4, z= 6 as before.