- #1

karush

Gold Member

MHB

- 3,269

- 5

$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