The point of intersection of the lines $\displaystyle 2x+3y=a$ and $\displaystyle x-2y=5$ is $\displaystyle (2,1)$. Use a matrix method to find $\displaystyle a$.
Thanks
You need to apply row reduction methods to this matrix:
$\displaystyle \left[\begin{array}{cc|c}2&3&a\\1&-2&5\end{array}\right]\xrightarrow{R_1\leftrightarrow R_2}{}\left[\begin{array}{cc|c}1&-2&5\\2&3&a\end{array}\right]\xrightarrow{-2R_1+R_2\rightarrow R_2}{}\left[\begin{array}{cc|c}1&-2&5\\0&7&a-10\end{array}\right]$ $\displaystyle \xrightarrow{\tfrac{1}{7}R_2\rightarrow R_2}\left[\begin{array}{cc|c}1&-2&5\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]\xrightarrow{2R_2+R_1\rightarrow R_1}{}\left[\begin{array}{cc|c}1&0&5+\tfrac{2}{7}\left(a-10\right)\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]$
Since the intersection point is (2,1), its matrix representation would be $\displaystyle \left[\begin{array}{cc|c}1&0&2\\0&1&1\end{array}\right]$, it follows that $\displaystyle \left[\begin{array}{cc|c}1&0&5+\tfrac{2}{7}\left(a-10\right)\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]=\left[\begin{array}{cc|c}1&0&2\\0&1&1\end{array}\right]\implies \tfrac{1}{7}\left(a-10\right)=1$ and $\displaystyle 5+\tfrac{2}{7}\left(a-10\right)=2$
However, if I solve the first equation for a, I get $\displaystyle a=17$, but when I solve the second equation for a, I get $\displaystyle a=-\frac{1}{2}$....
Did you copy the intersection pt correctly?
The problem is that the second line doesn't even pass through the point of intersection!
$\displaystyle f(x,y)=x-2y,\ \ f(2,1)=0,\ \ 0\neq5$
Basically I think the second equation should be $\displaystyle x-2y=0$ and then as far as I can see a=7 is the answer, which makes sense given that if you plug (2,1) into the first line equation you get ... 7. No real need for matrices here I fear...