# Matrix Question

• June 30th 2009, 11:00 PM
Stroodle
Matrix Question
The point of intersection of the lines $2x+3y=a$ and $x-2y=5$ is $(2,1)$. Use a matrix method to find $a$.

Thanks
• June 30th 2009, 11:32 PM
Chris L T521
Quote:

Originally Posted by Stroodle
The point of intersection of the lines $2x+3y=a$ and $x-2y=5$ is $(2,1)$. Use a matrix method to find $a$.

Thanks

You need to apply row reduction methods to this matrix:

$\left[\begin{array}{cc|c}2&3&a\\1&-2&5\end{array}\right]\xrightarrow{R_1\leftrightarrow R_2}{}\left[\begin{array}{cc|c}1&-2&5\\2&3&a\end{array}\right]\xrightarrow{-2R_1+R_2\rightarrow R_2}{}\left[\begin{array}{cc|c}1&-2&5\\0&7&a-10\end{array}\right]$ $\xrightarrow{\tfrac{1}{7}R_2\rightarrow R_2}\left[\begin{array}{cc|c}1&-2&5\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]\xrightarrow{2R_2+R_1\rightarrow R_1}{}\left[\begin{array}{cc|c}1&0&5+\tfrac{2}{7}\left(a-10\right)\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]$

Since the intersection point is (2,1), its matrix representation would be $\left[\begin{array}{cc|c}1&0&2\\0&1&1\end{array}\right]$, it follows that $\left[\begin{array}{cc|c}1&0&5+\tfrac{2}{7}\left(a-10\right)\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]=\left[\begin{array}{cc|c}1&0&2\\0&1&1\end{array}\right]\implies \tfrac{1}{7}\left(a-10\right)=1$ and $5+\tfrac{2}{7}\left(a-10\right)=2$

However, if I solve the first equation for a, I get $a=17$, but when I solve the second equation for a, I get $a=-\frac{1}{2}$....(Wondering)

Did you copy the intersection pt correctly?
• June 30th 2009, 11:47 PM
Stroodle
Awesome. Thanks for that. The text says the intersection point is $(2,1)$ and that the answer is $a=17$, but I also get $a=-\frac{1}{2}$...
• July 1st 2009, 03:31 AM
pfarnall
The problem is that the second line doesn't even pass through the point of intersection!
$f(x,y)=x-2y,\ \ f(2,1)=0,\ \ 0\neq5$
Basically I think the second equation should be $x-2y=0$ 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...