# Matrix Row Operations

• Nov 20th 2009, 08:40 AM
mhitch03
Matrix Row Operations
Could someone help me with this problem? Thanks!

6x+7y=13
x-y=0
x+y=12

Solve using matrix operations to get the coefficient matrix into reduced form. Write down the answer.
• Nov 20th 2009, 02:32 PM
tonio
Quote:

Originally Posted by mhitch03
Could someone help me with this problem? Thanks!

6x+7y=13
x-y=0
x+y=12

Solve using matrix operations to get the coefficient matrix into reduced form. Write down the answer.

The system is clearly incongruent (i.e., it has no solution) since by $\displaystyle x-y=0\,\Longrightarrow\,x=y$, but this doesn't solve $\displaystyle 6x+7y=13\,\,\,and\,\,\,also\,\,\,x+y=12$.

It's trivial to check that $\displaystyle x=y=1$ doesn't solve neither the first nor the third equation.

Tonio
• Nov 30th 2009, 09:43 AM
mhitch03
I'm sorry, that is a typo... the last equation is x+y=2
• Nov 30th 2009, 11:02 AM
tonio
Quote:

Originally Posted by mhitch03
I'm sorry, that is a typo... the last equation is x+y=2

Then you only need the second and third equation to get the solution at once, and using matrices:

$\displaystyle \left(\begin{array}{rrr}6&7&13\\1&-1&0\\1&1&2\end{array}\right) \rightarrow\left(\begin{array}{rrr}1&-1&0\\1&1&2\\6&7&13\end{array}\right)$ $\displaystyle \left(\begin{array}{rrr}1&-1&0\\0&2&2\\0&13&13\end{array}\right)\rightarrow\l eft(\begin{array}{rrr}1&-1&0\\0&1&1\\0&0&0\end{array}\right)$

From the second row, $\displaystyle y=1$ , and from the first one $\displaystyle x-1=0\rightarrow x=1$

Tonio