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Question from Kamal about Gaussian Elimination via email.

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Well-known member
MHB Math Helper
Jan 26, 2012
1,403
HI Sir,
I was having a bit of trouble with a question from the textbook. I have tried it multiple times with different methods and keep getting a different answer. The question wants us to solve it using Gaussian elimination.

2x+y−3z=−5
x − y + 2z = 12
7x − 2y + 3z = 3

The answer for Z is meant to be 2 but i keep getting bigger numbers for Z.

Thanks, Kamal
I'm not sure where you're getting the idea that z = 2, as this is not correct.

I'm assuming this is to be done without pivoting...

Set up your augmented matrix:

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & \phantom{-}1 & -3 & -5 \\ 1 & -1 & \phantom{-}2 & 12 \\ 7 & -2 & \phantom{-} 3 & \phantom{-}3 \end{matrix} \right] \end{align*}$

As we have to use Gaussian Elimination, we will use multiples of Row 1 to eliminate the terms under the main diagonal in the first column. So apply Row 2 - 1/2 Row 1 to Row 2 and Row 3 - 7/2 Row 1 to Row 3.

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & 1 & -3 & -5 \\ 1 - 1 & -1 - \frac{1}{2} & 2 - \left( -\frac{3}{2} \right) & 12 - \left( -\frac{5}{2} \right) \\ 7 - 7 & -2 - \frac{7}{2} & 3 - \left( -\frac{21}{2} \right) & 3 - \left( -\frac{35}{2} \right) \end{matrix} \right] &= \left[ \begin{matrix} 2 & \phantom{-}1 & -3 & -5 \\ 0 & -\frac{3}{2} & \phantom{-}\frac{7}{2} & \phantom{-}\frac{29}{2} \\ 0 & -\frac{11}{2} & \phantom{-}\frac{27}{2} & \phantom{-}\frac{41}{2} \end{matrix} \right] \end{align*}$

Now we have to use multiples of Row 2 to eliminate the term under the main diagonal in the second column. So we will have to apply Row 3 - 11/3 Row 2 to Row 3.

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & \phantom{-}1 & -3 & -5 \\ 0 & -\frac{3}{2} & \phantom{-}\frac{7}{2} & \phantom{-}\frac{29}{2} \\ 0 + 0 & -\frac{11}{2} - \left( -\frac{11}{2} \right) & \frac{27}{2} - \frac{77}{6} & \frac{41}{2} - \frac{319}{6} \end{matrix} \right] &= \left[ \begin{matrix} 2 & \phantom{-}1 & -3 & -5 \\ 0 & -\frac{3}{2} & \phantom{-}\frac{7}{2} & \phantom{-}\frac{29}{2} \\ 0 & \phantom{-}0 & \phantom{-}\frac{2}{3} & -\frac{98}{3} \end{matrix} \right] \end{align*}$

From here we can see that

$\displaystyle \begin{align*} \frac{2}{3} \,z &= -\frac{98}{3} \\ z &= -49 \end{align*}$.

Back substitution yields

$\displaystyle \begin{align*} -\frac{3}{2}\,y + \frac{7}{2}\,z &= \frac{29}{2} \\ -\frac{3}{2}\,y + \frac{343}{2} &= \frac{29}{2} \\ -\frac{3}{2}\,y &= -\frac{372}{2} \\ y &= -124 \end{align*}$

Back substitution again yields

$\displaystyle \begin{align*} 2\,x - y - 3\,z &= -5 \\ 2\,x - 124 + 147 &= -5 \\ 2\,x + 23 &= -5 \\ 2\,x &= -28 \\ x &= -14 \end{align*}$

So the solution to your system is $\displaystyle \begin{align*} \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] = \left[ \begin{matrix} -14 \\ -124 \\ -49 \end{matrix} \right] \end{align*}$. Substitution into any of your original equations will verify this to be correct.