1. ## Matricies, need help...

Have tried this a number of times and get different answers everytime. Could someone hold my hand through this because I'm taking functions online and my teacher is no help.

Says to solve the system of equations using matrices. Use Gauss-Jordan elimination.

$\displaystyle \begin{array}{l} {\rm{3x }} + {\rm{ 5y }} + {\rm{ 2w }} = {\rm{ }} - {\rm{12}}\\ {\rm{2x }} + {\rm{ 6z }} - {\rm{ w }} = {\rm{ }} - {\rm{5}}\\ - {\rm{2y }} + {\rm{ 3z }} - {\rm{ 3w }} = {\rm{ }} - {\rm{3}}\\ - {\rm{x }} + {\rm{ 2y }} + {\rm{ 4z }} + {\rm{ w }} = {\rm{ }} - {\rm{2}} \end{array}$

2. If you have tried a number of times, then please show what you have tried. It looks tedious but straightforward to me.

3. This is solution of your SLE fourth-order by Gauss–Jordan elimination step-by-step

1) interchange rows 1 and 4;
2) from elements of row 2 subtract elements of row 1, multiplied by -2;
3) from elements of row 4 subtract elements of row 1, multiplied by -3;
4) interchange rows 2 and 3;
5) from elements of row 3 subtract elements of row 2, multiplied by -2;
6) from elements of row 4 subtract elements of row 2, multiplied by -11/2;
7) from elements of row 4 subtract elements of row 3, multiplied by 57/40;
8) from elements of row 1 subtract elements of row 4 multiplied by -8/35;
9) from elements of row 2 subtract elements of row 4, multiplied by 24/35;
10) from elements of row 3 subtract elements of row 4, multiplied by 8/7;
11) from elements of row 1 subtract elements of row 3 multiplied by 1/5;
12) from elements of row 2 subtract elements of row 3 multiplied by 3/20;
13) from elements of row 1 subtract elements of row 2, multiplied by -1;
14) elements of row 1 divide by -1, row 2 – by -2, row 3 – by -20 and row 4 – by -35/8.

$\displaystyle {\left[\!\!\begin{array}{*{20}{r}}3&5&0&2&\!\vline\!\!&{-12}\\[3pt] 2&0&6&{-1}&\!\vline\!\!&{-5}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] {-1}&2&4&1&\!\vline\!\!&{-2} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&1&\!\vline\!\!&{-2}\\[3pt] 2&0&6&{-1}&\!\vline\!\!&{-5}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] 3&5&0&2&\!\vline\!\!&{-12} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&1&\!\vline\!\!&{-2}\\[3pt] 0&4&{14}&1&\!\vline\!\!&{-9}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] 3&5&0&2&\!\vline\!\!&{-12} \end{array}\!\!\right]\!\sim}$

$\displaystyle {\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&1&\!\vline\!\!&{-2}\\[3pt] 0&4&{14}&1&\!\vline\!\!&{-9}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] 0&{11}&{12}&5&\!\vline\!\!&{-18} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&1&\!\vline\!\!&{-2}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] 0&4&{14}&1&\!\vline\!\!&{-9}\\[3pt] 0&{11}&{12}&5&\!\vline\!\!&{-18} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&1&\!\vline\!\!&{-2}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] 0&0&{20}&{-5}&\!\vline\!\!&{-15}\\[3pt] 0&{11}&{12}&5&\!\vline\!\!&{-18} \end{array}\!\!\right]\!\sim}$

$\displaystyle {\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&1&\!\vline\!\!&{-2}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] 0&0&{20}&{-5}&\!\vline\!\!&{-15}\\[5pt] 0&0&{\dfrac{57}{2}}&{-\dfrac{23}{2}}&\!\vline\!\!&{-\dfrac{69}{2}} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&1&\!\vline\!\!&{-2}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] 0&0&{20}&{-5}&\!\vline\!\!&{-15}\\[5pt] 0&0&0&{-\dfrac{35}{8}}&\!\vline\!\!&{-\dfrac{105}{8}} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&0&\!\vline\!\!&{-5}\\[3pt] 0&{-2}&3&{-3}&\!\vline\!\!&{-3}\\[3pt] 0&0&{20}&{-5}&\!\vline\!\!&{-15}\\[5pt] 0&0&0&{-\dfrac{35}{8}}&\!\vline\!\!&{-\dfrac{105}{8}} \end{array}\!\!\right]\!\sim\!}$

$\displaystyle {\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&0&\!\vline\!\!&{-5}\\[3pt] 0&{-2}&3&0&\!\vline\!\!&6\\[3pt] 0&0&{20}&{-5}&\!\vline\!\!&{-15}\\[5pt] 0&0&0&{-\dfrac{35}{8}}&\!\vline\!\!&{-\dfrac{105}{8}} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&4&0&\!\vline\!\!&{-5}\\[3pt] 0&{-2}&3&0&\!\vline\!\!&6\\[3pt] 0&0&{20}&0&\!\vline\!\!&0\\[5pt] 0&0&0&{-\dfrac{35}{8}}&\!\vline\!\!&{-\dfrac{105}{8}} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&0&0&\!\vline\!\!&{-5}\\[3pt] 0&{-2}&3&0&\!\vline\!\!&6\\[3pt] 0&0&{20}&0&\!\vline\!\!&0\\[5pt] 0&0&0&{-\dfrac{{35}}{8}}&\!\vline\!\!&{-\dfrac{105}{8}} \end{array}\!\!\right]\!\sim}$

$\displaystyle {\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&2&0&0&\!\vline\!\!&{-5}\\[3pt] 0&{-2}&0&0&\!\vline\!\!&6\\[3pt] 0&0&{20}&0&\!\vline\!\!&0\\[5pt] 0&0&0&{-\dfrac{35}{8}}&\!\vline\!\!&{-\dfrac{105}{8}} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} {-1}&0&0&0&\!\vline\!\!&1\\[3pt] 0&{-2}&0&0&\!\vline\!\!&6\\[3pt] 0&0&{20}&0&\!\vline\!\!&0\\[5pt] 0&0&0&{-\dfrac{35}{8}&\!\vline\!\!&{-\dfrac{105}{8}} \end{array}\!\!\right]\!\sim\!\left[\!\!\begin{array}{*{20}{r}} 1&0&0&0&\!\vline\!\!&{-1}\\[3pt] 0&1&0&0&\!\vline\!\!&{-3}\\[3pt] 0&0&1&0&\!\vline\!\!&0\\[3pt] 0&0&0&1&\!\vline\!\!&3 \end{array}\!\!\right]}$

So, finally we have $\displaystyle x=-1,~y=-3,~z=0,~w=3$.

This solution is understandable to you?

4. ## Thank you!!

Wow. Ok so I started this problem switching rows 1 and 4 as you did. Then I multiplied everything in the top row by -1 to make the -1 a positive 1. Is this another way to start this problem or would that be wrong? I get so overwhelmed by these and thank you so much for doing that.

5. I keep trying to get ones in the diagonal as I go and I feel that may be what I'm doing wrong. You don't do that. You make everything 1 once you solved all your zeros.

6. Pay attention to the eighth matrix, which we have received after the seventh step – "from elements of row 4 subtract elements of row 3, multiplied by 57/40".