0
$\begingroup$

I have the following system of equations that I want to solve.

$R_1$ $2a-3b+5c+d =-41$,

$R_2$ $7a+2b-c=-28$,

$R_3$ $-a+2b-7c-2d=46$,

$R_4$ $3a+7b-6c+d=31$,

Now i can clearly see that I can reduce the system to three equations by eliminating $d$ variable.

$R'_3 = R_3 + 2R_1$

$(-a+2b-7c-2d=46) + (4a-6b+10c+2d =-82) = 3a-4b+3c=-36$

$R'_4 = R_4-R_1$

$(3a+7b-6c+d=31) - (2a-3b+5c+d =-41) = a + 10b-11c=72$

Now I have the following system of equations.

$R1$ $7a+2b-c=-28$

$R2$ $3a-4b+3c-36$

$R3$ $a+10b-11c=72$

Looking at this system of equations I can see that the easy variables to isolate are $c$ and $a$, but in the context of these equations, they will lead to fairly large numbers due to coefficients and I'd like to avoid that.

So I've decided to put this in Augmented Matrix and go from there.

\begin{bmatrix} 7 & 2 & -1 & -28 \\ 3 & -4 & 3 & -36 \\ 1 & 10 & -11 & 72 \end{bmatrix}

So based on what I understand the Gauss-Jordan Elimination requires me to manipulate the rows such that I get an identity matrix, which makes sense because I want to find the value of each variable.

$R'_2 = R_2 - 3R_3$

\begin{bmatrix} 7 & 2 & -1 & -28 \\ 0 & -34 & -30 & -252 \\ 1 & 10 & -11 & 72 \end{bmatrix}

$R'_3 = 7R_3-R_1$

\begin{bmatrix} 7 & 2 & -1 & -28 \\ 0 & -34 & -30 & -252 \\ 0 & 68 & -76 & 532 \end{bmatrix}

At this point it looks like A mess but I keep going.

$R'_3 = R_3+2R_2$

\begin{bmatrix} 7 & 2 & -1 & -28 \\ 0 & -34 & -30 & -252 \\ 0 & 0 & -106 & 28 \end{bmatrix}

Now at this point I need to do $-106c = 28 \therefore c = \frac{28}{-106}$ which gives a nonsensical number. I attempted to solve this problem via normal elimination, elimination and substitution. But I always arrive at nonsensical answers.

$\endgroup$
  • 1
    $\begingroup$ Why is your result non-sensical? $\endgroup$ – Maximilian Janisch Dec 8 '19 at 14:37
  • $\begingroup$ Because this question is supposed to be solved without a calculator according to the book. It's already difficult not to make mistakes with these numbers. Now the result is some small negative fraction and I'm not sure how to deal with this. $\endgroup$ – Vocaloidas Dec 8 '19 at 14:40
  • 1
    $\begingroup$ The solution is $a=-5,b=0,c=-7,d=4$. Maybe this helps you to check where your calculations went wrong $\endgroup$ – Maximilian Janisch Dec 8 '19 at 14:41
  • 2
    $\begingroup$ Check $R2-3R3$. $\endgroup$ – Empy2 Dec 8 '19 at 14:49
  • 1
    $\begingroup$ @Vocaloidas I used Mathematica. And Mathematica probably uses Cramer‘s Rule or Gaussian elimination $\endgroup$ – Maximilian Janisch Dec 8 '19 at 16:19
1
$\begingroup$

I changed the order of the rows, does not matter, and puts off the fractions until later

parisize = 4000000, primelimit = 500000
? m = [ -1,2,-7,-2,46; 2,-3,5,1,-41; 3,7,-6,1,31; 7,2,-1,0,-28]
%1 = 
[-1  2 -7 -2  46]

[ 2 -3  5  1 -41]

[ 3  7 -6  1  31]

[ 7  2 -1  0 -28]

? l1 = [ 1,0,0,0; 2,1,0,0; 3,0,1,0; 7,0,0,1]
%2 = 
[1 0 0 0]

[2 1 0 0]

[3 0 1 0]

[7 0 0 1]

? l1 * m
%3 = 
[-1  2  -7  -2  46]

[ 0  1  -9  -3  51]

[ 0 13 -27  -5 169]

[ 0 16 -50 -14 294]

? l2 = [ 1, -2,0,0; 0,1,0,0; 0,-13,1,0; 0, -16,0,1]
%4 = 
[1  -2 0 0]

[0   1 0 0]

[0 -13 1 0]

[0 -16 0 1]

? l2 * l1 * m
%5 = 
[-1 0 11  4  -56]

[ 0 1 -9 -3   51]

[ 0 0 90 34 -494]

[ 0 0 94 34 -522]

? l3 = [ 1,0,0,0; 0,1,0,0; 0,0,1/2,0; 0,0,0,1/2]
%6 = 
[1 0   0   0]

[0 1   0   0]

[0 0 1/2   0]

[0 0   0 1/2]

? l3 * l2 * l1 * m
%7 = 
[-1 0 11  4  -56]

[ 0 1 -9 -3   51]

[ 0 0 45 17 -247]

[ 0 0 47 17 -261]

? l4 = [ 1,0,0,0; 0,1,0,0; 0,0,1,-1; 0,0,0,1]
%8 = 
[1 0 0  0]

[0 1 0  0]

[0 0 1 -1]

[0 0 0  1]

? l4 * l3 * l2 * l1 * m
%9 = 
[-1 0 11  4  -56]

[ 0 1 -9 -3   51]

[ 0 0 -2  0   14]

[ 0 0 47 17 -261]

? l5 = [ 1,0,0,0; 0,1,0,0; 0,0,1/2,0; 0,0,0,1]
%10 = 
[1 0   0 0]

[0 1   0 0]

[0 0 1/2 0]

[0 0   0 1]

? l5 *l4 * l3 * l2 * l1 * m
%11 = 
[-1 0 11  4  -56]

[ 0 1 -9 -3   51]

[ 0 0 -1  0    7]

[ 0 0 47 17 -261]

? l6 = [ 1,0,0,0; 0,1,0,0; 0,0,1,0; 0,0,47,1]
%12 = 
[1 0  0 0]

[0 1  0 0]

[0 0  1 0]

[0 0 47 1]

? l6 * l5 *l4 * l3 * l2 * l1 * m
%13 = 
[-1 0 11  4 -56]

[ 0 1 -9 -3  51]

[ 0 0 -1  0   7]

[ 0 0  0 17  68]

? l7 = [ 1,0,0,0; 0,1,0,0; 0,0,1,0; 0,0,0,1/17]
%14 = 
[1 0 0    0]

[0 1 0    0]

[0 0 1    0]

[0 0 0 1/17]

? l7 * l6 * l5 *l4 * l3 * l2 * l1 * m
%15 = 
[-1 0 11  4 -56]

[ 0 1 -9 -3  51]

[ 0 0 -1  0   7]

[ 0 0  0  1   4]

? 
? 
? l8 = [ 1,0,11,0; 0,1,-9,0; 0,0,1,0; 0,0,0,1]
%16 = 
[1 0 11 0]

[0 1 -9 0]

[0 0  1 0]

[0 0  0 1]

? l8 * l7 * l6 * l5 *l4 * l3 * l2 * l1 * m
%17 = 
[-1 0  0  4  21]

[ 0 1  0 -3 -12]

[ 0 0 -1  0   7]

[ 0 0  0  1   4]

? l9 = [ 1,0,0,-4; 0,1,0,3; 0,0,1,0; 0,0,0,1]
%18 = 
[1 0 0 -4]

[0 1 0  3]

[0 0 1  0]

[0 0 0  1]

? l9 * l8 * l7 * l6 * l5 *l4 * l3 * l2 * l1 * m
%19 = 
[-1 0  0 0 5]

[ 0 1  0 0 0]

[ 0 0 -1 0 7]

[ 0 0  0 1 4]

? 
? 

All in all, we multiplied on the left by a nonsingular rational matrix,

? 
? L = l9 * l8 * l7 * l6 * l5 *l4 * l3 * l2 * l1 
%20 = 
[ -3/34 -11/68 -1/68  -7/68]

[-19/34 -16/17 -3/17   9/34]

[   1/2    3/4   1/4   -1/4]

[ 11/17 109/68 47/68 -45/68]

? matdet(L)
%21 = 1/136
? 
? 
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.