how to solve a 4x5 matrix with guassian elimination or gaussian-jordan

Question

$$ \left[ \begin{array}{ccccc} 2 & -1 & 3 & 4 & 9 \\ 1 & 0 & -2 & 7 & 11 \\ 3 &-3 & 1 & 5 & 8 \\ 2 & 1 & 4 & 4 & 10 \\ \end{array}\right] $$ I got the diagonal of leading 1s but only by making the bottom right number a large fraction.. and the answer in the solution are all small numbers without fractions.. I am not sure of a good method for doing 4x5, I tried doing the "forward" elimination to get the 0s under the diagonal.. but that just ended up in a large fraction on the bottom right.. im not sure how to attack this.. because it came out ugly...

here is my feeble soultion:

sitch R1 and R2

$$ \left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 2 & -1 & 3 & 4 & 9 \\ 3 &-3 & 1 & 5 & 8 \\ 2 & 1 & 4 & 4 & 10 \\ \end{array}\right] $$

R2 = R2 - 2R1

R2 = -1 x R2

$$ \left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 0 & 1 & -7 & 10 & -13 \\ 3 &-3 & 1 & 5 & 8 \\ 2 & 1 & 4 & 4 & 10 \\ \end{array}\right] $$

R3 += -3R1

$$ \left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 0 & 1 & -7 & 10 & -13 \\ 0 &-3 & 7 & -16 & -25 \\ 2 & 1 & 4 & 4 & 10 \\ \end{array}\right] $$

R3 += 3R2

$$ \left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 0 & 1 & -7 & 10 & -13 \\ 0 & 0 & -14 & 14 & -64 \\ 2 & 1 & 4 & 4 & 10 \\ \end{array}\right] $$

I actually can't figure out how I ever got any further than that...

EDIT: the very last 2 in the matrix was supposed to be a 10, but this doesnt change my solution thus far and I am still stuck... so if i could still have help please...

how can I change the first -14 into a 1?

Answer 1

\begin{align*} \begin{pmatrix} 2 & -1 & 3 & 4 & 9 \\ 1 & 0 & -2 & 7 & 11 \\ 3 &-3 & 1 & 5 & 8 \\ 2 & 1 & 4 & 4 & 10 \\ \end{pmatrix} &\overset{2II-I}{\leadsto} \begin{pmatrix} 2 & -1 & 3 & 4 & 9 \\ 0 & 1 & -7 & 10 & 13 \\ 3 &-3 & 1 & 5 & 8 \\ 2 & 1 & 4 & 4 & 10 \\ \end{pmatrix} \\ &\overset{2III-3I}{\leadsto} \begin{pmatrix} 2 & -1 & 3 & 4 & 9 \\ 0 & 1 & -7 & 10 & 13 \\ 0 &-3 & -7 & -2 & -11 \\ 2 & 1 & 4 & 4 & 10 \\ \end{pmatrix} \\ &\overset{IV -I}{\leadsto} \begin{pmatrix} 2 & -1 & 3 & 4 & 9 \\ 0 & 1 & -7 & 10 & 13 \\ 0 &-3 & -7 & -2 & -11 \\ 0 & 2 & 1 & 0 & 1 \\ \end{pmatrix} \\ &\overset{III+3II}{\leadsto} \begin{pmatrix} 2 & -1 & 3 & 4 & 9 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & -28 & 28 & 28 \\ 0 & 2 & 1 & 0 & 1 \\ \end{pmatrix} \\ &\overset{IV -2II}{\leadsto} \begin{pmatrix} 2 & -1 & 3 & 4 & 9 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & -28 & 28 & 28 \\ 0 & 0 & 15 & -20 & -25 \\ \end{pmatrix} \\ &\overset{28IV+15III}{\leadsto} \begin{pmatrix} 2 & -1 & 3 & 4 & 9 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & -28 & 28 & 28 \\ 0 & 0 & 0 & -140 & -280 \\ \end{pmatrix} \\ &\overset{\frac{III}{28}, \frac{IV}{140}}{\leadsto} \begin{pmatrix} 2 & -1 & 3 & 4 & 9 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & -1 & 1 & 1 \\ 0 & 0 & 0 & -1 & -2 \\ \end{pmatrix} \\ \end{align*}

Note that since the matrix multiplication is associative one can do any transformation when he wants, for example you could do the same step that i did in the opposite direction and get the same result, but to make it clearer i've wrote every single step. You can now solve the system just by evaluating the single components. Note that since your matrix has rang $4$ the solution is unique.

Does it help?

Answer 2

Jordan-Gauss elimination is convergent, meaning that however you proceed the normal form is unique. It is also always possible to reduce matrices of rank 4 (I assume yours is) to a normal form with the left 4x4 block being the identity, but the rightmost column cannot be reduced further. If your solution does not match the answer sheet, then you have made a computational error, which is frequent due to the sheer number of operations needed.

For a particular matrix, I could show you how I would reduce it, but I will let your friend WolframAlpha do the comuputation, since this is faster and safer. If you have an account, you can see the procedure step-by-step.

Edit: WolframAlpha removed the free feature, but I will still use it for checking the work. You made a mistake too, but I corrected it. Now for the show:

$$\left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & -14 & 14 & 8 \\ 2 & 1 & 4 & 4 & 10 \\ \end{array}\right]$$

$R_4 = R_4 + (-2)R_1$

$$\left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & -14 & 14 & 8 \\ 0 & 1 & 8 & -10 & -12 \\ \end{array}\right]$$

$R_4 = R_4 + (-1)R_2$

$$\left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & -14 & 14 & 8 \\ 0 & 0 & 15 & -20 & -25 \\ \end{array}\right]$$

$R_3 = R_3 + R_4$

$$\left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & 1 & -6 & -17 \\ 0 & 0 & 15 & -20 & -25 \\ \end{array}\right]$$

Now the 14 has disappeared. I will not write the rest because this is hard and long and I am not even sure I did not make mistakes, so you will want to check my work.

Edit 2: Actually, your corrected matrix is:

$$\left[ \begin{array}{ccccc} 1 & 0 & -2 & 7 & 11 \\ 0 & 1 & -7 & 10 & 13 \\ 0 & 0 & -14 & 14 & 14 \\ 2 & 1 & 4 & 4 & 10 \\ \end{array}\right]$$

You just have to divide by 14... I will leave the process above because the general "trick" has helped me when I did these sorts of exercises.