0
$\begingroup$

I am struggling with Gaussian elimination with matrices. I need to bring the following matrix

$$\left[\begin{array}{cccc|c} 1&-1&2&1 & 2\\ 2&-3&2&0 & 3\\ -1&1&2&3 & 6\\ -3&2&0&3 & 9\\ \end{array}\right]$$

The matrix after the = sign is supposed to be augmented wth the first matrix but I as unable to do the proper formatting.

I know the final answer is

$\begin{bmatrix} 1&0&0&-1 \\0&1&0&0 \\0&0&1&1 \\0&0&0&0 \end{bmatrix}=\begin{bmatrix}-5\\-3\\2\\0\end{bmatrix}$

The question asked if the columns of A are linearly independent. I am looking for some help with walking through the steps of reduction and whether or not this matrix needs to be in reduced row echelon form or just row echelon form. I know to bring this to row echelon form you can only swap rows, add/subtract rows, and multiply rows by scalars, or add/subtract rows when multiplied by scalers

$\endgroup$
2
  • $\begingroup$ just looking for some basic strategies for guassian elimination $\endgroup$ – jh123 Oct 15 '16 at 14:12
  • $\begingroup$ @ Rodrigo de Azevedo thanks for the edit, $\endgroup$ – jh123 Oct 15 '16 at 14:22
2
$\begingroup$

Let's walk through elimination: \begin{align} \left[ \begin{array}{cccc|c} 1 & -1 & 2 & 1 & 2\\ 2 & -3 & 2 & 0 & 3\\ -1 & 1 & 2 & 3 & 6\\ -3 & 2 & 0 & 3 & 9 \end{array}\right] &\to \left[ \begin{array}{cccc|c} 1 & -1 & 2 & 1 & 2 \\ 0 & -1 & -2 & -2 & -1 \\ 0 & 0 & 4 & 4 & 8 \\ 0 & -1 & 6 & 6 & 15 \end{array}\right] &&\begin{aligned} R_2&\gets R_2-2R_1\\R_3&\gets R_3+R_1\\R_4&\gets R_4+3R_1\end{aligned} \tag{1}\\[6px]&\to \left[ \begin{array}{cccc|c} 1 & -1 & 2 & 1 & 2 \\ 0 & 1 & 2 & 2 & 1 \\ 0 & 0 & 4 & 4 & 8 \\ 0 & 1 & 8 & 8 & 16 \end{array}\right] &&\begin{aligned} R_2&\gets -R_2\\R_4&\gets R_4+R_2\end{aligned} \tag{2}\\[6px]&\to \left[ \begin{array}{cccc|c} 1 & -1 & 2 & 1 & 2 \\ 0 & 1 & 2 & 2 & 1 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] &&\begin{aligned} R_3&\gets \tfrac{1}{4}R_3\\R_4&\gets R_4-8R_3\end{aligned} \tag{3}\\[6px]&\to \left[ \begin{array}{cccc|c} 1 & -1 & 0 & -1 & -2 \\ 0 & 1 & 0 & 0 & -3 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] &&\begin{aligned}R_2&\gets R_2-2R_3\\R_1&\gets R_1-2R_3\end{aligned} \tag{4}\\[6px]&\to \left[ \begin{array}{cccc|c} 1 & 0 & 0 & -1 & -5 \\ 0 & 1 & 0 & 0 & -3 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] &&R_1\gets R_1+R_2 \tag{5} \end{align}

Already after step $(3)$ you can conclude that the columns of the matrix $A$ are linearly dependent, because you get a null row in a row echelon form (just consider the first four columns); the fact that you get a full null row in the augmented matrix tells you that the linear system is solvable, because the last column cannot contain a leading $1$.

Steps $(4)$ and $(5)$ are backwards elimination, that lead to the reduced row echelon form, very useful to easily express all solutions.

Indeed, if we denote by $x_1$, $x_2$, $x_3$ and $x_4$ the unknowns, we see that $x_4$ is free and the equations read $$ \begin{cases} x_1-x_4=-5 \\[4px] x_2 = -3 \\[4px] x_3+x_4=2 \end{cases} $$ so the generic solution is $$ \begin{bmatrix} -5+h\\ -3\\ 2-h\\ h \end{bmatrix} $$

The advantage to go into this order for elimination is that it makes it possible to write the $LU$ decomposition of the matrix $A$. I use to denote the elementary operations we have performed up to step $(3)$ as $$ E_{21}(-2),\quad E_{31}(1),\quad E_{41}(3),\quad E_2(-1),\quad E_{42}(1),\quad E_3(\tfrac{1}{4}),\quad E_{43}(-8) $$ and this allows to write $$ L=E_{21}(2)E_{31}(-1)E_{41}(-3)E_2(-1)E_{42}(-1)E_3(4)E_{43}(8) =\begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ -1 & 0 & 4 & 0 \\ -3 & -1 & 8 & 1 \end{bmatrix} $$ and $$ U=\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$ You can check that $A=LU$.

$\endgroup$
6
  • $\begingroup$ @ egreg thanks a lot for your time, you ad a typo in your system of equations x1 - x4 = -5 not 5, also step 4 and 5 is this trying to get a 0 in row 1 column 2 and 3? if so why not get one for column 4? $\endgroup$ – jh123 Oct 15 '16 at 14:42
  • $\begingroup$ @javahelper123 Thanks for pointing up the typo. In steps 4 and 5 we use each leading 1 to get zeros above it, which is what the reduced row echelon form should look like. There is no leading 1 in column 4. $\endgroup$ – egreg Oct 15 '16 at 14:50
  • $\begingroup$ @ egreg okay so if a leading 1 has a number above it then I should get rid of that number above, and if it fdoesnt have a number above then that's fine, also if there is no leading 1 in a row and it has all 0s then I can not reduce anything above? $\endgroup$ – jh123 Oct 15 '16 at 14:54
  • $\begingroup$ and also does it have to be directly above? $\endgroup$ – jh123 Oct 15 '16 at 14:55
  • $\begingroup$ @javahelper123 In step 4 we use the leading 1 in column 3 to get rid of the nonzero coefficients in column 3 above the leading 1 (there are zeros below because we already have a row echelon form). $\endgroup$ – egreg Oct 15 '16 at 15:10
0
$\begingroup$

Call the rows $r_1 \ldots r_4$.. After the following 8 operations you arrive at your answer: $$r_4 \leftarrow r_4 - 3r_3$$ $$r_3 \leftarrow r_3 + r_1$$ $$r_2 \leftarrow r_2 - 2r_1$$ $$r_4 \leftarrow r_4 - r_2$$ $$r_4 \leftarrow r_4 + r_3$$ $$r_2 \leftarrow 2r_2 + r_3$$ $$r_1 \leftarrow 2r_1 - r_2$$ $$r_1 \leftarrow r_1 - r_3$$

$\endgroup$

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.