Solve the following linear system

[chwala]

Homework Statement

See attached

Relevant Equations

Gauss - jordan and related methods

Find the problem below with the solution indicated; The text approach is much clear to me... My way of tackling the problem is as follows; using echelon form (row reduction method) \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 2 & 5 & -1 & -9&-3 \\ 2 & 1 & -1 & 3&-11\\ 1& -3 & 2 & 7&-5 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 1 & 5 & -13&17\\ 0& 4& 0 & -12&8 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 0 & 12 & -40&60\\ 0& 0& -12 & -40&60 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 0 & 12 & -40&60\\ 0& 0& 0 & -80&120 \end{bmatrix}It follows that, $-80x_4=120$ $x_4=-1.5$ Also, $12x_3-40x_4=60$ $⇒x_3=0$ Also, $-3x_2-x_4=9$ $-3x_2=9-1.5$ $x_2=-2.5$ Also, $x_1+x_2-5x_4=3$ $x_1-2.5+7.5=3$ $x_1=-2$ I just checked the text solution and i could see that the solutions agree when $x_4=-1.5=t$ I would appreciate any other perspective.

 

[SammyS]

I think you have an error in Row 2 of the second matrix:

$\ \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 1 & 5 & -13&17\\ 0& 4& 0 & -12&8 \end{bmatrix}$

Added in Edit:

In fact the second row should be:

$\displaystyle \left[ \begin{matrix} 0 & -3 & 5 & -1 & 9 \end{matrix} \right]$

 

[chwala]

I think you have an error in Row 2 of the second matrix: $\begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 1 & 5 & -13&17\\ 0& 4& 0 & -12&8 \end{bmatrix}$

I'll check that and revert back...

 

[chwala]

Homework Statement:: See attached
Relevant Equations:: Gauss - jordan and related methods

Find the problem below with the solution indicated; View attachment 305359View attachment 305360 View attachment 305361 View attachment 305362 The text approach is much clear to me... My way of tackling the problem is as follows; using echelon form (row reduction method) \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 2 & 5 & -1 & -9&-3 \\ 2 & 1 & -1 & 3&-11\\ 1& -3 & 2 & 7&-5 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 1 & 5 & -13&17\\ 0& 4& 0 & -12&8 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 0 & 12 & -40&60\\ 0& 0& -12 & -40&60 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 0 & 12 & -40&60\\ 0& 0& 0 & -80&120 \end{bmatrix}It follows that, $-80x_4=120$ $x_4=-1.5$ Also, $12x_3-40x_4=60$ $⇒x_3=0$ Also, $-3x_2-x_4=9$ $-3x_2=9-1.5$ $x_2=-2.5$ Also, $x_1+x_2-5x_4=3$ $x_1-2.5+7.5=3$ $x_1=-2$ I just checked the text solution and i could see that the solutions agree when $x_4=-1.5=t$ I would appreciate any other perspective.

Amended;

\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
2 & 5 & -1 & -9&-3 \\
2 & 1 & -1 & 3&-11\\
1& -3 & 2 & 7&-5
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
0 & -3 & 5 & -1&9 \\
0 & 1 & 5 & -13&17\\
0& 4& 0 & -12&8
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
0 & -3 & 5 & -1&9 \\
0 & 0 & 10 & -40&60\\
0& 0& 20 & -40&60
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
0 & -3 & 5 & -1&9 \\
0 & 0 & 10 & -40&60\\
0& 0& 0 & -40&60
\end{bmatrix}
It follows that,
$-40x_4=60$
$x_4=-1.5$
Also,
$10x_3-40x_4=60$
$⇒x_3=0$
Also,
$-3x_2-x_4=9$
$-3x_2=9-1.5$
$x_2=-2.5$
Also,
$x_1+x_2-5x_4=3$
$x_1-2.5+7.5=3$
$x_1=-2$
I just checked the text solution and i could see that the solutions agree when $x_4=-1.5=t$
I would appreciate any other perspective.

 

[SammyS]

Amended;

$$\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
2 & 5 & -1 & -9&-3 \\
2 & 1 & -1 & 3&-11\\
1& -3 & 2 & 7&-5
\end{bmatrix}$$
$$\begin{bmatrix}
1 & 1 & 2 & -5 & 3\\
0 & -3 & 5 & -1 & 9 \\
0 & 1 & 5 & -13 & 17\\
0& 4& 0 & -12 & 8
\end{bmatrix}$$
$$\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
0 & -3 & 5 & -1&9 \\
0 & 0 & 10 & -40&60\\
0& 0& 20 & -40&60
\end{bmatrix}$$

No that is not correct.

The following is now correct for the second matrix.

$\begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & 5 & -1&9 \\ 0 & 1 & 5 & -13&17\\ 0& 4& 0 & -12&8 \end{bmatrix}$

Take 3 times Row 3 added to Row 2 and use this for Row 3 in the third matrix. This gives the same result as what you have correctly given for Row 4 of the third matrix.

The third matrix should be:

$\begin{bmatrix} 1 & 1 & 2 & -5 & 3 \\ 0 & -3 & 5 & -1 & 9 \\ 0 & 0 & 20 & -40 & 60 \\ 0& 0& 20 & -40 & 60 \end{bmatrix}$

 

[chwala]

No that is not correct.

The following is now correct for the second matrix.

$\begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & 5 & -1&9 \\ 0 & 1 & 5 & -13&17\\ 0& 4& 0 & -12&8 \end{bmatrix}$

Take 3 times Row 3 added to Row 2 and use this for Row 3 in the third matrix. This gives the same result as what you have correctly given for Row 4 of the third matrix.

The third matrix should be:

$\begin{bmatrix} 1 & 1 & 2 & -5 & 3 \\ 0 & -3 & 5 & -1 & 9 \\ 0 & 0 & 20 & -40 & 60 \\ 0& 0& 20 & -40 & 60 \end{bmatrix}$

What do you mean by saying my steps are not correct? The amended steps in post $4$ are correct...argghhhhhh true ... thanks @SammyS

 

[SammyS]

What do you mean by saying my steps are not correct? The amended steps in post $4$ are correct...argghhhhhh true ... thanks @SammyS

$x_4$ can take on any value.

 

[chwala]

$x_4$ can take on any value.

True we shall end up with $20x_3-40x_4=60$ implying that we shall have infinetely many solutions...by either setting $x_3=t$ or $x_4=t$ and not necessarily $x_4$ only.

$x_3$ or $x_4$ can take any value...on setting $x_3=t$ we shall end up with,

$x_1=-2-t$
$x_2=\dfrac{9t-15}{6}=\dfrac{3t-5}{2}$
$x_3=t$
$x_4=\dfrac{t-3}{2}$ where $t$ is the independent variable.

 

[pbuk]

$x_3$ or $x_4$ can take any value...on setting $x_3=t$ we shall end up with,

$x_1=-2-t$
$x_2=\dfrac{9t-15}{6}=\dfrac{3t-5}{2}$
$x_3=t$
$x_4=\dfrac{t-3}{2}$ where $t$ is the independent variable.

Show your workings.

 

[chwala]

Show your workings.

Ok I will later...

 

[chwala]

Show your workings.

Let $x_3=t$

then using the equation;

$x_3-2x_4=3$

we shall have

$t-2x_4=3$

$x_4= \dfrac{t-3}{2}$

and from
$-3x_2+5x_3- x_4=9$

$-3x_2+5x_3-\left[\dfrac{t-3}{2}\right]=9$

$-6x_2+10t-t+3=18$

$-6x_2+9t=15$

$6x_2=9t-15$

$x_2$=$\dfrac{9t-15}{6}$=$\dfrac{3t-5}{2}$

And from;

$x_1+x_2+2x_3-5x_4=3$

We shall have;

$2x_1+3t-5+4t-5t+15=6$

$2x_1+2t+10=6$

$2x_1=-2t-4$

$x_1=-t-2$

 

[chwala]

Just thinking...I will need to try and see if we can solve the problem by using determinant and Cramer's rule...I guess an exercise for the weekend

 

[nuuskur]

Cramer's rule works if the system matrix is nonsingular. I would advise not using Cramer's method. It's computationally inefficient. If Step 2 in the first post is correct, then Cramer's method is inapplicable, anyway.

You should check whether solutions to your system even exist in the first place. Verify that the ranks of the system matrix and the augmented matrix are equal. If this is not the case, then there's no reason to do any further work. (cf Kronecker-Capelli criterion)

Usually, this check is performed at Step 2. It's easy to see that the system matrix is of rank $3$. Thus, solutions exist.

 

[chwala]

Cramer's rule works if the system matrix is nonsingular. I would advise not using Cramer's method. It's computationally inefficient. If Step 2 in the first post is correct, then Cramer's method is inapplicable, anyway.

You should check whether solutions to your system even exist in the first place. Verify that the ranks of the system matrix and the augmented matrix are equal. If this is not the case, then there's no reason to do any further work. (cf Kronecker-Capelli criterion)

Usually, this check is performed at Step 2. It's easy to see that the system matrix is of rank $3$. Thus, solutions exist.

Thanks I will try...and also do a refresher on rank, dimension of matrix form...let me see what comes out of my trial...

 

[chwala]

Cramer's rule works if the system matrix is nonsingular. I would advise not using Cramer's method. It's computationally inefficient. If Step 2 in the first post is correct, then Cramer's method is inapplicable, anyway.

You should check whether solutions to your system even exist in the first place. Verify that the ranks of the system matrix and the augmented matrix are equal. If this is not the case, then there's no reason to do any further work. (cf Kronecker-Capelli criterion)

Usually, this check is performed at Step 2. It's easy to see that the system matrix is of rank $3$. Thus, solutions exist.

Thanks I will try...and also do a refresher on rank, dimension of matrix form...let me see what comes out of my trial

Cramer's rule works if the system matrix is nonsingular. I would advise not using Cramer's method. It's computationally inefficient. If Step 2 in the first post is correct, then Cramer's method is inapplicable, anyway.

You should check whether solutions to your system even exist in the first place. Verify that the ranks of the system matrix and the augmented matrix are equal. If this is not the case, then there's no reason to do any further work. (cf Kronecker-Capelli criterion)

Usually, this check is performed at Step 2. It's easy to see that the system matrix is of rank $3$. Thus, solutions exist.

aaaargh @nuuskur ... the determinant of this matrix is $0$ i.e
from

\begin{bmatrix}
1 & 1 & 2 & -5 \\
2 & 5 & -1 & -9 \\
2 & 1 & -1 & 3\\
1& -3 & 2 & 7
\end{bmatrix}

using
1. $R_2-2R1$
2. $R_3-2R_1$
3. $R_4-R_1$

We shall have;
\begin{bmatrix}
1 & 1 & 2 & -5 \\
0 & 3 & -5 & 1 \\
0 & -1 & -5 & 13\\
0& -4 & 0& 12
\end{bmatrix}

The determinant is given by;

$1[3(-60)+5(-12+52)+1(0-20)]=-180+180=0$ implying existence of dependent solutions as earlier seen.

 

[FactChecker]

My way of tackling the problem is as follows; using echelon form (row reduction method) \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 2 & 5 & -1 & -9&-3 \\ 2 & 1 & -1 & 3&-11\\ 1& -3 & 2 & 7&-5 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 & -5 &3\\ 0 & -3 & -3 & -1&9 \\ 0 & 1 & 5 & -13&17\\ 0& 4& 0 & -12&8 \end{bmatrix}

If you want help, it would be nice if you indicated what row operations you were doing on each row.
Like your r2 row operation to get the second matrix is -(r2-r1) => r2. And your r3 row operation to get your third matrix is (r2+3*r3) => r3.
That way, we can help and don't have to spend time fishing around.

 

[chwala]

If you want help, it would be nice if you indicated what row operations you were doing on each row.
Like your r2 row operation to get the second matrix is -(r2-r1) => r2. And your r3 row operation to get your third matrix is (r2+3*r3) => r3.
That way, we can help and don't have to spend time fishing around.

You made a mistake (I guess a typo mistake) on the second row.

$4--1=5$ and not $-3$.

 

[FactChecker]

You made a mistake (I guess a typo mistake) on the second row.

$4--1=5$ and not $-3$.

Right. My next guess would be -(r2-2*r1) => r2. No, that only works for the first two columns. I think that makes my point. Suppose you tell me what you did instead of making me guess.

 

[chwala]

Right. My next guess would be -(r2-2*r1) => r2. No, that only works for the first two columns. I think that makes my point. Suppose you tell me what you did instead of making me guess.

\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
2 & 5 & -1 & -9&-3 \\
2 & 1 & -1 & 3&-11\\
1& -3 & 2 & 7&-5
\end{bmatrix}

Using row-reduction echelon form i.e

$2R_1-R_2$
$2R_1-R_3$
$R_1-R_4$

We get;

\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
0 & -3 & 5 & -1&9 \\
0 & 1 & 5 & -13&17\\
0& 4& 0 & -12&8
\end{bmatrix}

On using;

$R_2+3R_3$
$4R_2+3R_3$

We get;
\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
0 & -3 & 5 & -1&9 \\
0 & 0 & 20 & -40&60\\
0& 0& 20 & -40&60
\end{bmatrix}

From this the steps to solution will follow as indicated in my posts $8$ and $11$.

 

[FactChecker]

\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
2 & 5 & -1 & -9&-3 \\
2 & 1 & -1 & 3&-11\\
1& -3 & 2 & 7&-5
\end{bmatrix}

Using row-reduction echelon form i.e

$2R_1-R_2$
$2R_1-R_3$
$R_1-R_4$

We get;

\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
0 & -3 & 5 & -1&9 \\
0 & 1 & 5 & -13&17\\
0& 4& 0 & -12&8
\end{bmatrix}

On using;

$R_2+3R_3$
$4R_2+3R_3$

We get;
\begin{bmatrix}
1 & 1 & 2 & -5 &3\\
0 & -3 & 5 & -1&9 \\
0 & 0 & 20 & -40&60\\
0& 0& 20 & -40&60
\end{bmatrix}

From this the steps to solution will follow as indicated in my posts $8$ and $11$.

Thanks! That helped me a lot to follow your work.

 

[nuuskur]

@chwala
Be careful of your choice of free variables. In this instance you have chosen correctly, because the system is of the form
$$\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & b\\ 1 & 1 & 2 & -5 &3\\ 0 & -3 & 5 & -1&9 \\ 0 & 0 & 2 & -4&6\\ \end{bmatrix}$$
The minor corresponding to $x_1,x_2,x_4$ (i.e rows 123 and columns 124) is nonzero, therefore $x_3$ maybe regarded as free. Similarly, $x_4$ can also be free. In fact, anything here is free.

If the situation were something like this instead
$$\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & b\\ 1 & 1 & 2 & -5 &3\\ 0 & -3 & -6 & -1&9 \\ 0 & 1 & 2 & -4&6\\ \end{bmatrix}$$
then $x_4$ cannot be regarded as free. (consider the Kronecker-Capelli theorem)

In general, one determines the rank of the system and finds a nonzero minor of that order. All variables that are "outside" of the minor can be regarded as free.

 

[chwala]

@chwala
Be careful of your choice of free variables. In this instance you have chosen correctly, because the system is of the form
$$\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & b\\ 1 & 1 & 2 & -5 &3\\ 0 & -3 & 5 & -1&9 \\ 0 & 0 & 2 & -4&6\\ \end{bmatrix}$$
The minor corresponding to $x_1,x_2,x_4$ (i.e rows 123 and columns 124) is nonzero, therefore $x_3$ maybe regarded as free. Similarly, $x_4$ can also be free. In fact, anything here is free.

If the situation were something like this instead
$$\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & b\\ 1 & 1 & 2 & -5 &3\\ 0 & -3 & -6 & -1&9 \\ 0 & 1 & 2 & -4&6\\ \end{bmatrix}$$
then $x_4$ cannot be regarded as free. (consider the Kronecker-Capelli theorem)

In general, one determines the rank of the system and finds a nonzero minor of that order. All variables that are "outside" of the minor can be regarded as free.

Thanks! Let me look at the Kronecker - Capelli theorem. I am hearing it for the 1st time...