Solving Linear Algebra Problem 8: Gauss-Jordan Method

[cookiemnstr510510]

Hello All,

I have a question regarding the wording of this problem and my method of solving. (Problem and directions attached in Linear.jpg) PROBLEM 8 NOT 7! :)

Here is my thought process:
Keep doing elementary row operations until we have it it gauss-jordan form, then we have our answers?! I am new to Linear algebra and am still trying to figure differences in "directions" of problems.

I haven't figured out how to use the LaTex commands with matricies yet (looked in the latex help link and can't figure it out) for now the best I can do is write my work neatly and attach. You will find my work in Problem8.jpg attached.
will learn to use Latex soon!

Thanks

 

[cookiemnstr510510]

Edit: Just figured out how to write problems in matrix form with LaTex

 

[Mark44]

Here is my thought process:
Keep doing elementary row operations until we have it it gauss-jordan form, then we have our answers?! I am new to Linear algebra and am still trying to figure differences in "directions" of problems.

Your answer to #8 looks fine, and yes that's the right strategy.
Regarding augmented matrices, for this particular problem, since the right-most column of your augmented matrix consisted solely of 0's, you really didn't need it. None of the row operations could possibly have changed these values. Once you get down to the 4 x 4 identity matrix by row operations, then your solution is $x_1 = 0, x_2 = 0, x_3 = 0, x_4 = 0$, and this is the unique solution.

 

[Ray Vickson]

Hello All,

I have a question regarding the wording of this problem and my method of solving. (Problem and directions attached in Linear.jpg) PROBLEM 8 NOT 7! :)

Here is my thought process:
Keep doing elementary row operations until we have it it gauss-jordan form, then we have our answers?! I am new to Linear algebra and am still trying to figure differences in "directions" of problems.

I haven't figured out how to use the LaTex commands with matricies yet (looked in the latex help link and can't figure it out) for now the best I can do is write my work neatly and attach. You will find my work in Problem8.jpg attached.
will learn to use Latex soon!

Thanks

Yes, your answer is correct, and it is easy to see without doing any work! If
$${\bf A} = \begin{bmatrix} 1 & -5 & 4 & 0 \\ 0 & 1 & 0 & 1 \\
0 & 0 & 3 & 0 \\ 0& 0& 0& 2
\end{bmatrix}, \: {\bf x} = \begin{bmatrix} x_1 \\x_2 \\x_3 \\x_4 \end{bmatrix}\; \text{and} \;
{\bf 0} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} $$
then your equations read as ${\bf A x} = {\bf 0}.$

Notice that all the elements of ${\bf A}$ below the diagonal are zero; that means that your matrix has a nonzero determinant, hence has a unique inverse ${\bf A}^{-1}$. The solution is ${\bf x} = {\bf A}^{-1} {\bf 0} = {\bf 0}.$