matrix solution assistance PLEEEASE

[kmille]

Can someone assist me with the proper solution and how obtained for the following?

The following Matrix is obtained from a system of equations.


[1 0 6 5]
[0 1 -4 3]
[0 0 2 -6]


The solution to the system is


A. No solution
B. (23, -9, -3)
C. (5, 3, -6)
D. (3+4k, 5-6k, k)

 

[tkhunny]

It's almost done.

Divide the third by 2, giving [0 0 1 -3] ==> (something, something else, -3)

That nicely discards C and D. B is looking good.

Just for fun, and if you have time, go a bit more - Twice the third added to the second, giving [0 1 0 -9] ==> (something, -9, -3)

If you need a complete solution, keep going. If you are on a panic timed test, you may wish to mark B and move on.

 

[soroban]

Hello, kmille!

Exactly where is your difficulty?
As Tkhunny said, "It's almost done."


We have: \(\displaystyle \;\begin{vmatrix} 1 & 0 & 6 & | & 5 \\
0 & 1 & -4 & | & 3 \\ 0 & 0 & 2 & | & -6\end{vmatrix}\)


. . . . . \(\displaystyle \begin{array}{c} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \frac{1}{2}R_3\end{array}\;
\begin{vmatrix}1 & 0 & 6 & | & 5\\
0 & 1 & -4 & | & 3 \\
0 & 0 & 1 & | & -3\end{vmatrix}\)


. \(\displaystyle \begin{array}{c} \\ \\ \\ \\ \\ \\ \\ R_1-6\cdot R_3 \\ \\ \\ R_2+4\cdot R_3\end{array}\;
\begin{vmatrix}1 & 0 & 0 & | & 23 \\
0 & 1 & 0 & | &-9 \\
0 & 0 & 1 & | & -3\end{vmatrix}\)
. . . . . . . . . . . . . .\(\displaystyle \uparrow\)
. . . . . . . . .\(\displaystyle B.\23,\,-9,\,-3)\)

 

[kmille]

Thanks for the assistance. I still do not fully understand how you figured the equasion solution such as where the 1/2 r3 came from, but I will head back to the textbook to try and figure it out. This is an online course so I do not have a teacher to turn to, just a very hard to understand textbook. Being out of the math loop for 30 years before returning to online classes is real difficult.

Keith :?

 

[tkhunny]

30 years will do it.

Do you know that ueac row of the matrix can translate into an equation?

[2 3 4 5] might represent 2x + 2y + 4z = 5

In this case: [0 0 2 -6] gives 0x + 0y + 2z = -6 or just 2z = -6. Solve this for 'z' and see how it relates to the matrix operations.