Difficult Matrix Equation

[Tala]

Hi

I'm supposed to solve this matrix equation by GaussJordan elimination and I find it very difficult to see and carry out the row operations mainly because of the z's. If someone is able to help me I would truly appreciate it.

zx1 + x2 + x3 = 1

x1 + zx2 + x3 = 1

x1 + x2 + zx3 = 1


z has a real value.

 

[Deleted member 4993]

Hi

I'm supposed to solve this matrix equation by GaussJordan elimination and I find it very difficult to see and carry out the row operations mainly because of the z's. If someone is able to help me I would truly appreciate it.

zx1 + x2 + x3 = 1

x1 + zx2 + x3 = 1

x1 + x2 + zx3 = 1


z has a real value.

Iam not sure why that (z) should create any problem!!

zx1 + x2 + x3 = 1

x1 + zx2 + x3 = 1

x1 + x2 + zx3 = 1

z*R2 - R1

zx1+ .........x2 + .........x3 ....= 1

0 + ..(z²- 1)x2 + (z-1)*x3 ....= z -1

x1 +...........x2 + z*.....x3 ....= 1

Now continue (z*R3 - R1) and so forth...

 

[soroban]

Helo, Tala!

I modified the problem.

I'm supposed to solve this matrix equation by Gauss-Jordan elimination.

. . \(\displaystyle \begin{array}{ccc}ax + y + z &=& 1 \\ x + ay + z &=& 1 \\ x+y+az &=& 1\end{array}\)

We have: .\(\displaystyle \left|\begin{array}{ccc|c} a &1&1&1 \\ 1&a&1&1 \\ 1&1&a&1 \end{array}\right|\)

. .\(\displaystyle \begin{array}{c}\text{Switch} \\ R_1 \& R_3 \\ \end{array}\:\left|\begin{array}{ccc|c} 1&1&a&1 \\ 1&a&1&1 \\ a&1&1&1 \end{array}\right|\)

\(\displaystyle \begin{array}{c}\\ \\\\ R_2-R_1 \\ R_3-aR_1 \end{array}\:\left|\begin{array}{ccc|c}1&1&a&1 \\ 0& a\!-\!1 & 1\!-\!a & 0 \\ 0&1\!-\!a & 1\!-\!a^2 & 1\!-\!a \end{array}\right|\)

. . \(\displaystyle \begin{array}{c} \\ \\ \frac{1}{a\!-\!1}R_2 \\ \frac{1}{a\!-\!1}R_3\end{array}\: \left|\begin{array}{ccc|c} 1&1&a&1 \\ 0&1&\text{-}1&0 \\ 0&\text{-}1 & \text{-}(a\!+\!1) & \text{-}1 \end{array}\right|\)

\(\displaystyle \begin{array}{c}R_1-R_2 \\ \\ \\ R_3+R_2 \end{array}\:\left|\begin{array}{ccc|c}1 &0& a+1 & 1 \\ 0&1&\text{-}1&0 \\ 0&0&\text{-}(a+2) & \text{-}1 \end{array}\right|\)

\(\displaystyle \begin{array}{c} \\ \\ \\ \\ \text{-}\frac{1}{a+2}R_3 \end{array}\:\left|\begin{array}{ccc|c} 1&0&a\!+\!1 & 1 \\ 0&1&\text{-}1 & 0 \\ 0&0&1&\frac{1}{a\!+\!2} \end{array}\right|\)

\(\displaystyle \begin{array}{c}R_1 - (a\!+\!1)R_3 \\ R_2+R_3 \\ \\ \\ \end{array}\:\left|\begin{array}{ccc|c} 1&0&0 & \frac{1}{a+2} \\ 0&1&0 & \frac{1}{a+2} \\ 0&0&1 & \frac{1}{a+2} \end{array}\right|\)


Therefore: .\(\displaystyle x\;=\;y\;=\;z\;=\;\dfrac{1}{a+2}\)

 

[mmm4444bot]

I find it very difficult to see and carry out the row operations mainly because of the z's.

:idea: Try using a large sheet of paper, so that you may write clear, organized algebraic expressions when you write out the rows of the matrices.

Otherwise, I'm not sure what the meaning is for "very difficult to see … the row operations".

For example, the leading coefficient in the first equation is z, and the leading coefficient in the second equation is 1. In other words, in the coefficient matrix, the first elements in R1 and R2 are z and 1, respectively.

To obtain a 1 in the upper-left corner of the matrix, you would need to divide z by z, yes? That is, replace R1 with 1/z*R1. Is this the sort of reasoning to which you refer when you state that the row operations are very difficult "to see"?

We would like to see your beginning efforts. Otherwise, how could we envision what you're doing? :cool:

 

[Tala]

Thank you everybody it makes sense now.

 

[Deleted member 4993]

Hi

I'm supposed to solve this matrix equation by GaussJordan elimination and I find it very difficult to see and carry out the row operations mainly because of the z's. If someone is able to help me I would truly appreciate it.

zx1 + x2 + x3 = 1

x1 + zx2 + x3 = 1

x1 + x2 + zx3 = 1


z has a real value.

Now that Soroban posted the work, there is a shortcut to the solution.

We can see that if we exchange x1 and x2 and x3 - the equations do not change.

Thus

x1 = x2 = x3

then

zx1 + x2 + x3 = 1 →

x1(z + 2) = 1 →

x1 = x2 = x3 = 1/(z+2) for z <>-2

 

[Deleted member 4993]

Hi

I'm supposed to solve this matrix equation by GaussJordan elimination and I find it very difficult to see and carry out the row operations mainly because of the z's. If someone is able to help me I would truly appreciate it.

zx1 + x2 + x3 = 1 ..............................................(1)

x1 + zx2 + x3 = 1

x1 + x2 + zx3 = 1


z has a real value.

Another method - I was hoping student would discover these (particular) methods while working through the drudgery of row elimination.

Add all the equations to get

(2+z)(x1 + x2 + x3) = 3

x1 + x2 + x3 = 3/(2+z) .....................z<>-2

subtract from eqn (1)

(z-1)x1 = 1 - 3/(2+z)

x1 = 1/(2+z)...................................z<>-2 & z<>1

z<>1 is an implied constraint for the previous solutions also - because at z =1 the system is no longer linearly independent.