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So this is my homework :

Let $$ A= \begin{bmatrix} 1 & 0 & 1 & 3 \\ 2 & 0 & \lambda & 6 \\ 1 & 1 & 1 & 1 \\ \end{bmatrix} \\$$ and $b=\begin{pmatrix} 1\\ 2\\ 1\\ \end{pmatrix}\\$

Solution of linear equation system $Ax=b$ for every $\lambda € R$?

My Solution:

I brought the matrix to row echelon Form ==>

\begin{bmatrix} 1 & 0 & 1 & 3 |1\\ 0 & 1 & 0 & 2 |0 \\ 0 & 0 & \lambda-2 & 0 |0 \\ \end{bmatrix}

For every $\lambda$ $rank(A)=rank(A|b)\\$ Which means this Linear equation system is soluble. But what i m confused is we have for every lambda more Variable then equations so this LES has only infinite solutions? and for example for $\lambda =2\\$

we have $\begin{pmatrix} 1-x_3-3x_4\\ 2x_4\\ x_3\\ x_4\end{pmatrix}\\$

But for every diffrent $\lambda$ we get different combination of variables for the Solution which is not equal to other infinite solution.So what is incorrect in my solution ?

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  • $\begingroup$ Unless $\lambda=2$, that's not reduced row-echelon form, and $x_3$ is not arbitrary, it must be zero. $\endgroup$ – Gerry Myerson Nov 9 '13 at 10:58
  • $\begingroup$ @GerryMyerson so unless lambda =2 the solution is (1-3x4,2x4,0,x4) ? $\endgroup$ – nbdip Nov 9 '13 at 11:05
  • $\begingroup$ Yes, assuming you've done the row reduction correctly --- I didn't check your arithmetic. $\endgroup$ – Gerry Myerson Nov 9 '13 at 11:11
  • $\begingroup$ @nbdip: Did you still need an answer to this? $\endgroup$ – Amzoti Nov 9 '13 at 23:15
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The $(2,4)$ entry of your reduced matrix should be $-2$ rather than $2$.

Then you distinguish between two cases: Case I, $\lambda =2$; and case II, $\lambda\neq2$.

Case I: $\lambda\neq2$, rank is 3, one free parameter. Write the equations down $x_4=s$ (free parameter):

  • $(\lambda -2)x_3=0$ hence $x_3=0$
  • $x_2=2s$
  • $x_1=1-3s$.

Case II: $\lambda=2$, rank is 2, set two free parameters and solve as before.

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