Determine the value of A & B for this system ?!!!

Determine the value of A & B for which system

$$\begin{pmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & A \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}=\begin{pmatrix} B \\ 3 \\ -1 \\ \end{pmatrix}$$ 1) has a unique solution.
2) has infinite number of solutions.
3) has no solution.

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What did you try?!!! – Asaf Karagila Jan 11 '13 at 17:50
I tried to reduce to echelon form 2A + 3B = 5 but I'm not sure that is right or not !!! – eagle93 Jan 11 '13 at 17:59

3 Answers

Note that an equation of the form $\mathbf{A}\vec{x}=\vec{b}$ has a unique solution iff $\det\mathbf{A}\not=0$, therefore, we can find the value of $\mathbf{A}$ for which there is a unique solution:

$$\det\mathbf{A}=\begin{vmatrix}3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & A \end{vmatrix}\not= 0\implies-14A-42\not=0 \therefore A\not=-3$$

If $A\not=3$ then $B\in\mathbb{R}$ is a sufficient condition to ensure a unique solution.

If there are an infinite number of solutions, then $A=-3$ and $3x-2y+z=B$ is some linear combination of $5x-8y+9z=3$ and $2x+y-3z=-1$. Solving the following system:

$$\begin{pmatrix}3 \\ -2 \\ 1\end{pmatrix}=\lambda\begin{pmatrix}5 \\ -8 \\ 9\end{pmatrix}+\mu\begin{pmatrix}2 \\ 1 \\ -3\end{pmatrix}$$

Gives the solutions $\lambda=\frac{1}{3}$ and $\mu=\frac{2}{3}$, therefore $B=1-\frac{2}{3}=\frac{1}{3}$ for infinite solutions.

If there are no solutions then the three equations are inconsistent and therefore $B\not=\frac{1}{3}$.

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compare the ranks of the matrices $\begin{pmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 &A \end{pmatrix}$ and $\begin{pmatrix} 3 & -2 & 1 & B \\ 5 & -8 & 9 &3 \\ 2 & 1 &A&-1 \end{pmatrix}$.

If the ranks are equal, there is a solution. If they are not, there is none. There is a unique solution iff the ranks are full, i.e. $3$.

The interisting values should be $A=-3$ and $B=\frac 13$.

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but how did you get the value of B ? <br/> I tried REF and got $2A + 3B = 5$ <br/> Is it right ?!!! – eagle93 Jan 11 '13 at 18:20
@Babak Sorouh: Thanks for your help. I want A,B in the 3 conditions simultaneously :) – eagle93 Jan 11 '13 at 18:22
@eagle93: Your welcome. $\ddot\smile$ – Babak S. Jan 11 '13 at 18:27
@eagle93 There is a hi probability, i made a mistake in my compuatation. But I just added the integral multiplicities of the lines to each other and so there was no mixture of A and B in a row. I received that the extended matrix is equivalent to $\begin{pmatrix} 1 &-3 &1-A & B+1 \\ 0 &7&4+5A & -2-5B \\ 0&0& -6-2A & -1+3B \end{pmatrix}$. – Greyfox Jan 11 '13 at 18:32

For 1, it is proved that the non-homogeneous system of $n$ equations with $n$ variables has unique solution if and only if the matrix including all coefficient of the system has a non zero determinant. So evaluate the determine of $$S:=\begin{pmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & A \\ \end{pmatrix}$$ and put it equal to zero. The resulted values should not be given to $A$. Here we have $\det(S)=-14A-42$ and so for all $A\neq\frac{42}{-14}$ the system has unique solution.

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Should of ended with "just one solution!!!" to fit to the title and comments:P – Belgi Jan 11 '13 at 18:05
@Belgi: I am in a bad doubt if the OP is looking foe $A,B$ which satisfy the conditions above simultaneously or not. I meant unique solution. – Babak S. Jan 11 '13 at 18:08
+1 for helpfulness! :-) – amWhy Feb 18 '13 at 16:22