# Find exactly one solution

• October 1st 2012, 06:14 AM
nivek0078
Find exactly one solution
I'm confussed on how to solve this because of the constants a for the given value of s.
Given:
x1 + 0x2 + x3 = a
5x1 + (s-1)x2 + 3x3 = a
sx1 + 0x2 + 4x3 = 1

Find what value of s does the system have on solution?
• October 1st 2012, 10:43 AM
johnsomeone
Re: Find exactly one solution
For the general case, a makes no difference:

Have $MX = B$, which is $\left( \begin{matrix} 1 & 0 & 1 \\ 5 & (s-1) & 3 \\ s & 0 & 4 \end{matrix} \right)\left( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right) = \left( \begin{matrix} a \\ a \\ 1 \end{matrix} \right)$,

so, when $M^{-1}$ exists, have $X = M^{-1}B$ and it's solved. But $M^{-1}$ exists iff $det(M) \ne 0$.

Thus, the general condition that will make that system solvable has nothing to do with the B. It's decided by entirely by M, via $det(M) \ne 0$.

Using cofactors:

$det(M) = (1)det \left( \begin{matrix} (s-1) & 3 \\ 0 & 4 \end{matrix} \right) - (0)det \left( \begin{matrix} 5 & 3 \\ s & 4 \end{matrix} \right) + (1)det \left( \begin{matrix} 5 & (s-1) \\ s & 0 \end{matrix} \right)$.

$det(M) = (1)((s-1)(4)-(3)(0)) - (0)(whatever) +(1)((5)(0) - (s)(s-1))$

$= (4s-4) + (- s^2 + s) = -s^2 +5s -4$

So $det(M) =0$ when $-s^2 +5s -4 = 0$ when $s^2 -5s +4= 0$

Factoring gives $s^2 -5s +4= (s-4)(s-1)$, so $s^2 -5s +4=0$ when $s\in \{1, 4\}$.

Thus this is solveable whenever $s \notin \{ 1, 4 \}$, no matter what a is.

Now, when $s \in \{ 1, 4 \}$, it *might* have solutions, but then the value of $a$ will decide that.

Check by doing row reduction on: $\left( \begin{matrix} 1 & 0 & 1 & | & a \\ 5 & (s-1) & 3 & | & a\\ s & 0 & 4 & | & 1 \end{matrix} \right)$

When $s = 1$ you get:

$\left( \begin{matrix} 1 & 0 & 1 & | & a \\ 5 & 0 & 3 & | & a\\ 1 & 0 & 4 & | & 1 \end{matrix} \right)$

so $\left( \begin{matrix} 1 & 0 & 1 & | & a \\ 0 & 0 & -2 & | & -4a\\ 0 & 0 & 3 & | & 1-a \end{matrix} \right)$ (used row 1 to clear column 1)

so $\left( \begin{matrix} 1 & 0 & 1 & | & a \\ 0 & 0 & 1 & | & 2a\\ 0 & 0 & 1 & | & \frac{1-a}{3} \end{matrix} \right)$

so $\left( \begin{matrix} 1 & 0 & 0 & | & -a \\ 0 & 0 & 1 & | & 2a\\ 0 & 0 & 0 & | & (\frac{1-a}{3}-2a) \end{matrix} \right)$ (used row 2 to clear column 3)

Thus $x_1 = -a, x_3 = 2a,$ and $0 = \frac{1-a}{3}-2a$.

The last equation puts a restriction on a: $\frac{1-a}{3}-2a = 0$ implies $1-a = 6a$ implies $a = 1/7$.

Thus when s = 1, it has a solution, but only if a = 1/7.

The s = 1 and a = 1/7 solution is: $x_1 = -1/7, x_2 = anything, x_3 = 2/7$.

I'll leave it to you to see what happens when s = 4.