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Consider the system of linear equations: \begin{align} \begin{cases} x+ay=1\\[2ex] bx+5y=2, \end{cases} \end{align} where $a$ and $b$ are parameters.

(a) Determine the conditions on $a$ and $b$ to get a unique solution.
(b) Determine the conditions on $a$ and $b$ to get infinitely many solutions.
(c) Determine the conditions on $a$ and $b$ such that the system has no solutions.

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Let's write the equations as follows: $$x+ay=1\\ bx+5y=2$$ So, one has $$\begin{pmatrix}x\\ y\end{pmatrix}\begin{pmatrix}1 & a\\ b & 5\end{pmatrix}=\begin{pmatrix}1\\ 2\end{pmatrix}\\ \implies \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1 & a\\ b & 5\end{pmatrix}^{-1}\begin{pmatrix}1\\ 2\end{pmatrix}=\dfrac{1}{5-ab}\begin{pmatrix}5 & -a\\ -b & 1\end{pmatrix}\begin{pmatrix}1\\ 2\end{pmatrix}$$ This should considerably simplify your solution. As a hint, look at the term $\dfrac{1}{5-ab}$ on the RHS.

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Answer:

a) $$ab\not=5$$. The determinant of the matrix not equal to zero. There is a unique solution

b)$$a = \frac{5}{2}, b=2$$ The determinant is zero, but the two lines are identical, There are infinitely many solution

c) $$ab = 5 \land \lnot (a = \frac{5}{2} \land b=2)$$ The determinant is zero, but the two lines are parallel. There are no solutions

The determinant => (5-ab)

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  • $\begingroup$ c) should be $ab = 5 \land \lnot(a = \frac{5}{2} \land b = 2)$. $\endgroup$ – Keba Mar 12 '14 at 23:23
  • $\begingroup$ You are right, I do not know the syntax. If you don't mind, could you edit my answer $\endgroup$ – Satish Ramanathan Mar 12 '14 at 23:24
  • $\begingroup$ @Keba, Thanks. I appreciate it!! $\endgroup$ – Satish Ramanathan Mar 12 '14 at 23:29
  • $\begingroup$ Done. :) Btw: $\land$ is land, $\lnot$ is lnot. Most times l (for logical, I assume) + "that what you assume" does the job for these symbols. There is also Detextify. $\endgroup$ – Keba Mar 12 '14 at 23:29

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