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$$ \left[ \begin{array}{cc|c} 1&-2&4\\ a&4&5 \end{array} \right] $$

I came across this question on one of my course slides, and I am having trouble understanding the whole concept of an equation having no solution, one solution or infinitely many solutions.

$$ \left[ \begin{array}{cc|c} 1&-2&4\\ 0&4+2a&5-4a \end{array} \right] $$

this was the resulting matrix. What I don't understand is how you get $(4+2a)$ and $(5-4a)$? And what needs to be done so that the same equation has no solution, one unique solution, and infinitely many solutions. Can anybody please help?

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  • $\begingroup$ for example if $a=-2$ ? $\endgroup$ – percusse Nov 20 '17 at 23:54
  • $\begingroup$ yes, -2. But how do we know that we should put -2 and not some other number? $\endgroup$ – Lame Person Nov 20 '17 at 23:58
  • $\begingroup$ multiply first row with $-a$ and add it to the second one. $\endgroup$ – percusse Nov 20 '17 at 23:59
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To answer your first question, about how you're getting that matrix, recall that you can apply operations involving the lines of a matrix (excuse my weak expression regarding English).

Namingly, you get that matrix when you do the following :

$$-aL_1 + L_2 \to L_2$$

which means that you multiply the first line of the initial matrix with $-a$ and then add it to the second line. This steps are usually carried out when you want to form a lower triangular matrix and it's one of the most basic actions in Linear Algebra (recall Gauss Elimination method).

Now, let's check the second part of your question.

  • If $4+2a = 0 \Leftrightarrow a = -2$ then you'd get an equation of the form : $0x + 0y=13$ which does not stand. Thus, for $a=-2$ the linear system does not hold.

  • If $5-4a=0\Leftrightarrow a=5/4$ then you'd get an equation of the form : $0x + (4+5/2)y=0$ which means $y=0$ and thus substituting in the first line equation $x=4$. Thus, for this value of $a$, there is a unique solution to your given system of equations.

  • In order for the system to have infinitely many solutions, one of the two equations should be : $0x + 0y = 0$ which holds for any $(x,y) \in \mathbb R^2$ and means that any pair (infinite pairs) that satisfy the first line equation : $x-2y = 4$ is a solution to the system, thus it has infinite solutions. Can you figure that part out ?

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The first formula (augmented matrix) stands for the following system of linear eqations:

$$x-2y=4\tag 1$$ $$ax+4y=5.\tag 2$$

From the first equation$$x=4+2y.$$

Substituting this into the second equation we get $$0x+(2a+4)y=5-4a.\tag 3$$

$(1)$ and $(3)$ together can be expressed as the following augmented matrix

$$ \left[ \begin{array}{cc|c} 1&-2&4\\ 0&4+2a&5-4a \end{array} \right] .$$

From $(3)$ the solution for $y$ is

$$\frac{5-4a}{4+2a}.$$

If $a=-2$ then there is no solution as the denominator is zero then. This can be seen, as well, if we substitute $a=-2$ in $(1)$ and $(2)$:

$$x-2y=4\tag 1$$ $$-2x+4y=5\tag 2$$ or $$x-2y=4\tag 1$$ $$2(x-2y)=-5\tag 2$$

which is impossible.

If $a\not =-2$ then there is always one unique solution.

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