[SOLVED] Need help on matricies

SPOIDUR · January 31st 2008, 12:00 PM

I'm doing homework and there is one question that I am suck on, the example in the text does not explain it enough to really get, for me so I figured if someone could go through it and show / explain me the steps, it would be much appreciated.

Anyways the question itself is

In each of the following, find conditions on a and b such that the system has no solution, one solution, and infinitely many solutions

x + by = -1
ax + 2y = 5

The Solution in the book says
If ab =/= 2, unique solution x = (-2-5b)/(2-ab) y= (a+5)/(2-ab)
If ab=2: no solution if a =/= -5 ; if a = -5, the solutions are x=-1+2/5t, y=t.

Could someone please go through this problem explaining how I find this answer?

**mr fantastic** · January 31st 2008, 02:51 PM

Originally Posted by SPOIDUR

I'm doing homework and there is one question that I am suck on, the example in the text does not explain it enough to really get, for me so I figured if someone could go through it and show / explain me the steps, it would be much appreciated.

Anyways the question itself is

In each of the following, find conditions on a and b such that the system has no solution, one solution, and infinitely many solutions

x + by = -1
ax + 2y = 5

The Solution in the book says
If ab =/= 2, unique solution x = (-2-5b)/(2-ab) y= (a+5)/(2-ab)
If ab=2: no solution if a =/= -5 ; if a = -5, the solutions are x=-1+2/5t, y=t.

Could someone please go through this problem explaining how I find this answer?

I'll assume you know:
1. How to express your linear equations in the matrix form AX = B.
2. You know how to find the determinant of A when A is a 2x2 matrix.
3. You understand that the equations have no unique solution when the determinant of A, det(A), is equal to zero.

In your problem, det(A) = 2 - ab.

det(A) = 0 => 2 - ab = 0 => ab = 2.

So there will be NO unique solution when ab = 2. Therefore there WILL be a unique solution when $ab \neq 2$ .

When there's no unique solution, that means that either there is an infinite number of solutions or no solution at all. To test which of these happens when ab = 2, you might do something like the following:

$ab = 2 \Rightarrow b = \frac{2}{a}$ . Substitute this into you equations and do some re-arranging:

Equation (1): $x + by = -1 \Rightarrow x + \frac{2}{a} y = -1 \Rightarrow ax + 2y = -a$ .

Equation (2): $ax + 2y = 5$ .

Now compare equations (1) and (2). You can see that when $-a = 5 \Rightarrow a = -5$ the equations are the same. One equation and two unknowns => infinite number of solutions.

So when ab = 2 AND a = -5, there an infinite number of solutions.

It follows that there will be no solution at all when ab = 2 AND $a \neq -5$ .

When a = -5 and ab = 2, $b = -\frac{2}{5}$ . So the equations boil down to -5x + 2y = 5. The form of the infinite number of solutions can be expressed in several ways. To see where the books answer came from, note that if you let y = t, where t can be any real number:

$-5x + 2t = 5 \Rightarrow -5x = 5 - 2t \Rightarrow 5x = 2t - 5 \Rightarrow x = \frac{2}{5}t - 1$ .

As to where the unique solution comes from when $ab \neq 2$ :

x + by = -1 .... (1)

ax + 2y = 5 .... (2)

Solve using the elimination method:

Multiply equation (1) by a, subtract it from equation (2) and then solve for y:

(2 - ab)y = 5 + a => ......

Then sub y into either equation and solve for x.

**mr fantastic** · January 31st 2008, 04:56 PM

Alternatively, if you're NOT familiar with matrices, here's an edited version of my earlier reply that doesn't use matrices:

Originally Posted by mr fantastic

Unique solution:

x + by = -1 .... (1)

ax + 2y = 5 .... (2)

Solve using the elimination method:

Multiply equation (1) by a, subtract it from equation (2) and then solve for y:

$(2 - ab)y = 5 + a \Rightarrow y = \frac{5 + a}{2 - ab}$ .

Note that this solution is OK provided the denominator is NOT equal to zero, that is, $2 - ab \neq 0 \Rightarrow ab \ne 2$ .

Now sub $y = \frac{5 + a}{2 - ab}$ into either equation and solve for x. Using equation (1):

$x + by = -1 \Rightarrow x = -1 - by = -1 - b \left( \frac{5 + a}{2 - ab} \right)$ $= \frac{-(2 - ab)}{2 - ab} - \frac{b(5 + a)}{2 - ab} = \frac{-2 + ab - 5b - ab}{2 - ab} = \frac{-2 - 5b}{2 - ab}$ .

So there's a unique solution, given by the above, when $ab \neq 2$ .

--------------------------------------------------------------------------

No unique solution: $ab = 2$ .

When there's no unique solution, that means that either there is an infinite number of solutions or no solution at all. To test which of these happens when ab = 2, you might do something like the following:

$ab = 2 \Rightarrow b = \frac{2}{a}$ . Substitute this into you equations and do some re-arranging:

Equation (1): $x + by = -1 \Rightarrow x + \frac{2}{a} y = -1 \Rightarrow ax + 2y = -a$ .

Equation (2): $ax + 2y = 5$ .

Now compare equations (1) and (2). You can see that when $-a = 5 \Rightarrow a = -5$ the equations are the same. One equation and two unknowns => infinite number of solutions.

So when ab = 2 AND a = -5, there an infinite number of solutions.

It follows that there will be no solution at all when ab = 2 AND $a \neq -5$ .

When a = -5 and ab = 2, $b = -\frac{2}{5}$ . So the equations boil down to -5x + 2y = 5. The form of the infinite number of solutions can be expressed in several ways. To see where the books answer came from, note that if you let y = t, where t can be any real number:

$-5x + 2t = 5 \Rightarrow -5x = 5 - 2t \Rightarrow 5x = 2t - 5 \Rightarrow x = \frac{2}{5}t - 1$ .

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