System of linear equations question.

I have an exam tomorrow on linear and discrete maths and there's an exercise question example in my book with no answer because i missed the lecture we went through it, and i'm certain I need to know how to do it, if any one can do this, please answer or at least have a go as it will help me greatly! thank you :)

The question is:
Solve completely the following system of linear equations:

x + 2y + λz = 0
2x + 3y -2z = λ
λx + y + λ²z = 3

I know this is a homogeneous system so you have to put it in the form:

1 2 λ | 0
2 3 -2 | λ
λ 1 λ² | 3

when it says solve, it means you have to find out what x y and z are.

Actually, your system is non-homogeneous. For it to be homogenous, all the terms on the right-hand sides would need to be 0. Perhaps the most convenient way to solve it is to row-reduce it to triangular form. I'll use t instead of lambda.

1 2 t | 0
2 3 -2 | t
t 1 t^2 | 3

...subtract t * R1 from R3:

1 2 t | 0
2 3 -2 | t
0 (1-2t) 0 | 3

...subtract 2 * R1 from R2:

1 2 t | 0
0 -1 -(2+2t) | t
0 (1-2t) 0 | 3

...add (1-2t) * R2 to R3:

1 2 t | 0
0 -1 -(2+2t) | t
0 0 (2+2t)(2t - 1) | 3+(1-2t)t


...divide R3 by (2t-1); note from the third row of the second step that this is non-zero, since otherwise 0=3 and the system has no solution:

1 2 t | 0
0 -1 -(2+2t) | t
0 0 (2+2t) | 3/(2t-1) - t

...add R3 to R2, and afterwards multiply R2 by -1:

1 2 t | 0
0 1 0 | -3/(2t-1)
0 0 (2+2t) | 3/(2t-1) - t

...subtract 2*R2 from R1:

1 0 t | 6/(2t-1)
0 1 0 |-3/(2t-1)
0 0 (2+2t) | 3/(2t-1) - t

Now if (2+2t) = 0, t=-1, and the third equation is 0=0. The second equation is y = 1, and the first is x - z = -2, which generates an infinite number of solutions. Otherwise...

...divide R3 by (2+2t) and simplify:

1 0 t | 6/(2t-1)
0 1 0 | -3/(2t-1)
0 0 1 | (3-2t)/(4t-2)

...subtract t*R3 from R1 and simplify:

1 0 0 | (2t^2 - 3t + 12)/(4t-2)
0 1 0 | -3/(2t-1)
0 0 1 | (3-2t)/(4t-2)


In summary, if t != -1 and if t != 1/2, we have solutions at
x = (2t^2 - 3t + 12)/(4t-2)
y = -3/(2t-1)
z = (3-2t)/(4t-2)

If t = -1, we have solutions whenever
y=1
x-z = -2

If t=1/2, we have no solutions.

"Is 1/(a+bi) ever equal to i/a + 1/bi ?"

> I did the following:

> 1/(a + bi) = i / a + 1 / bi
> 1/(a + bi) = i / a + (-bi / b^2)
> 1/(a + bi) = i / a + (-i / b)
> 1/(a + bi) = ((-a + b) i ) / ab
> 1/(a + bi) = (-1(a - b) i ) / ab
> (a - bi) / (a^2 + b^2) = (-1(a - b) i ) / ab

i have solved this .Is this right .............Is there any way ti solve it by Algebra - Solver, Help and Calculator

You say "i have solved this" though your final step is not a solution. Indeed, most of your manipulations are pointless and need to be reversed to arrive at a solution.

The right side of your third line is useful, though you can arrive at it more quickly by not multiplying and then dividing by b. In any case, it gives

i / a + (-i / b)
= i * (1/a - 1/b)

Using the left side of your final line, we get
1/(a + bi)
= (a - bi) / (a^2 + b^2)
= a / (a^2 + b^2) - bi / (a^2 + b^2)
= i * (1/a - 1/b)

Since the real parts must be equal, we get
a / (a^2 + b^2) = 0
=> a = 0

However, we assumed implicitly that a is non-zero (so that i/a is defined). There is then no solution.


An easier way to do this problem is to start from
1/(a + bi) = i / a + 1 / (bi)
= i/a - i/b

and multiply both sides by (a+bi). With FOILing, that gives
1 = (i/a - i/b)(a + bi)
= i - b/a - ai/b + 1

Subtracting 1 from both sides and comparing real parts, we find b/a = 0, but then b=0 so 1/(bi) wasn't defined. Hence there is no solution (again).