# Solving equation

1. Jun 16, 2009

### Gregg

1. The problem statement, all variables and given/known data

Express the determinant as a product of 4 linear factors

$\left( \begin{array}{ccc} a & \text{bc} & b+c \\ b & \text{ac} & a+c \\ c & \text{ab} & a+b \end{array} \right)$

b hence or otherwise find the values of a for which the sumultaneous equations.

$ax+2t+3z=0$

$2x+ay+(1+a)z=0$

$x+2ay+(2+a)z=0$

have a solution other than x=y=z=0

ii) solve the equations when a=-3

3. The attempt at a solution
$\text{Det}\left[\left( \begin{array}{ccc} a & \text{bc} & b+c \\ b & \text{ac} & a+c \\ c & \text{ab} & a+b \end{array} \right)\right] = (a-b)(c-b)(a-c)(a+b+c)$

I get the right answers for part 2 of 1,2,-3. I don't know why that determinant of 0 implies that solution.

for the last part i get

$x=\frac{5\lambda}{3}$
$y=\lambda$
$z=\lambda$

$r=\lambda\left( \begin{array}{c} \frac{5}{3} \\ 1 \\ 1 \end{array} \right)$

$r = \lambda\left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right)$ ???

2. Jun 16, 2009

### Gregg

$3x-18y-3z=0$

$-3x+2y+3z=0$

$y=0$

$x=z$

$z=\lambda$

$r=\lambda\left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right)$

I wonder why it messes up when I use the other equations

3. Jun 16, 2009

### HallsofIvy

Staff Emeritus
I presume you mean "y" rather than "t" here.

Determinant 0 does not imply any particular solution. It implies that there is NOT a unique solution. Since the given matrix equation has (0 0 0) as a solution, that means that there must be other solutions- in fact an infinite number of solutions, forming a subspace of R3. Since you have shown that the determinant is (a- b)(c- b)(a- c)(a+ b+ c), it will be 0 when any one of those factors is 0. Further, the determinant in (b) is the same as in (a) with b= 2 and c= 1. So its determinant is (a-2)(a-1)(a+ 3)= 0. The determinant will be 0 when a= 2, a= 1 or a= -3, making those factors 0.

With a= 3, the equations become 3x+ 2y+ 3z= 0, 2x+ 3y+ 4z= 0, and x+ 6y+ 5z= 0. $x= \lambda$, y= 0, $z= \lambda$ makes the first equation $3\lambda+ 3\lambda= 6\lambda= 0$ which is true only for $\lambda$ equals 0 and that is the "trivial" solution.
$r = \lambda\left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right)$
is definitely NOT a solution.

4. Jun 16, 2009

### HallsofIvy

Staff Emeritus
Where did you get these three equations? They are NOT the equation from yyour first post.

5. Jun 16, 2009

### Gregg

(3) x-6y-z=0
3x-18y-3z=0
from (1) -3x+2y+3z=0

(1)+(3) = -16y=0
y=0

x=z
z=x
etc?