Solving Equations: Det as Product of 4 Linear Factors

[Gregg]

Homework Statement

Express the determinant as a product of 4 linear factors

<br /> \left(<br /> \begin{array}{ccc}<br /> a &amp; \text{bc} &amp; b+c \\<br /> b &amp; \text{ac} &amp; a+c \\<br /> c &amp; \text{ab} &amp; a+b<br /> \end{array}<br /> \right)

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

<br /> ax+2t+3z=0<br />

<br /> 2x+ay+(1+a)z=0<br />

<br /> x+2ay+(2+a)z=0<br />

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

ii) solve the equations when a=-3

The Attempt at a Solution

<br /> \text{Det}\left[\left(<br /> \begin{array}{ccc}<br /> a &amp; \text{bc} &amp; b+c \\<br /> b &amp; \text{ac} &amp; a+c \\<br /> c &amp; \text{ab} &amp; a+b<br /> \end{array}<br /> \right)\right] = (a-b)(c-b)(a-c)(a+b+c)<br />


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(<br /> \begin{array}{c}<br /> \frac{5}{3} \\<br /> 1 \\<br /> 1<br /> \end{array}<br /> \right)

The answer is

r = \lambda\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 1<br /> \end{array}<br /> \right) ?

 

[Gregg]

3x-18y-3z=0

-3x+2y+3z=0

y=0

x=z

z=\lambda

r=\lambda\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 1<br /> \end{array}<br /> \right)

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

 

[HallsofIvy]

Homework Statement

Express the determinant as a product of 4 linear factors

<br /> \left(<br /> \begin{array}{ccc}<br /> a &amp; \text{bc} &amp; b+c \\<br /> b &amp; \text{ac} &amp; a+c \\<br /> c &amp; \text{ab} &amp; a+b<br /> \end{array}<br /> \right)

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

<br /> ax+2t+3z=0<br />

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

<br /> 2x+ay+(1+a)z=0<br />

<br /> x+2ay+(2+a)z=0<br />

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

ii) solve the equations when a=-3

The Attempt at a Solution

<br /> \text{Det}\left[\left(<br /> \begin{array}{ccc}<br /> a &amp; \text{bc} &amp; b+c \\<br /> b &amp; \text{ac} &amp; a+c \\<br /> c &amp; \text{ab} &amp; a+b<br /> \end{array}<br /> \right)\right] = (a-b)(c-b)(a-c)(a+b+c)<br />


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

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 anyone 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.

for the last part i get

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

r=\lambda\left(<br /> \begin{array}{c}<br /> \frac{5}{3} \\<br /> 1 \\<br /> 1<br /> \end{array}<br /> \right)

The answer is

r = \lambda\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 1<br /> \end{array}<br /> \right) ?

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(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 1<br /> \end{array}<br /> \right)
is definitely NOT a solution.

 

[HallsofIvy]

3x-18y-3z=0

-3x+2y+3z=0

y=0

x=z

z=\lambda

r=\lambda\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 1<br /> \end{array}<br /> \right)

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

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

 

[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?