MHB Solver for Value of $a$: Unique, Infinite, No Solution

[evinda]

Hello! (Wave)

I want to find the value of $a$ for which the system

$\begin{pmatrix}
a & 1 & 1\\
1 & a & 1\\
1 & 1 & a
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
1\\
1\\
1\end{pmatrix}$

1. has a unique solution
2. has infinite solutions
3. has no solution

The determinant is equal to $a^3-3a+2$.

The system

1. has a unique solution if $a^3-3a+2 \neq 0 \Rightarrow a \neq 1,-2$
2. has infinite solutions if $a^3-3a+2 = 0 \Rightarrow a =1,-2$
3. has no solution for no value of $a$

Am I right? (Thinking)

 

[Deveno]

If the determinant is $0$ it doesn't necessarily mean we have any solutions-but if we do, there are infinitely many (since the null space of the matrix is a subspace of dimension at least one).

If the determinant is $0$, you must first confirm that $(1,1,1)^t$ lies in the column space of the matrix for your system. This is obvious for $a = 1$ (in fact, any vector $(x,y,z)^t$ where $x+y+z = 1$ will be a solution), but not so obvious for $a = -2$.

 

[evinda]

If the determinant is $0$ it doesn't necessarily mean we have any solutions-but if we do, there are infinitely many (since the null space of the matrix is a subspace of dimension at least one).

If the determinant is $0$, you must first confirm that $(1,1,1)^t$ lies in the column space of the matrix for your system. This is obvious for $a = 1$ (in fact, any vector $(x,y,z)^t$ where $x+y+z = 1$ will be a solution), but not so obvious for $a = -2$.

So if $a=1$ we know that there are infinitely many solutions, since we get three times the same equation, right?

For $a=-2$, we get:

$$-2x+y+z=1 \\ x-2y+z=1 \\ x+y-2z=1$$

From the last two equations we get $y=z$ and then from the first two we get that $\frac{1}{2}=2$ that is a contradiction and thus we conclude that for $a=-2$ the system has no solution, right? (Thinking)

 

[Deveno]

Yes! In this case, when $a = -2$ we have $(1,1,1) \not \in \text{im }A$, where $A$ is our matrix with $a = -2$ used.

Moral of the story, in problems like these, when substitution of a certain value into a system gives $\det = 0$, you have to check such substitutions individually for consistency.

 

[evinda]

Yes! In this case, when $a = -2$ we have $(1,1,1) \not \in \text{im }A$, where $A$ is our matrix with $a = -2$ used.

Moral of the story, in problems like these, when substitution of a certain value into a system gives $\det = 0$, you have to check such substitutions individually for consistency.

Nice... Thank you! (Smile)