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$$\begin{pmatrix}1+a & 1 & 1 & 1\\\ 1 & 1-a & 1 & 1\\1 & 1 & 1 & a\\1 & 2 & 2 & 1\\\end{pmatrix}$$

As the title says, find all values of a for which the system has trivial solutions.

I have tried row reduction but I haven't gotten anywhere with it. So what I have done is calculate the determinant, which is:

2 a^3 - a

From what I understand system has trivial solutions if and only if det(A) != 0.

So the solution for this problem would be every ${\rm I\!R}$ number except where det(A) == 0 ?

IronCanTaco
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    Your reasoning seems correct. – Hanul Jeon Feb 03 '19 at 01:29
  • What do you mean by “trivial solutions?” That usually means that the zero vector is a solution, which is true for every homogeneous system. Perhaps you meant “unique” instead of “trivial?” – amd Feb 03 '19 at 02:08

1 Answers1

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Your determinant is correct, so we get some values to check

$$\det A = a \left(2 a^2-1\right) \implies a = 0, \pm \dfrac{1}{\sqrt{2}}$$

If $a = 0$, the RREF is

$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

If $a = \dfrac{1}{\sqrt{2}}$, the RREF is

$$\begin{bmatrix} 1 & 0 & 0 & \sqrt{2}-1 \\ 0 & 1 & 0 & 1-\sqrt{2} \\ 0 & 0 & 1 & \frac{1}{\sqrt{2}} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

If $a = -\dfrac{1}{\sqrt{2}}$, the RREF is

$$\begin{bmatrix} 1 & 0 & 0 & -\sqrt{2}-1 \\ 0 & 1 & 0 & \sqrt{2}+1 \\ 0 & 0 & 1 & -\frac{1}{\sqrt{2}} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

If $a \ne 0, a \ne \pm \dfrac{1}{\sqrt{2}}$, the RREF is

$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

Hopefully you can complete the question now.

Moo
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