## Problem on nontrivial solutions to a linear system

For what values of $\lambda$ are there nontrivial solutions to $$\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \lambda \begin{pmatrix}x\\y\\z\end{pmatrix}$$
For each value, find a nontrivial solution.
• ## Solution

Recall that
The unknowns in the linear equations are $x$, $y$, and $z$.
Because the right-hand side involves these unknowns, we subtract it from both sides. The equation becomes
$$\begin{pmatrix}1-\lambda&0&0\\0&2-\lambda&0\\0&0&3-\lambda\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$
Recall that
In order for there to be a nonzero solution, $\text{det } \mathbf{A}$ must be zero. Hence, we seek the values of $\lambda$ for which
$$\text{det }\begin{pmatrix}1-\lambda&0&0\\0&2-\lambda&0\\0&0&3-\lambda\end{pmatrix} = 0.$$
That is to say, there are nontrivial solutions when $\lambda = 1$ or $\lambda = 2$ or $\lambda = 3$.

#### Nontrivial solution for $\lambda = 1$

We now try to find a nontrivial solution to $$\begin{pmatrix}0&0&0\\0&1&0\\0&0&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$
The second and third rows say that $y=0$, $z=0$. The first row provides no constraint on $x$. Hence, any multiple of $\begin{pmatrix}1\\0\\0\end{pmatrix}$ is a nontrivial solution for $\lambda = 1$.

#### Nontrivial solution for $\lambda =2$

Similarly, a nontrivial solution to $$\begin{pmatrix}-1&0&0\\0&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$ is given by any multiple of $\begin{pmatrix}0\\1\\0\end{pmatrix}$.

#### Nontrivial solution for $\lambda = 3$

Similarly, a nontrivial solution to $$\begin{pmatrix}-2&0&0\\0&-1&0\\0&0&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$ is given by any multiple of $\begin{pmatrix}0\\0\\1\end{pmatrix}$.