Calculate the determinant of $D_n$ and calculate $D_3$ when $D_3=D_4$

Question

I was asked the following question:

a. Calculate the determinant over $\mathbb{R}$ (n - order of the matrix): b. We're given that $t \neq 0$ and $D_3=D_4$. Find $D_3$.

My solution:

a. It's obvious that if:

1. $n=1$ then: $D_1=det(D_1)=t+x$.
2. $n=2$ then $D_2=\begin{bmatrix} t&x\\-1&t+x \end{bmatrix}$ and $det(D_2)=t(t+x)-x(-1)=t(t+x)+x$.
3. $n=3$ then $D_3=\begin{bmatrix} t&0&x\\-1&t&x\\0&-1&t+x \end{bmatrix}$ and $det(D_3)=t(t(t+x))-tx(-1)+x=t(t(t+x)+x)+x$.

We can definitely see a pattern, so now we need to find a formula to calculate the determinant of the matrix when $n>3$, we can do it with recursion: $det(D_n)=t(det(D_{n-1}))+x$. (I'm supposing they wanted me to do it with Laplace Expansion, but I couldn't get my head around it in this case.)

b. First, Lets find $D_3$ and $D_4$:

$D_3=t(t(t+x)+x)+x$

$D_4=t(t(t(t+x)+x)+x)+x$

And since we're given that $D_3=D_4$:

$t(t(t+x)+x)+x=t(t(t(t+x)+x)+x)+x \Rightarrow t^3 (-t - x + 1)=0$

I'm not so sure about my answer for b. Thanks for the help!

Answer 1

For a). You can easily finish by induction. Expanding with respect to the first row, you have $D_n= t D_{n-1} + x\cdot(-1)^{n+1}\det(M_{n-1})$, where $M_{n-1}$ is an upper triangular matrix with only $(-1)$ on the diagonal, whence $\det(M_{n-1})=(-1)^{n-1}$, and we are done: $D_n=tD_{n-1}+x$.

For b). Your computation is correct, and as $t\neq 0$, you get $t=1-x$. Plug this into your formula for $D_3$ and get $D_3=t(t(t+x)+x)+x=(1-x)((1-x) (1-x+x)+x)+x=1$.