0
$\begingroup$

\begin{vmatrix} 3 & 2 & 0 & 0 & . &. & . & . &0 &0 \\ 1 & 3 & 2 & 0 & . &. & . & . &0 &0 \\ 0 & 1 & 3 & 2 & . &. & . & . &0 &0 \\ 0 & 0 & 1 & 3 & . &. & . & . &0 &0 \\ . & . & . & . & . &. & . & . &. &. \\ 0 & 0 & 0 & 0 & . &. & . & . &1 &3 \end{vmatrix}

Given this determinant, how do I find its value?

$\endgroup$
  • $\begingroup$ You could do row-operations to make it triangular. The triangular matrix's determinant is easy, and the row-operations change the determinant in a predictable way. $\endgroup$ – TravisJ Apr 14 '15 at 0:59
  • $\begingroup$ @TravisJ. I think for this case using expansion will be better $\endgroup$ – Vim Apr 14 '15 at 1:10
  • $\begingroup$ @TravisJ I have been thinking for the past 15 minutes, I can't seem to figure out what the operations should be. Should I do operations by rows or by columns? $\endgroup$ – Andrej Naumovski Apr 14 '15 at 1:11
  • $\begingroup$ AndrejNaumovski, I think @Vim is right. You "could" do the row operations, starting with adding $-1/3$ the first row to the second. That changes the determinant by (multiplication) $-1/3$. I think this will become messy though. You'll probably want to try the direct expansion, perhaps with induction. I.e., try the small cases first $3\times 3$, $4\times 4$, $5\times 5$ and look for a pattern. $\endgroup$ – TravisJ Apr 14 '15 at 1:43
  • $\begingroup$ The thing about doing row/column operations is you end up computing the determinant of the smaller matrices in the process anyway, i.e. you compute det($A_2$) after the first row operation, det($A_3$) after the second and so on. So a recurrence relation may be better since you do not computer $A_m$ for all $m$ less than whatever $n$ you are looking for. $\endgroup$ – ET93 Apr 14 '15 at 1:49
2
$\begingroup$

we will derive a three term recurrence relations for the determinant $a_n$ of the matrix of size $n.$ expanding by the first row, we get $$a_n = 3a_{n-1} - 2a_{n-2}, a_0 = 1, a_1 = 3, a_2 = 7, a_3= 15, \cdots$$

the characteristic equation is $$\lambda^2 - 3\lambda + 2= 0\to \lambda = 1, 2$$ therefore $$ a_n = C \, 2^n + D, C + D = 1, 2C + D = 3$$ gives $C = 2, D = -1.$ that is $$a_n = 2^{n+1} - 1. $$

$\endgroup$
1
$\begingroup$

Let $A_n$ be the $n \times n$ matrix of the above type. Expanding the determinant along the first row gives two terms: $det(A_n)=3\cdot det(A_{n-1})-2\cdot det(B_{n-1})$, where $B_{n-1}$ is the below matrix.

\begin{vmatrix} 1 & 2 & 0 & . &. & . & . &0 &0 \\ 0 & 3 & 2 & . &. & . & . &0 &0 \\ 0 & 1 & 3 & . &. & . & . &0 &0 \\ . & . & . & . &. & . & . &. &. \\ 0 & 0 & 0 & . &. & . & . &1 &3 \end{vmatrix}

But expanding along the first column gives $det(B_{n-1})=det(A_{n-2})$ so altogether, $det(A_n)=3\cdot det(A_{n-1})-2\cdot det(A_{n-2})$. $det(A_1)=3$ and $det(A_2)=7$. Letting $a_n=det(A_n)$, we have the recurrence:

$$a_n=3a_{n-1}-2a_{n-2}$$ with $a_1=3,a_2=7$. The recurrence is linear so its not hard to solve.

$\endgroup$
1
$\begingroup$

Denote by $D_k$ the $k$-th principal minor of the matrix, i.e. the minor made up of the first $k$ rows and columns, and develop $D_n$ along the last row. You obtain the relation: $$D_n=3D_{n-1}-1\cdot2D_{n-2}$$ The first values are: $$D_1=3,\enspace D_2=7, \enspace D_3=15,\enspace D_4=31.$$ Thus we conjecture $D_n=2^{n+1}-1$.

Induction step:

Suposse the formula is true for orders $n$ and $n-1$. Let's use the recurrence relation: \begin{align*} D_{n+1}&=3D_n-2D_{n-1}=3\cdot2^{n+1}-3 -2\cdot2^n+2\\ &=2\cdot 2^{n+1}-1=2^{n+2}-1.\end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.