# Find determinant value

\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?

• 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. – TravisJ Apr 14 '15 at 0:59
• @TravisJ. I think for this case using expansion will be better – Vim Apr 14 '15 at 1:10
• @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? – Andrej Naumovski Apr 14 '15 at 1:11
• 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. – TravisJ Apr 14 '15 at 1:43
• 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. – ET93 Apr 14 '15 at 1:49

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 $$a_n = 2^{n+1} - 1.$$

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.

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*}