5,327 views

Let $A$ be an $n \times n$ matrix of the following form.

$$A = \begin{bmatrix}3&1&0&0&0&\ldots&0&0&0\\ 1&3&1&0&0&\ldots&0&0&0\\ 0&1&3&1&0&\ldots&0&0&0\\ 0&0&1&3&1&\ldots&0&0&0\\ \ldots\\ \ldots \\ 0&0&0&0&0&\ldots&1&3&1\\ 0&0&0&0&0&\ldots&0&1&3\\ \end{bmatrix}_{n\times n}$$
What is the value of the determinant of $A$?

1. $\left(\frac{5+\sqrt3}{2}\right)^{n-1} \left(\frac{5\sqrt3 + 7}{2 \sqrt 3}\right) +\left(\frac{5-\sqrt3}{2}\right)^{n-1} \left(\frac{5\sqrt3 - 7}{2 \sqrt 3}\right)$
2. $\left(\frac{7+\sqrt5}{2}\right)^{n-1} \left(\frac{7\sqrt5 + 3}{2 \sqrt 5}\right) +\left(\frac{7-\sqrt5}{2}\right)^{n-1} \left(\frac{7\sqrt5 - 3}{2 \sqrt 5}\right)$
3. $\left(\frac{3+\sqrt7}{2}\right)^{n-1} \left(\frac{3\sqrt7 + 5}{2 \sqrt 7}\right) +\left(\frac{3-\sqrt7}{2}\right)^{n-1} \left(\frac{3\sqrt7 - 5}{2 \sqrt 7}\right)$
4. $\left(\frac{3+\sqrt5}{2}\right)^{n-1} \left(\frac{3\sqrt5 + 7}{2 \sqrt 5}\right) + \left(\frac{3-\sqrt5}{2}\right)^{n-1} \left(\frac{3\sqrt5 - 7}{2 \sqrt 5}\right)$

### 1 comment

This kind of matrix is called  TRIDIAGONAL MATRIX.

Answer verification is very easy in this question.
Just put $n=1$, you'll get a matrix like $[3]$.
Find its determinant.
Determinant $= 3$. Now check options.

By putting $n=1$, we get the following results.
$A. 5$
$B. 7$
$C. 3$
$D. 3$

$A, B$ cant be the answer.
Now check for $n=2$.
Determinant $= 9-1 =8$

Put $n=2$ in $C,D$.
$C. 7$
$D. 8$

So, $D$ is the answer.

most easy way to arrive at the option  D.

@ Sir Thanks

I am only correcting the answer given by Sandeep_Uniyal

We can express this determinant as recurrence relation $|A_n|=3|A_{n-1}|-|A_{n-2}|$

Characteristic equation for this recurrence relation is $t^2-3t+1=0$

solving the equation we get,

$$t = \frac{3\pm\sqrt{5}}{2}$$

Hence the solution for recurrence realtion is,

$$|A_n|=C_1t^n+C_2t^n$$

$$|A_n|=C_1\left(\frac{3\pm\sqrt{5}}{2}\right)^n+C_2\left(\frac{3\pm\sqrt{5}}{2}\right)^n$$

Looking at this solution, we can easily say only option D match the form of the equation.

But if you want complete solution, you have to dig some more.

solve the following two equation, which we got after putting $n=1,2$

$$|A_1|=3=C_1\left(\frac{3\pm\sqrt{5}}{2}\right)+C_2\left(\frac{3\pm\sqrt{5}}{2}\right)$$

$$|A_2|=8=C_1\left(\frac{7\pm3\sqrt{5}}{2}\right)+C_2\left(\frac{7\pm3\sqrt{5}}{2}\right)$$

by

can someone explain why we are writing n-2 subscript in recurrence relation
@Kaluti

Try solving the determinant, you will realize it. :)

@lambda

One small correction.

Instead of ±$\sqrt{5}$ we need to have +$\sqrt{5}$ and -$\sqrt{5}$ in the last two equations.

Please see below image for more details. (wasn't able to type it in latex :P)

|An| can be written as a recurrence relation using Det. of lower order matrices as |An|= 3*|An-1| + 1*|An-2|

This I have done by expanding along the first row of the given matrix. You will find that the remaining sub-matrix is same as the given matrix with a LOWER order.

say |An| is denoted as T(n).

Then recurrence relation is T(n) =3. T(n-1)+ T(n-2)  -->  T(n)- 3.T(n-1) -T(n-2)

Divide this by T(n-2) we get    t^2 + 3.t- 1=0 .

Solving for general solution we get roots of this equation as (3+52)/2 and (3-52)/2

Therefore general solution is     Tg = C1 . ((3(+/-)52)/2)^n

where C1 is some constant.

To get this take |A1|=3  (Det. for n=1 is 3). Put this in general solution  to get C1

put the value of C1 in general solution Tg. Arrange the answer to match with the options.

plz explain how you got the recurrence?
I am getting recurrence as T(n)=T(n-1)*3-T(n-2)... how you got +?

Hi @Sandeep_Uniyal ji Good approach. But please correct your recurrence relation and use LaTex writting for equations. I will be great help.