This kind of matrix is called TRIDIAGONAL MATRIX.

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37 votes

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 $A$?

- $\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) $
- $\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) $
- $\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) $
- $\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) $

144 votes

Best answer

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.

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.

24 votes

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.

So answer is option D.

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

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

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

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

14 votes

**(D ) is answer.**

|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+5√2)/2 and (3-5√2)/2

Therefore general solution is Tg = C1 . ((3(+/-)5√2)/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.

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

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