Easiest way to calculate determinant 5x5 witx x

Question

I would like to calculate this determinant: \begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix}

I agree with @voldemort, there is no "easiest way" than applying the Laplace expansion method, and indeed expanding along the first row (which contains the most zeros). You should get a determinant $\Delta=x^5-20x^3+64x$. — Demosthene, Jan 18 '15 at 20:30
Should I add some rows to first row before applaying the Laplace expansion method? — USB_MAT, Jan 18 '15 at 20:32
Nope, adding any of the rows will give you fewer zeros at this point. — user141592, Jan 18 '15 at 20:33

score 4 · Accepted Answer · answered Jan 18 '15 at 20:33

Hint By adding all the rows to first you get

$$\begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} = \begin{vmatrix}x+4&x+4&x+4&x+4&x+4\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} \\ =(x+4)\begin{vmatrix}1&1&1&1&1\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix}=(x+4)\begin{vmatrix}1&1&1&1&1\\0&x-4&-2&-4&-4\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} \\ =(x+4)\begin{vmatrix}x-4&-2&-4&-4\\3&x&3&0\\0&2&x&4\\0&0&1&x\end{vmatrix}$$

Now use expansion by a row or column.

score 1 · Answer 2 · answered Jan 18 '15 at 22:01

The matrix is tridiagonal, and the determinant can be computed recursively as follows.

Let $f_n$ be the determinant of the $n\times n$ submatrix in the upper left of your matrix, and let $f_0=1$. Write the matrix as

\begin{vmatrix}a_1&b_1&0&0&0\\c_1&a_2&b_2&0&0\\0&c_2&a_3&b_3&0\\0&0&c_3&a_4&b_4\\0&0&0&c_4&a_5\end{vmatrix}

and this recurrence holds: $f_n=a_nf_{n-1}-c_{n-1}b_{n-1}f_{n-2}$. (The procedure is described on this Wikipedia page.) For this particular matrix,

$f_2=x^2-4$,

$f_3=x(x^2-4)-6x=x^3-10x$,

$f_4=x(x^3-10x)-6(x^2-4)=x^4-16x^2+24$, and

$f_5=x(x^4-16x^2+24)-4(x^3-10x)=x^5-20x^3+64x$.

Easiest way to calculate determinant 5x5 witx x

2 Answers2